# Solving trig functions...

• Apr 29th 2010, 11:38 PM
KAY03
Solving trig functions...
I am having trouble with this problem...It goes as follows:

Find the exact value of:
sin(2x), cos(2x), and tan(2x) if cos(x) = -4/5, [pi]/2 <x<[pi].

It gives me a hint saying( sin2x=2sinxcosx and cos(2x)=2cos^2 (X-1)

I have no idea how to do this (Worried) Anyone care to help?
• Apr 30th 2010, 12:04 AM
Hello KAY03
Quote:

Originally Posted by KAY03
I am having trouble with this problem...It goes as follows:

Find the exact value of:
sin(2x), cos(2x), and tan(2x) if cos(x) = -4/5, [pi]/2 <x<[pi].

It gives me a hint saying( sin2x=2sinxcosx and cos(2x)=2cos^2 (X-1)

I have no idea how to do this (Worried) Anyone care to help?

First, note that, if $\pi/2 , then $\sin x$ is positive.

Then, using $\sin^2x =1-\cos^2x$, plug in the value $\cos x = -\tfrac45$:
$\sin^2x=1 - \tfrac{16}{25}$
$=\tfrac{9}{25}$
$\Rightarrow\sin x = \tfrac35$ (taking the positive square root, since, as we said, $\sin x > 0$)
Now you can simply plug in the values $\sin x = \tfrac35$ and $\cos x = -\tfrac45$ into the formulae that you've been given. The first one is:
$\sin 2x =2\sin x \cos x$
$=2\cdot\tfrac35\cdot(-\tfrac45)$

$= ...$ ?
Do $\cos 2x$ in the same way, using the formula:
$\cos2x=2\cos^2x -1$
Finally, use $\tan2x = \frac{\sin2x}{\cos2x}$ to find $\tan2x$.

Can you complete them now?

• Apr 30th 2010, 12:16 AM
KAY03
Just to make sure I did it right..my final answers are

sin2x= -24/25
cos2x=7/25

Am I correct?
• Apr 30th 2010, 05:49 AM
Hello KAY03
Quote:

Originally Posted by KAY03
Just to make sure I did it right..my final answers are

sin2x= -24/25
cos2x=7/25