Does the trigonometric identity $cos^2(theta)+sin^2(theta)=1$ apply even when $theta$ is not in radians or...
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Does the trigonometric identity $cos^2(theta)+sin^2(theta)=1$ apply even when $theta$ is not in radians or degrees but simply a fraction?
The 2019 Stack Overflow Developer Survey Results Are InTrigonometry Identity: $tan thetasin theta + cos theta = sec theta$$cos^2(theta)-sin^2(theta)=1+sin(theta)$ over the interval $(0,2pi)$When finding $sintheta$ and $costheta$ given $tantheta = frac35$, shouldn't I get positive and negative answers?Maximum value of trigonometric equation $cos^2(costheta) + sin^2(sintheta)$Proving the identity $frac{cos^2theta+tan^2theta-1}{sin^2theta}=tan^2theta$Find the coordinates of the vector in quadrant 4, in terms of $sin(theta)$ and $cos(theta)$. Assume the hypotenuse is equal to 1.Find the coordinates of the vector in quadrant 4, in terms of $sin(theta)$ and $cos(theta)$. Assume the hypotenuse is equal to 1. #2If $ sintheta + costheta = frac 1 2$, what does $tantheta + cottheta$ equal?Solving trigonometric equations like $1-ssin^2theta=asin^6theta+bcos^6theta$What proves this trigonometric identity $cos2 theta=cos^2theta-sin^2theta$?
$begingroup$
I have been trying to solve this question but have so far been unable to do so as the question does not seem to be "cohesive throughout". Here is my reasoning:
The question is: given that $cos A=−3/5$, $sin B=−5/13$, and both $A$ and $B$ are in the 3rd quadrant, find $cos^2(A)+sin^2(A)$.
I know of the trigonometric identity $cos^2(theta)+sin^2(theta)=1$. In this identity however, $theta$ is in place of $A$. cos $A$ is a fraction in the case of the question, however I often see $theta$ as in the radian or degree form. Does this mean that the trigonometric identity does not apply to the question or is my assumption based on familiarity incorrect?
Also, if the identity were to apply to the question, does the question have an answer or not? When I attempted to solve this question, I did not get $1$ as the answer.
trigonometry
asked Mar 21 at 6:14
JamesJames
425
$endgroup$
|
show 3 more comments
$begingroup$
I have been trying to solve this question but have so far been unable to do so as the question does not seem to be "cohesive throughout". Here is my reasoning:
The question is: given that $cos A=−3/5$, $sin B=−5/13$, and both $A$ and $B$ are in the 3rd quadrant, find $cos^2(A)+sin^2(A)$.
I know of the trigonometric identity $cos^2(theta)+sin^2(theta)=1$. In this identity however, $theta$ is in place of $A$. cos $A$ is a fraction in the case of the question, however I often see $theta$ as in the radian or degree form. Does this mean that the trigonometric identity does not apply to the question or is my assumption based on familiarity incorrect?
Also, if the identity were to apply to the question, does the question have an answer or not? When I attempted to solve this question, I did not get $1$ as the answer.
trigonometry
asked Mar 21 at 6:14
JamesJames
425
$endgroup$
$begingroup$
Why can't a fraction be the measure of some angle in radians?
$endgroup$
– Hrit Roy
Mar 21 at 6:21
$begingroup$
The fraction however in this case is not in radians.
$endgroup$
– James
Mar 21 at 6:22
$begingroup$
You have not mentioned what that fraction is, but it doesn't matter. It will indeed be the measure of SOME angle in radians. (And also some angle in degrees.)
$endgroup$
– Hrit Roy
Mar 21 at 6:26
1
$begingroup$
Of course it does.
$endgroup$
– David G. Stork
Mar 21 at 6:26
$begingroup$
By the way, you are saying that $A$ is a fraction, but do you mean $cos A$ is a fraction (note that it is $color{blue}{cos}(A)$ that is $-3/5$, not $A$ itself)?
