sin,cos problems

[Warwick]

I'm doing a review for my math class and came upon some that I cannot figure out.I don't know why I cannot get the right answer. I know all the rules, I've had tests on this stuff and received A's but these are different for some reason and I don't know why. If anyone can do even one of them I would so greatly appreciate it! my exam is tomarrow :cry: thanks




let sin(x)=5/13, cos(x)=12/13, sin(y)=4/5 and y is in Quadrant II and 0<or=to x <2pi

81. sin(x-y)=____

83. sin 2x =___

84. sin x/2 =___

now the rules above don't apply to these below

85. sin(pheta)=2/3 and 0<pheta<90 then cos2(pheta)=___

86.sin(2cos^-1(1/4))=___

[arildno]

1.
Determine the cosine value of y:
You are given that y is in Quadrant 2, which means that the cosine value of y is negative (Agreed?)

In order to determine the cosine value of y, use the fundamental identity:
\cos^{2}y+\sin^{2}y=1.
Knowing that the cosine value has to be negative, you should be able to figure out the answer.
2. 81,82,83:
Knowing the summation, double angle, half-angle formulae for the trigonometric functions should now give you the answers.
3.
85:
\sin^{2}\theta=\frac{1-\cos(2\theta)}{2}
4.86:

\sin(2\phi)=2\sin\phi\cos\phi, \sin\phi=\pm\sqrt{1-\cos^{2}\phi}