How do you evaluate $sin(pi/5)$ ?

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Answer 1 · 2016-07-02T04:10:42

$sin(pi/5)=sqrt(10-2sqrt5)/4$

Let $theta=pi/5$ , then $5theta=pi$

and $3theta=pi-2theta$ . Note #theta) is an acute angle.

Hence $sin3theta=sin(pi-2theta)$ but as $sin(pi-A)=sinA$

This can be written as

$sin3theta=sin2theta$ expanding them

or $3sintheta-4sin^3theta=2sinthetacostheta$

as $theta=pi/5$ we have $sintheta!=0$ and dividing by it we get

$3-4sin^2theta=2costheta$ or

$3-4(1-cos^2theta)=2costheta$

or $4cos^2theta-2costheta-1=0$

and using quadratic formula $costheta=(2+-sqrt(2^2-4*4*(-1)))/(2*4)$

= $(2+-sqrt(20))/8=(1+-sqrt5)/4$ .

But as $(1-sqrt5)/4$ is negative and $costheta$ cannot take this value, hence

$costheta=(1+sqrt5)/4$ and

$sintheta=sqrt(1-((1+sqrt5)/4)^2)=sqrt(1-((1+5+2sqrt5)/16))$

= $sqrt((16-6-2sqrt5)/16)=sqrt(10-2sqrt5)/4$

As $theta=pi/5$

$sin(pi/5)=sqrt(10-2sqrt5)/4$

How do you evaluate sin(pi/5)sin(pi/5)?