$endgroup$
– Minus One-Twelfth
Mar 21 at 6:28
|
show 3 more comments
$begingroup$
I have been trying to solve this question but have so far been unable to do so as the question does not seem to be "cohesive throughout". Here is my reasoning:
The question is: given that $cos A=−3/5$, $sin B=−5/13$, and both $A$ and $B$ are in the 3rd quadrant, find $cos^2(A)+sin^2(A)$.
I know of the trigonometric identity $cos^2(theta)+sin^2(theta)=1$. In this identity however, $theta$ is in place of $A$. cos $A$ is a fraction in the case of the question, however I often see $theta$ as in the radian or degree form. Does this mean that the trigonometric identity does not apply to the question or is my assumption based on familiarity incorrect?
Also, if the identity were to apply to the question, does the question have an answer or not? When I attempted to solve this question, I did not get $1$ as the answer.
trigonometry
asked Mar 21 at 6:14
JamesJames
425
$endgroup$
I have been trying to solve this question but have so far been unable to do so as the question does not seem to be "cohesive throughout". Here is my reasoning:
The question is: given that $cos A=−3/5$, $sin B=−5/13$, and both $A$ and $B$ are in the 3rd quadrant, find $cos^2(A)+sin^2(A)$.
I know of the trigonometric identity $cos^2(theta)+sin^2(theta)=1$. In this identity however, $theta$ is in place of $A$. cos $A$ is a fraction in the case of the question, however I often see $theta$ as in the radian or degree form. Does this mean that the trigonometric identity does not apply to the question or is my assumption based on familiarity incorrect?
Also, if the identity were to apply to the question, does the question have an answer or not? When I attempted to solve this question, I did not get $1$ as the answer.
trigonometry
trigonometry
asked Mar 21 at 6:14
JamesJames
425
asked Mar 21 at 6:14
JamesJames
425
edited Mar 21 at 6:34
James
asked Mar 21 at 6:14
JamesJames
425
asked Mar 21 at 6:14
JamesJames
425
asked Mar 21 at 6:14
JamesJames
425
425
$begingroup$
Why can't a fraction be the measure of some angle in radians?
$endgroup$
– Hrit Roy
Mar 21 at 6:21
$begingroup$
The fraction however in this case is not in radians.
$endgroup$
– James
Mar 21 at 6:22
$begingroup$
You have not mentioned what that fraction is, but it doesn't matter. It will indeed be the measure of SOME angle in radians. (And also some angle in degrees.)
$endgroup$
– Hrit Roy
Mar 21 at 6:26
1
$begingroup$
Of course it does.
$endgroup$
– David G. Stork
Mar 21 at 6:26
$begingroup$
By the way, you are saying that $A$ is a fraction, but do you mean $cos A$ is a fraction (note that it is $color{blue}{cos}(A)$ that is $-3/5$, not $A$ itself)?
$endgroup$
– Minus One-Twelfth
Mar 21 at 6:28
|
show 3 more comments
$begingroup$
Why can't a fraction be the measure of some angle in radians?
$endgroup$
– Hrit Roy
Mar 21 at 6:21
$begingroup$
The fraction however in this case is not in radians.
$endgroup$
– James
Mar 21 at 6:22
$begingroup$
You have not mentioned what that fraction is, but it doesn't matter. It will indeed be the measure of SOME angle in radians. (And also some angle in degrees.)
$endgroup$
– Hrit Roy
Mar 21 at 6:26
1
$begingroup$
Of course it does.
$endgroup$
– David G. Stork
Mar 21 at 6:26
$begingroup$
By the way, you are saying that $A$ is a fraction, but do you mean $cos A$ is a fraction (note that it is $color{blue}{cos}(A)$ that is $-3/5$, not $A$ itself)?
$endgroup$
– Minus One-Twelfth
Mar 21 at 6:28
$begingroup$
Why can't a fraction be the measure of some angle in radians?
$endgroup$
– Hrit Roy
Mar 21 at 6:21
$begingroup$
Why can't a fraction be the measure of some angle in radians?
$endgroup$
– Hrit Roy
Mar 21 at 6:21
$begingroup$
The fraction however in this case is not in radians.
