Finding the possible values of sin and tan given cosx = 3/5Find $sin theta$ and $cos theta$ given $tan...
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Finding the possible values of sin and tan given cosx = 3/5
Find $sin theta$ and $cos theta$ given $tan 2theta$Tirgo problem knowing roots are given find the value of : $sin^2(alpha +beta) +psin(alpha +beta) cos(alpha +beta)+qcos^2(alpha +beta)$Given $n tan b = tan a$ and $ m sin b = sin a$ to find out the value of $frac{m^2 - 1}{n^2-1}$Working with sin, cos and tan.Finding exact values of trig functionsIt is given $(tan theta +1)(sin^² theta -3cos^² theta)=0$, find the possible values of $tan theta$converting cos to sin and tan in specific quadrantsWhen finding $sintheta$ and $costheta$ given $tantheta = frac35$, shouldn't I get positive and negative answers?Find possible values of $tan(B)tan(C)$Given $180°<{theta}<360°$and$frac{1+tan{theta}}{1-tan{theta}}=7$, compute the value of $sin{theta}+cos{theta}$
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Find the possible values of $sintheta$ and $tantheta$, given that $costheta = 3/5$ and $0 leq theta leq pi/2$.
Here is my working out:
Using a right hand triangle:
$$costheta =a/c$$
$$costheta =3/5$$
$$a^2 = b^2 + c^2$$
$$5^2= 3^2+ b^2$$
Therefore $b=+ 4$ or $- 4$.
Compute the values of $b$ and $c$ to find $sintheta$
$$sintheta = b/ c = 4/5$$
$$tantheta = sin/cos = (3/5) / (4/5)$$
$$tantheta = 4/5.$$
Is this right?
Or is it better to do this is in a graph?
trigonometry
$endgroup$
add a comment |
$begingroup$
Find the possible values of $sintheta$ and $tantheta$, given that $costheta = 3/5$ and $0 leq theta leq pi/2$.
Here is my working out:
Using a right hand triangle:
$$costheta =a/c$$
$$costheta =3/5$$
$$a^2 = b^2 + c^2$$
$$5^2= 3^2+ b^2$$
Therefore $b=+ 4$ or $- 4$.
Compute the values of $b$ and $c$ to find $sintheta$
$$sintheta = b/ c = 4/5$$
$$tantheta = sin/cos = (3/5) / (4/5)$$
$$tantheta = 4/5.$$
Is this right?
Or is it better to do this is in a graph?
trigonometry
$endgroup$
$begingroup$
Actually the result is $frac 4 3$. Also you can use directly $sin^2theta + cos^2theta = 1$
$endgroup$
– user261263
Nov 2 '16 at 17:34
$begingroup$
For writing future questions, note that you should put entire equations between$signs rather than putting$s around every single symbol you want to format. Line endings in your input do not cause new lines to occur in the output, but$$...$$does put the math on line(s) by itself. Don't use spacing to indicate grouping of operands, use parentheses.
$endgroup$
– David K
Nov 2 '16 at 17:42
$begingroup$
Also for working problems in general, if you're going to use symbols (and often you should), be consistent. You wrote $cos theta = a/c$ but a few lines later you substitute $5$ for $a$ and $3$ for $c$ in a different equation, which would have meant $cos theta = 5/3$, obviously not what you intended to say.
$endgroup$
– David K
Nov 2 '16 at 17:47
add a comment |
$begingroup$
Find the possible values of $sintheta$ and $tantheta$, given that $costheta = 3/5$ and $0 leq theta leq pi/2$.
Here is my working out:
Using a right hand triangle:
$$costheta =a/c$$
$$costheta =3/5$$
$$a^2 = b^2 + c^2$$
$$5^2= 3^2+ b^2$$
Therefore $b=+ 4$ or $- 4$.
Compute the values of $b$ and $c$ to find $sintheta$
$$sintheta = b/ c = 4/5$$
$$tantheta = sin/cos = (3/5) / (4/5)$$
$$tantheta = 4/5.$$
Is this right?
