show by algebraic means the value of sin^4(15) - cos^4(15_

**koolaid123** · January 23rd 2013, 12:40 PM

i know they are difference of squares but i dont know what to do after....the answer is -root3over2

**Plato** · January 23rd 2013, 01:03 PM

Originally Posted by koolaid123

i know they are difference of squares but i dont know what to do after....the answer is -root3over2

$\begin{align*} \sin^4(15^o)- \cos^4 (15^o) &= [\sin^2(15^o)- \cos^2 (15^o)][ \sin^2(15^o)+\cos^2 (15^o)] \\ &=\sin^2(15^o)- \cos^2 (15^o) \\ &= -\cos(30^o)\end{align*}$

**MarkFL** · January 23rd 2013, 01:06 PM

We are given to evaluate:

$\sin^4(15^{\circ})-\cos^4(15^{\circ})$

You are right, recognizing this as a difference of squares is a good start:

$(\sin^2(15^{\circ})+\cos^2(15^{\circ}))(\sin^2(15^ {\circ})-\cos^2(15^{\circ}))$

Next, with the angle sum/difference identities for cosine in mind, let's rewrite this as:

$-(\sin^2(15^{\circ})+\cos^2(15^{\circ}))(\cos^2(15^ {\circ})-\sin^2(15^{\circ}))$

The first factor should look familiar as a Pythagorean identity, and the second factor should look familiar as a double-angle identity for cosine...

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