$endgroup$
– James
Mar 21 at 6:22
$begingroup$
The fraction however in this case is not in radians.
$endgroup$
– James
Mar 21 at 6:22
$begingroup$
You have not mentioned what that fraction is, but it doesn't matter. It will indeed be the measure of SOME angle in radians. (And also some angle in degrees.)
$endgroup$
– Hrit Roy
Mar 21 at 6:26
$begingroup$
You have not mentioned what that fraction is, but it doesn't matter. It will indeed be the measure of SOME angle in radians. (And also some angle in degrees.)
$endgroup$
– Hrit Roy
Mar 21 at 6:26
1
1
$begingroup$
Of course it does.
$endgroup$
– David G. Stork
Mar 21 at 6:26
$begingroup$
Of course it does.
$endgroup$
– David G. Stork
Mar 21 at 6:26
$begingroup$
By the way, you are saying that $A$ is a fraction, but do you mean $cos A$ is a fraction (note that it is $color{blue}{cos}(A)$ that is $-3/5$, not $A$ itself)?
$endgroup$
– Minus One-Twelfth
Mar 21 at 6:28
$begingroup$
By the way, you are saying that $A$ is a fraction, but do you mean $cos A$ is a fraction (note that it is $color{blue}{cos}(A)$ that is $-3/5$, not $A$ itself)?
$endgroup$
– Minus One-Twelfth
Mar 21 at 6:28
|
show 3 more comments
2 Answers
2
active
oldest
votes
$begingroup$
The question tells you that $cos A$ is a fraction, not that $A$ itself is a fraction. Like many of the cases I imagine you've encountered, $A$ is just some angle that can be expressed in degrees or radians.
While you could calculate the value of $cos^2hspace{-0.9mm}A+sin^2hspace{-0.9mm}A$ by hand, I can assure you that it will be $1$ in this case. In general, $cos^2(text{something})+sin^2(text{something})=1$ will be true unless the "something" happens to be an expression that involves division by zero, taking the square root of a negative number, or something else similarly problematic.
Here's how you could calculate the value of $cos^2hspace{-0.9mm}A+sin^2hspace{-0.9mm}A$ without relying on that identity:
Since you know that $A$ is some angle in quadrant III, and you know that $cos A=-3/5$, consider the triangle I've drawn in Desmos here, and let $A$ be the angle at the origin. By design, $$cos A=frac{text{adjacent}}{text{hypotenuse}}=frac{-3}{5}.$$ From looking at the triangle, it's also evident that $$sin A=frac{text{opposite}}{text{hypotenuse}}=frac{-4}{5}.$$ Both of those negative signs come from the fact that we're moving in the negative $x$ and $y$ directions, since of course it's impossible for a side of a triangle to have a negative length. At this point it's just calculation: $$cos^2hspace{-0.9mm}A+sin^2hspace{-0.9mm}A=left(frac{-3}{5}right)^2+left(frac{-4}{5}right)^2=frac{9}{25}+frac{16}{25}=frac{25}{25}=1$$
answered Mar 21 at 6:30
Robert HowardRobert Howard
2,3033935
$endgroup$
$begingroup$
If possible could you show me how it is done by hand just so I can see the truth behind this identity? I have attempted to do it by hand but it comes out to be approximately 0.5, not 1.
$endgroup$
– James
Mar 21 at 6:33
$begingroup$
Of course; give me a minute to edit my answer.
$endgroup$
– Robert Howard
Mar 21 at 6:38
$begingroup$
In the case of the question, sin A= -5/13. With such a value for sin A, the result is not 1 according to my calculations.
$endgroup$
– James
Mar 21 at 6:58
$begingroup$
You said in your question that $sin B=-5/13$, not $sin A=-5/13$, suggesting that there are two separate angles $A$ and $B$.
$endgroup$
– Robert Howard
Mar 21 at 7:00
1
$begingroup$
Thank you Mr. Howard. Your detailed, step-by-step explanations are undoubtedly appreciated by the Mathematics Stack Exchange community.
