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Trigonometry identities help

Section B

Number 9) II

How do go about answering this?

Please see attached pdf

What i did

________

cos2x=1-2sin^{2}75

cos 2x = cos (x+x)

cos (x+x) = cosA*CosB-SinA*SinB

cos( 60 + 90)= cos60*cos90 - sin60*cos90

answer i got was wrong

section B part III number 9

____________________

cos 2x = cos^{2}x - sin^{2}x

cos( x + x) = cosA*CosB-SinA*SinB

cos(90 + 45)= cos90*cos45-sin90*sin45

I attempted this one similarly.. wrong answer... help!

Re: Trigonometry identities help

9. II: Note that $\displaystyle 1 - 2 \sin^2 {75^{\circ}} = \cos(150^{\circ})$.

Re: Trigonometry identities help

I know, sorry i made a mistake in labelling and typing , now fixed,

Re: Trigonometry identities help

Oh okay, I was a little confused with the 45/60 thing.

Also, $\displaystyle \sin(60^{\circ}) = \frac{\sqrt{3}}{2}$, not $\displaystyle \frac{2}{\sqrt{3}}$. The sine/cosine of an angle can only be from -1 to 1 inclusive. Since $\displaystyle -\frac{2}{\sqrt{3}}$ is outside the set [-1,1], this answer cannot be correct.

Re: Trigonometry identities help

Re: Trigonometry identities help

what about 9) III? my answer is -1/square root of 2