Or is it better to do this is in a graph?
trigonometry
$endgroup$
Find the possible values of $sintheta$ and $tantheta$, given that $costheta = 3/5$ and $0 leq theta leq pi/2$.
Here is my working out:
Using a right hand triangle:
$$costheta =a/c$$
$$costheta =3/5$$
$$a^2 = b^2 + c^2$$
$$5^2= 3^2+ b^2$$
Therefore $b=+ 4$ or $- 4$.
Compute the values of $b$ and $c$ to find $sintheta$
$$sintheta = b/ c = 4/5$$
$$tantheta = sin/cos = (3/5) / (4/5)$$
$$tantheta = 4/5.$$
Is this right?
Or is it better to do this is in a graph?
trigonometry
trigonometry
edited Nov 2 '16 at 17:35
David K
55.6k345121
55.6k345121
asked Nov 2 '16 at 17:31
MaggieMaggie
238
238
$begingroup$
Actually the result is $frac 4 3$. Also you can use directly $sin^2theta + cos^2theta = 1$
$endgroup$
– user261263
Nov 2 '16 at 17:34
$begingroup$
For writing future questions, note that you should put entire equations between$signs rather than putting$s around every single symbol you want to format. Line endings in your input do not cause new lines to occur in the output, but$$...$$does put the math on line(s) by itself. Don't use spacing to indicate grouping of operands, use parentheses.
$endgroup$
– David K
Nov 2 '16 at 17:42
$begingroup$
Also for working problems in general, if you're going to use symbols (and often you should), be consistent. You wrote $cos theta = a/c$ but a few lines later you substitute $5$ for $a$ and $3$ for $c$ in a different equation, which would have meant $cos theta = 5/3$, obviously not what you intended to say.
$endgroup$
– David K
Nov 2 '16 at 17:47
add a comment |
$begingroup$
Actually the result is $frac 4 3$. Also you can use directly $sin^2theta + cos^2theta = 1$
$endgroup$
– user261263
Nov 2 '16 at 17:34
$begingroup$
For writing future questions, note that you should put entire equations between$signs rather than putting$s around every single symbol you want to format. Line endings in your input do not cause new lines to occur in the output, but$$...$$does put the math on line(s) by itself. Don't use spacing to indicate grouping of operands, use parentheses.
$endgroup$
– David K
Nov 2 '16 at 17:42
$begingroup$
Also for working problems in general, if you're going to use symbols (and often you should), be consistent. You wrote $cos theta = a/c$ but a few lines later you substitute $5$ for $a$ and $3$ for $c$ in a different equation, which would have meant $cos theta = 5/3$, obviously not what you intended to say.
$endgroup$
– David K
Nov 2 '16 at 17:47
$begingroup$
Actually the result is $frac 4 3$. Also you can use directly $sin^2theta + cos^2theta = 1$
$endgroup$
– user261263
Nov 2 '16 at 17:34
$begingroup$
Actually the result is $frac 4 3$. Also you can use directly $sin^2theta + cos^2theta = 1$
$endgroup$
– user261263
Nov 2 '16 at 17:34
$begingroup$
For writing future questions, note that you should put entire equations between
$ signs rather than putting $s around every single symbol you want to format. Line endings in your input do not cause new lines to occur in the output, but $$...$$ does put the math on line(s) by itself. Don't use spacing to indicate grouping of operands, use parentheses.$endgroup$
– David K
Nov 2 '16 at 17:42
$begingroup$
For writing future questions, note that you should put entire equations between
$ signs rather than putting $s around every single symbol you want to format. Line endings in your input do not cause new lines to occur in the output, but $$...$$ does put the math on line(s) by itself. Don't use spacing to indicate grouping of operands, use parentheses.$endgroup$
– David K
Nov 2 '16 at 17:42
$begingroup$
Also for working problems in general, if you're going to use symbols (and often you should), be consistent. You wrote $cos theta = a/c$ but a few lines later you substitute $5$ for $a$ and $3$ for $c$ in a different equation, which would have meant $cos theta = 5/3$, obviously not what you intended to say.