$endgroup$
– James
Mar 21 at 7:16
|
show 5 more comments
$begingroup$
The fraction is never in Radians or degrees it is a dimensionless quantity. The input to the trigonometric function is in radians or degrees. So the "identity" $sin^2 x+cos^2x =1$ always holds which is why it is called an identity (something that always holds).
$$begin{aligned}sintheta &=dfrac{overbrace{text{side opposite}}^{text{units}}}{underbrace{text{hypotenuse}}_{text{units}}}\costheta&=dfrac{overbrace{text{side adjacent}}^{text{units}}}{underbrace{text{hypotenuse}}_{text{units}}}end{aligned}$$
answered Mar 21 at 6:30
Paras KhoslaParas Khosla
3,051625
$endgroup$
$begingroup$
Just to clarify, you mean to say that even though x is a fraction that is neither in the degree or radian form that the trigonometric identity will always hold true?
$endgroup$
– James
Mar 21 at 6:37
$begingroup$
Also, why did you say that the input is in radians or degrees?
$endgroup$
– James
Mar 21 at 6:37
$begingroup$
Since the angle has to be defined in terms of some units, its either radians or degrees but the ratio is a dimensionless quantity.
$endgroup$
– Paras Khosla
Mar 21 at 7:13
add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The question tells you that $cos A$ is a fraction, not that $A$ itself is a fraction. Like many of the cases I imagine you've encountered, $A$ is just some angle that can be expressed in degrees or radians.
While you could calculate the value of $cos^2hspace{-0.9mm}A+sin^2hspace{-0.9mm}A$ by hand, I can assure you that it will be $1$ in this case. In general, $cos^2(text{something})+sin^2(text{something})=1$ will be true unless the "something" happens to be an expression that involves division by zero, taking the square root of a negative number, or something else similarly problematic.
Here's how you could calculate the value of $cos^2hspace{-0.9mm}A+sin^2hspace{-0.9mm}A$ without relying on that identity:
Since you know that $A$ is some angle in quadrant III, and you know that $cos A=-3/5$, consider the triangle I've drawn in Desmos here, and let $A$ be the angle at the origin. By design, $$cos A=frac{text{adjacent}}{text{hypotenuse}}=frac{-3}{5}.$$ From looking at the triangle, it's also evident that $$sin A=frac{text{opposite}}{text{hypotenuse}}=frac{-4}{5}.$$ Both of those negative signs come from the fact that we're moving in the negative $x$ and $y$ directions, since of course it's impossible for a side of a triangle to have a negative length. At this point it's just calculation: $$cos^2hspace{-0.9mm}A+sin^2hspace{-0.9mm}A=left(frac{-3}{5}right)^2+left(frac{-4}{5}right)^2=frac{9}{25}+frac{16}{25}=frac{25}{25}=1$$
answered Mar 21 at 6:30
Robert HowardRobert Howard
2,3033935
$endgroup$
$begingroup$
If possible could you show me how it is done by hand just so I can see the truth behind this identity? I have attempted to do it by hand but it comes out to be approximately 0.5, not 1.
$endgroup$
– James
Mar 21 at 6:33
$begingroup$
Of course; give me a minute to edit my answer.
$endgroup$
– Robert Howard
Mar 21 at 6:38
$begingroup$
In the case of the question, sin A= -5/13. With such a value for sin A, the result is not 1 according to my calculations.
$endgroup$
– James
Mar 21 at 6:58
$begingroup$
You said in your question that $sin B=-5/13$, not $sin A=-5/13$, suggesting that there are two separate angles $A$ and $B$.
$endgroup$
– Robert Howard
Mar 21 at 7:00
1
$begingroup$
Thank you Mr. Howard. Your detailed, step-by-step explanations are undoubtedly appreciated by the Mathematics Stack Exchange community.
$endgroup$
– James
Mar 21 at 7:16
|
show 5 more comments
$begingroup$
The question tells you that $cos A$ is a fraction, not that $A$ itself is a fraction. Like many of the cases I imagine you've encountered, $A$ is just some angle that can be expressed in degrees or radians.
