Question

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

Answer 1

Hey :)

Answer 2

hello

Answer 3

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

Answer 4

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

Answer 5

So I'm drawing a triangle in quadrant 3.

Answer 6

We can label the sides like this, since sine = opposite / hypotenuse.

Note: Never put the negative sign on the hypotenuse length.

Answer 7

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

Answer 8

12

Answer 9

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

Answer 10

-12

Answer 11

Ok good good.

Answer 12

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.

Answer 13

Remember that one? :)

Answer 14

would it be -25/78

Answer 15

Hmmm no :o

Answer 16

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

Answer 17

120/169?

Answer 18

Mmm ya that looks right! :)

Answer 19

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