$endgroup$
– David K
Nov 2 '16 at 17:47
$begingroup$
Also for working problems in general, if you're going to use symbols (and often you should), be consistent. You wrote $cos theta = a/c$ but a few lines later you substitute $5$ for $a$ and $3$ for $c$ in a different equation, which would have meant $cos theta = 5/3$, obviously not what you intended to say.
$endgroup$
– David K
Nov 2 '16 at 17:47
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
That's right. I think you should specify what's wrong with $b=-4$ though. The question says that $theta$ has to be between 0 and $pi/2$. That's the problem. Also just to clarify, it should be sin $theta$ and tan $theta$, not just sin and tan. The way I see it is that it's like writing the square root sign without anything underneath. Actually yes, tan$theta$ should be 4/3.
$endgroup$
$begingroup$
Thank you, I will do that.
$endgroup$
– Maggie
Nov 2 '16 at 17:35
add a comment |
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$begingroup$
That's right. I think you should specify what's wrong with $b=-4$ though. The question says that $theta$ has to be between 0 and $pi/2$. That's the problem. Also just to clarify, it should be sin $theta$ and tan $theta$, not just sin and tan. The way I see it is that it's like writing the square root sign without anything underneath. Actually yes, tan$theta$ should be 4/3.
$endgroup$
$begingroup$
Thank you, I will do that.
$endgroup$
– Maggie
Nov 2 '16 at 17:35
add a comment |
$begingroup$
That's right. I think you should specify what's wrong with $b=-4$ though. The question says that $theta$ has to be between 0 and $pi/2$. That's the problem. Also just to clarify, it should be sin $theta$ and tan $theta$, not just sin and tan. The way I see it is that it's like writing the square root sign without anything underneath. Actually yes, tan$theta$ should be 4/3.
$endgroup$
$begingroup$
Thank you, I will do that.
$endgroup$
– Maggie
Nov 2 '16 at 17:35
add a comment |
$begingroup$
That's right. I think you should specify what's wrong with $b=-4$ though. The question says that $theta$ has to be between 0 and $pi/2$. That's the problem. Also just to clarify, it should be sin $theta$ and tan $theta$, not just sin and tan. The way I see it is that it's like writing the square root sign without anything underneath. Actually yes, tan$theta$ should be 4/3.
$endgroup$
That's right. I think you should specify what's wrong with $b=-4$ though. The question says that $theta$ has to be between 0 and $pi/2$. That's the problem. Also just to clarify, it should be sin $theta$ and tan $theta$, not just sin and tan. The way I see it is that it's like writing the square root sign without anything underneath. Actually yes, tan$theta$ should be 4/3.
edited Nov 2 '16 at 17:35
answered Nov 2 '16 at 17:33
Grad studentGrad student
1,169612
1,169612
$begingroup$
Thank you, I will do that.
$endgroup$
– Maggie
Nov 2 '16 at 17:35
add a comment |
$begingroup$
Thank you, I will do that.
$endgroup$
– Maggie
Nov 2 '16 at 17:35
$begingroup$
Thank you, I will do that.
$endgroup$
– Maggie
Nov 2 '16 at 17:35
$begingroup$
Thank you, I will do that.
$endgroup$
– Maggie
Nov 2 '16 at 17:35
add a comment |
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$begingroup$
Actually the result is $frac 4 3$. Also you can use directly $sin^2theta + cos^2theta = 1$
$endgroup$
– user261263
Nov 2 '16 at 17:34
$begingroup$
For writing future questions, note that you should put entire equations between
$signs rather than putting$s around every single symbol you want to format. Line endings in your input do not cause new lines to occur in the output, but$$...$$does put the math on line(s) by itself. Don't use spacing to indicate grouping of operands, use parentheses.$endgroup$
– David K
Nov 2 '16 at 17:42
$begingroup$
Also for working problems in general, if you're going to use symbols (and often you should), be consistent. You wrote $cos theta = a/c$ but a few lines later you substitute $5$ for $a$ and $3$ for $c$ in a different equation, which would have meant $cos theta = 5/3$, obviously not what you intended to say.
$endgroup$
– David K
Nov 2 '16 at 17:47