While you could calculate the value of $cos^2hspace{-0.9mm}A+sin^2hspace{-0.9mm}A$ by hand, I can assure you that it will be $1$ in this case. In general, $cos^2(text{something})+sin^2(text{something})=1$ will be true unless the "something" happens to be an expression that involves division by zero, taking the square root of a negative number, or something else similarly problematic.
Here's how you could calculate the value of $cos^2hspace{-0.9mm}A+sin^2hspace{-0.9mm}A$ without relying on that identity:
Since you know that $A$ is some angle in quadrant III, and you know that $cos A=-3/5$, consider the triangle I've drawn in Desmos here, and let $A$ be the angle at the origin. By design, $$cos A=frac{text{adjacent}}{text{hypotenuse}}=frac{-3}{5}.$$ From looking at the triangle, it's also evident that $$sin A=frac{text{opposite}}{text{hypotenuse}}=frac{-4}{5}.$$ Both of those negative signs come from the fact that we're moving in the negative $x$ and $y$ directions, since of course it's impossible for a side of a triangle to have a negative length. At this point it's just calculation: $$cos^2hspace{-0.9mm}A+sin^2hspace{-0.9mm}A=left(frac{-3}{5}right)^2+left(frac{-4}{5}right)^2=frac{9}{25}+frac{16}{25}=frac{25}{25}=1$$
answered Mar 21 at 6:30
Robert HowardRobert Howard
2,3033935
$endgroup$
$begingroup$
If possible could you show me how it is done by hand just so I can see the truth behind this identity? I have attempted to do it by hand but it comes out to be approximately 0.5, not 1.
$endgroup$
– James
Mar 21 at 6:33
$begingroup$
Of course; give me a minute to edit my answer.
$endgroup$
– Robert Howard
Mar 21 at 6:38
$begingroup$
In the case of the question, sin A= -5/13. With such a value for sin A, the result is not 1 according to my calculations.
$endgroup$
– James
Mar 21 at 6:58
$begingroup$
You said in your question that $sin B=-5/13$, not $sin A=-5/13$, suggesting that there are two separate angles $A$ and $B$.
$endgroup$
– Robert Howard
Mar 21 at 7:00
1
$begingroup$
Thank you Mr. Howard. Your detailed, step-by-step explanations are undoubtedly appreciated by the Mathematics Stack Exchange community.
$endgroup$
– James
Mar 21 at 7:16
|
show 5 more comments
$begingroup$
The question tells you that $cos A$ is a fraction, not that $A$ itself is a fraction. Like many of the cases I imagine you've encountered, $A$ is just some angle that can be expressed in degrees or radians.
While you could calculate the value of $cos^2hspace{-0.9mm}A+sin^2hspace{-0.9mm}A$ by hand, I can assure you that it will be $1$ in this case. In general, $cos^2(text{something})+sin^2(text{something})=1$ will be true unless the "something" happens to be an expression that involves division by zero, taking the square root of a negative number, or something else similarly problematic.
Here's how you could calculate the value of $cos^2hspace{-0.9mm}A+sin^2hspace{-0.9mm}A$ without relying on that identity:
Since you know that $A$ is some angle in quadrant III, and you know that $cos A=-3/5$, consider the triangle I've drawn in Desmos here, and let $A$ be the angle at the origin. By design, $$cos A=frac{text{adjacent}}{text{hypotenuse}}=frac{-3}{5}.$$ From looking at the triangle, it's also evident that $$sin A=frac{text{opposite}}{text{hypotenuse}}=frac{-4}{5}.$$ Both of those negative signs come from the fact that we're moving in the negative $x$ and $y$ directions, since of course it's impossible for a side of a triangle to have a negative length. At this point it's just calculation: $$cos^2hspace{-0.9mm}A+sin^2hspace{-0.9mm}A=left(frac{-3}{5}right)^2+left(frac{-4}{5}right)^2=frac{9}{25}+frac{16}{25}=frac{25}{25}=1$$
answered Mar 21 at 6:30
Robert HowardRobert Howard
2,3033935
$endgroup$
The question tells you that $cos A$ is a fraction, not that $A$ itself is a fraction. Like many of the cases I imagine you've encountered, $A$ is just some angle that can be expressed in degrees or radians.
