# Math Help - evaluating trig ratios

1. ## evaluating trig ratios

hello, i am trying to evaluate a trig ratio but the example is quite hard to follow. can someone please explain how this question is done,

given that $sinx=\frac{12}{13},$and that $\frac{\pi}{2}< x<\pi$
show that $cos x= \frac{-5}{13}$
hence without calculators evaluate $sin (x +\pi)$ and $cos2x$

2. Originally Posted by sigma1
hello, i am trying to evaluate a trig ratio but the example is quite hard to follow. can someone please explain how this question is done,

given that $sinx=\frac{12}{13},$and that $\frac{\pi}{2}< x<\pi$
show that $cos x= \frac{-5}{13}$
hence without calculators evaluate $sin (x +\pi)$ and $cos2x$
Dear sigma1,

For the first part use, $cosx=\pm\sqrt{1-sin^{2}x}$ but since theta is in the second quadrant the cosine must be negative hence you can leave out the positive sign.

3. Originally Posted by Sudharaka
Dear sigma1,

For the first part use, $cosx=\pm\sqrt{1-sin^{2}x}$ but since theta is in the second quadrant the cosine must be negative hence you can leave out the positive sign.

thanks i now see the identity butcan you show me how it is applied to prove the question and later evaluate the given trig ratios. thanks.. didnt understand it in class.

4. Originally Posted by sigma1
thanks i now see the identity butcan you show me how it is applied to prove the question and later evaluate the given trig ratios. thanks.. didnt understand it in class.
Dear sigma1,

All you got here is to substitute the values. I think you can do that. For the second part, what you should know is that $sin(x+\pi)=-sinx$ and $cos2x=cos^{2}x-sin^{2}x$

A more detailed discription about trignometric identities could be found at, List of trigonometric identities - Wikipedia, the free encyclopedia

Hope this helps.

5. ok from what i can understand i

substituted $
\sqrt{1-sin^2x} = \frac{-5}{3}

$

$
1-sin^2x =\frac {-5}{13}^2
$

simplifying that should prove the answer right. but i cant seem to simplify it to prove the answer. i must be doing something wrong

because i get $sinx= \frac{8}{13}$

6. Originally Posted by sigma1
hello, i am trying to evaluate a trig ratio but the example is quite hard to follow. can someone please explain how this question is done,

given that $sinx=\frac{12}{13},$and that $\frac{\pi}{2}< x<\pi$
show that $cos x= \frac{-5}{13}$
hence without calculators evaluate $sin (x +\pi)$ and $cos2x$
hi

Since x falls in the second quadrant , sin is positive and cos is negative

Draw a triangle with sides labelled according to the trigo ratio , and use phythagoras theorem to evaluate the other side .

in this case , hypothenus=13 , opposite side =12, adjacent=5

cos x=adjacent/hypothenus=5/13 but look at red , so cos x=-5/13

for the remaining examples ,just expand and substitute appropriately .

7. thanks alot i evaluated the others easily from that...