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

• Jan 23rd 2013, 12:40 PM
koolaid123
show by algebraic means the value of sin^4(15) - cos^4(15_
i know they are difference of squares but i dont know what to do after....the answer is -root3over2
• Jan 23rd 2013, 01:03 PM
Plato
Re: show by algebraic means the value of sin^4(15) - cos^4(15_
Quote:

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*}
• Jan 23rd 2013, 01:06 PM
MarkFL
Re: show by algebraic means the value of sin^4(15) - cos^4(15_
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...