While you could calculate the value of $cos^2hspace{-0.9mm}A+sin^2hspace{-0.9mm}A$ by hand, I can assure you that it will be $1$ in this case. In general, $cos^2(text{something})+sin^2(text{something})=1$ will be true unless the "something" happens to be an expression that involves division by zero, taking the square root of a negative number, or something else similarly problematic.
Here's how you could calculate the value of $cos^2hspace{-0.9mm}A+sin^2hspace{-0.9mm}A$ without relying on that identity:
Since you know that $A$ is some angle in quadrant III, and you know that $cos A=-3/5$, consider the triangle I've drawn in Desmos here, and let $A$ be the angle at the origin. By design, $$cos A=frac{text{adjacent}}{text{hypotenuse}}=frac{-3}{5}.$$ From looking at the triangle, it's also evident that $$sin A=frac{text{opposite}}{text{hypotenuse}}=frac{-4}{5}.$$ Both of those negative signs come from the fact that we're moving in the negative $x$ and $y$ directions, since of course it's impossible for a side of a triangle to have a negative length. At this point it's just calculation: $$cos^2hspace{-0.9mm}A+sin^2hspace{-0.9mm}A=left(frac{-3}{5}right)^2+left(frac{-4}{5}right)^2=frac{9}{25}+frac{16}{25}=frac{25}{25}=1$$
answered Mar 21 at 6:30
Robert HowardRobert Howard
2,3033935
edited Mar 21 at 6:51
answered Mar 21 at 6:30
Robert HowardRobert Howard
2,3033935
answered Mar 21 at 6:30
Robert HowardRobert Howard
2,3033935
answered Mar 21 at 6:30
Robert HowardRobert Howard
2,3033935
2,3033935
$begingroup$
If possible could you show me how it is done by hand just so I can see the truth behind this identity? I have attempted to do it by hand but it comes out to be approximately 0.5, not 1.
$endgroup$
– James
Mar 21 at 6:33
$begingroup$
Of course; give me a minute to edit my answer.
$endgroup$
– Robert Howard
Mar 21 at 6:38
$begingroup$
In the case of the question, sin A= -5/13. With such a value for sin A, the result is not 1 according to my calculations.
$endgroup$
– James
Mar 21 at 6:58
$begingroup$
You said in your question that $sin B=-5/13$, not $sin A=-5/13$, suggesting that there are two separate angles $A$ and $B$.
$endgroup$
– Robert Howard
Mar 21 at 7:00
1
$begingroup$
Thank you Mr. Howard. Your detailed, step-by-step explanations are undoubtedly appreciated by the Mathematics Stack Exchange community.
$endgroup$
– James
Mar 21 at 7:16
|
show 5 more comments
$begingroup$
If possible could you show me how it is done by hand just so I can see the truth behind this identity? I have attempted to do it by hand but it comes out to be approximately 0.5, not 1.
$endgroup$
– James
Mar 21 at 6:33
$begingroup$
Of course; give me a minute to edit my answer.
$endgroup$
– Robert Howard
Mar 21 at 6:38
$begingroup$
In the case of the question, sin A= -5/13. With such a value for sin A, the result is not 1 according to my calculations.
$endgroup$
– James
Mar 21 at 6:58
$begingroup$
You said in your question that $sin B=-5/13$, not $sin A=-5/13$, suggesting that there are two separate angles $A$ and $B$.
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– Robert Howard
Mar 21 at 7:00
1
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Thank you Mr. Howard. Your detailed, step-by-step explanations are undoubtedly appreciated by the Mathematics Stack Exchange community.
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– James
Mar 21 at 7:16
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If possible could you show me how it is done by hand just so I can see the truth behind this identity? I have attempted to do it by hand but it comes out to be approximately 0.5, not 1.
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– James
Mar 21 at 6:33
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If possible could you show me how it is done by hand just so I can see the truth behind this identity? I have attempted to do it by hand but it comes out to be approximately 0.5, not 1.
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– James
Mar 21 at 6:33
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Of course; give me a minute to edit my answer.
