Losing my mind

I have searched my book front to back to find just one example of this type of problem and fond nothing. I
have also looked through both this forum and yahoo answers and saw nothing this ridiculous.

PLZ help me and save me from an anurism

Sin x = -2/5 x in quad 3

exact value of cos (x+Pi/4)




and the second seems easier but im afraid to try it

Exact value of sin 2x

I already did something i though helped, but got me no closer to the answer

sin x^2 + cos x^2 = 1
cos x^2 = 1+ 2/5^2=1 + 4/25=29/25=sqrt-29/25=sqrt-29/5


and this is where i contemplate my lifes worth..... This problem was on my final at OSU and i need a good grade! Help!

x in quad III

$\displaystyle \sin{x} = -\frac{2}{5}$

sketch a reference triangle in quad III ... opposite side = -2, hypotenuse = 5

adjacent side = $\displaystyle -\sqrt{5^2 - 2^2} = -\sqrt{21}$

$\displaystyle \cos{x} = -\frac{\sqrt{21}}{5}$

now that you have the necessary values ...

$\displaystyle \cos\left(x + \frac{\pi}{4}\right) = \cos{x}\cos\left(\frac{\pi}{4}\right) - \sin{x}\sin\left(\frac{\pi}{4}\right)$

$\displaystyle \sin(2x) = 2\sin{x}\cos{x}$