find cos2x and sin2x given sinx=-5/13 and x is in quadrant III

Question

zepdrix · Answer

Hey :)

anonymous · Answer

hello

zepdrix · Answer

Sorry gimme couple mins +_+ wrappin up a phone call.

zepdrix · Answer

Ok ok ok sorry bout that.
Where we at? We gotta evaluate the cos2x and sin2x?

zepdrix · Answer

|dw:1393810435940:dw|

zepdrix · Answer

So I'm drawing a triangle in quadrant 3.

zepdrix · Answer

We can label the sides like this, since sine = opposite / hypotenuse.

Note: Never put the negative sign on the hypotenuse length.

zepdrix · Answer

We can use the Pythagorean Theorem to find the missing leg.
What do you get? :D

anonymous · Answer

12

zepdrix · Answer

Ok good.
Since we're in quadrant 3, will it be 12 or -12?

anonymous · Answer

-12

zepdrix · Answer

|dw:1393810623870:dw|Ok good good.

zepdrix · Answer

So we've successfully setup our triangle. Yay!

Now let's get to the meat and potatoes of this problem.
We need to take our expression,$$\Large\bf\sf \sin2x$$And apply the `Sine Double Angle Identity` to it.

zepdrix · Answer

Remember that one? :)

anonymous · Answer

would it be -25/78

zepdrix · Answer

Hmmm no :o

zepdrix · Answer

$$\Large\bf\sf \sin2x\quad=\quad 2 \sin x \cos x\quad=\quad 2\left(-\frac{5}{13}\right)\left(-\frac{12}{13}\right)$$

anonymous · Answer

120/169?

zepdrix · Answer

Mmm ya that looks right! :)

zepdrix · Answer

Do you remember your Cosine Double Angle Identity?
You'll need to use that for the other one.