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– Robert Howard
Mar 21 at 6:38
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Of course; give me a minute to edit my answer.
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– Robert Howard
Mar 21 at 6:38
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In the case of the question, sin A= -5/13. With such a value for sin A, the result is not 1 according to my calculations.
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– James
Mar 21 at 6:58
$begingroup$
In the case of the question, sin A= -5/13. With such a value for sin A, the result is not 1 according to my calculations.
$endgroup$
– James
Mar 21 at 6:58
$begingroup$
You said in your question that $sin B=-5/13$, not $sin A=-5/13$, suggesting that there are two separate angles $A$ and $B$.
$endgroup$
– Robert Howard
Mar 21 at 7:00
$begingroup$
You said in your question that $sin B=-5/13$, not $sin A=-5/13$, suggesting that there are two separate angles $A$ and $B$.
$endgroup$
– Robert Howard
Mar 21 at 7:00
1
1
$begingroup$
Thank you Mr. Howard. Your detailed, step-by-step explanations are undoubtedly appreciated by the Mathematics Stack Exchange community.
$endgroup$
– James
Mar 21 at 7:16
$begingroup$
Thank you Mr. Howard. Your detailed, step-by-step explanations are undoubtedly appreciated by the Mathematics Stack Exchange community.
$endgroup$
– James
Mar 21 at 7:16
|
show 5 more comments
$begingroup$
The fraction is never in Radians or degrees it is a dimensionless quantity. The input to the trigonometric function is in radians or degrees. So the "identity" $sin^2 x+cos^2x =1$ always holds which is why it is called an identity (something that always holds).
$$begin{aligned}sintheta &=dfrac{overbrace{text{side opposite}}^{text{units}}}{underbrace{text{hypotenuse}}_{text{units}}}\costheta&=dfrac{overbrace{text{side adjacent}}^{text{units}}}{underbrace{text{hypotenuse}}_{text{units}}}end{aligned}$$
answered Mar 21 at 6:30
Paras KhoslaParas Khosla
3,051625
$endgroup$
$begingroup$
Just to clarify, you mean to say that even though x is a fraction that is neither in the degree or radian form that the trigonometric identity will always hold true?
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– James
Mar 21 at 6:37
$begingroup$
Also, why did you say that the input is in radians or degrees?
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– James
Mar 21 at 6:37
$begingroup$
Since the angle has to be defined in terms of some units, its either radians or degrees but the ratio is a dimensionless quantity.
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– Paras Khosla
Mar 21 at 7:13
add a comment |
$begingroup$
The fraction is never in Radians or degrees it is a dimensionless quantity. The input to the trigonometric function is in radians or degrees. So the "identity" $sin^2 x+cos^2x =1$ always holds which is why it is called an identity (something that always holds).
$$begin{aligned}sintheta &=dfrac{overbrace{text{side opposite}}^{text{units}}}{underbrace{text{hypotenuse}}_{text{units}}}\costheta&=dfrac{overbrace{text{side adjacent}}^{text{units}}}{underbrace{text{hypotenuse}}_{text{units}}}end{aligned}$$
answered Mar 21 at 6:30
Paras KhoslaParas Khosla
3,051625
$endgroup$
$begingroup$
Just to clarify, you mean to say that even though x is a fraction that is neither in the degree or radian form that the trigonometric identity will always hold true?
$endgroup$
– James
Mar 21 at 6:37
$begingroup$
Also, why did you say that the input is in radians or degrees?
$endgroup$
– James
Mar 21 at 6:37
$begingroup$
Since the angle has to be defined in terms of some units, its either radians or degrees but the ratio is a dimensionless quantity.
