
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=sqrt29/25=sqrt29/5
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 ...
$\displaystyle \cos\left(x + \frac{\pi}{4}\right) = \cos{x}\cos\left(\frac{\pi}{4}\right)  \sin{x}\sin\left(\frac{\pi}{4}\right)$
$\displaystyle \sin(2x) = 2\sin{x}\cos{x}$