Question

If f(x)=cosx and x=2pie\15
Prove that 16f(x)f(2x)f(4x)f(7x)=1

Answer 1

Answer on Question #83946 - Math - Trigonometry

Problem . If $f(x)=cos(x)$ and $x=2\pi/15$. Prove that $16f(x)f(2x)f(4x)f(7x)=1$

Proof. $f(x)f(2x)f(4x)f(7x)=cos(2\pi/15)cos(4\pi/15)cos(8\pi/15)cos(14\pi/15)$

Note that $cos(14\pi/15)=-cos(\pi/15)$

Therefore, we have

$f(x)f(2x)f(4x)f(7x)=-cos(\pi/15)cos(2\pi/15)cos(4\pi/15)cos(8\pi/15)$

$cos(\pi/15)cos(4\pi/15)=1/2(cos(\pi/3)+cos(\pi/5))=1/2(1/2+cos(\pi/5))$

$cos(2\pi/15)cos(8\pi/15)=1/2(cos(2\pi/3)+cos(2\pi/5))=1/2(-1/2+cos(2\pi/5))$

So, $f(x)f(2x)f(4x)f(7x)=-1/4(1/2+cos(\pi/5))(-1/2+cos(2\pi/5))=$

$=-1/4(-1/4-1/2cos(\pi/5)+1/2cos(2\pi/5)+cos(\pi/5)cos(2\pi/5))=$

$=-1/4(-1/4-1/2cos(\pi/5)+1/2cos(2\pi/5)+1/2cos(\pi/5)+1/2cos(3\pi/5))=$

$=-1/4(-1/4+1/2cos(2\pi/5)+1/2cos(3\pi/5))$

since $cos(2\pi/5)=cos(\pi-3\pi/5)=-cos(3\pi/5)$

$f(x)f(2x)f(4x)f(7x)=-1/4(-1/4)=1/16$

So, $16f(x)f(2x)f(4x)f(7x)=1$ $\Box$

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