$endgroup$
– Paras Khosla
Mar 21 at 7:13
add a comment |
$begingroup$
The fraction is never in Radians or degrees it is a dimensionless quantity. The input to the trigonometric function is in radians or degrees. So the "identity" $sin^2 x+cos^2x =1$ always holds which is why it is called an identity (something that always holds).
$$begin{aligned}sintheta &=dfrac{overbrace{text{side opposite}}^{text{units}}}{underbrace{text{hypotenuse}}_{text{units}}}\costheta&=dfrac{overbrace{text{side adjacent}}^{text{units}}}{underbrace{text{hypotenuse}}_{text{units}}}end{aligned}$$
answered Mar 21 at 6:30
Paras KhoslaParas Khosla
3,051625
$endgroup$
The fraction is never in Radians or degrees it is a dimensionless quantity. The input to the trigonometric function is in radians or degrees. So the "identity" $sin^2 x+cos^2x =1$ always holds which is why it is called an identity (something that always holds).
$$begin{aligned}sintheta &=dfrac{overbrace{text{side opposite}}^{text{units}}}{underbrace{text{hypotenuse}}_{text{units}}}\costheta&=dfrac{overbrace{text{side adjacent}}^{text{units}}}{underbrace{text{hypotenuse}}_{text{units}}}end{aligned}$$
answered Mar 21 at 6:30
Paras KhoslaParas Khosla
3,051625
answered Mar 21 at 6:30
Paras KhoslaParas Khosla
3,051625
answered Mar 21 at 6:30
Paras KhoslaParas Khosla
3,051625
answered Mar 21 at 6:30
Paras KhoslaParas Khosla
3,051625
3,051625
$begingroup$
Just to clarify, you mean to say that even though x is a fraction that is neither in the degree or radian form that the trigonometric identity will always hold true?
$endgroup$
– James
Mar 21 at 6:37
$begingroup$
Also, why did you say that the input is in radians or degrees?
$endgroup$
– James
Mar 21 at 6:37
$begingroup$
Since the angle has to be defined in terms of some units, its either radians or degrees but the ratio is a dimensionless quantity.
$endgroup$
– Paras Khosla
Mar 21 at 7:13
add a comment |
$begingroup$
Just to clarify, you mean to say that even though x is a fraction that is neither in the degree or radian form that the trigonometric identity will always hold true?
$endgroup$
– James
Mar 21 at 6:37
$begingroup$
Also, why did you say that the input is in radians or degrees?
$endgroup$
– James
Mar 21 at 6:37
$begingroup$
Since the angle has to be defined in terms of some units, its either radians or degrees but the ratio is a dimensionless quantity.
$endgroup$
– Paras Khosla
Mar 21 at 7:13
$begingroup$
Just to clarify, you mean to say that even though x is a fraction that is neither in the degree or radian form that the trigonometric identity will always hold true?
$endgroup$
– James
Mar 21 at 6:37
$begingroup$
Just to clarify, you mean to say that even though x is a fraction that is neither in the degree or radian form that the trigonometric identity will always hold true?
$endgroup$
– James
Mar 21 at 6:37
$begingroup$
Also, why did you say that the input is in radians or degrees?
$endgroup$
– James
Mar 21 at 6:37
$begingroup$
Also, why did you say that the input is in radians or degrees?
$endgroup$
– James
Mar 21 at 6:37
$begingroup$
Since the angle has to be defined in terms of some units, its either radians or degrees but the ratio is a dimensionless quantity.
$endgroup$
– Paras Khosla
Mar 21 at 7:13
$begingroup$
Since the angle has to be defined in terms of some units, its either radians or degrees but the ratio is a dimensionless quantity.
$endgroup$
– Paras Khosla
Mar 21 at 7:13
add a comment |
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$begingroup$
Why can't a fraction be the measure of some angle in radians?
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– Hrit Roy
Mar 21 at 6:21
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The fraction however in this case is not in radians.
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– James
Mar 21 at 6:22
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You have not mentioned what that fraction is, but it doesn't matter. It will indeed be the measure of SOME angle in radians. (And also some angle in degrees.)
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– Hrit Roy
Mar 21 at 6:26
1
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Of course it does.
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– David G. Stork
Mar 21 at 6:26
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By the way, you are saying that $A$ is a fraction, but do you mean $cos A$ is a fraction (note that it is $color{blue}{cos}(A)$ that is $-3/5$, not $A$ itself)?
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– Minus One-Twelfth
Mar 21 at 6:28