# Trigonometry identities help

• Nov 3rd 2012, 02:52 PM
Benja303
Trigonometry identities help
Section B

Number 9) II

What i did
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cos2x=1-2sin275

cos 2x = cos (x+x)

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

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

section B part III number 9
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cos 2x = cos2x - sin2x

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

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

I attempted this one similarly.. wrong answer... help!
• Nov 3rd 2012, 02:57 PM
richard1234
Re: Trigonometry identities help
9. II: Note that $\displaystyle 1 - 2 \sin^2 {75^{\circ}} = \cos(150^{\circ})$.
• Nov 3rd 2012, 03:02 PM
Benja303
Re: Trigonometry identities help
I know, sorry i made a mistake in labelling and typing , now fixed,
• Nov 3rd 2012, 03:09 PM
richard1234
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.
• Nov 3rd 2012, 03:46 PM
Benja303
Re: Trigonometry identities help
got it! thanks !
• Nov 3rd 2012, 04:02 PM
Benja303
Re: Trigonometry identities help