Find sin72 if you know the sines of 30 45 60?

1 Answer
Narad T.
Jul 20, 2018

The answer is #=sqrt(10+2sqrt5)/4#

Explanation:

We need

#sin(2x)=2sinxcosx#

#cos(2x)=1-2sin^2x#

#sinx=cos(90-x)#

#sin^2x+cos^2x=1#

Let's start

#sin(72)=sin(2*36)#

#=2sin36cos36#

#=2sin(2*18)cos(2*18)#

#=2sin(2sin18cos18)(1-2sin^2 18)#

#=4sin18cos18(1-2sin^2 18)#

Therefore,

#sin72=cos(90-72)=4sin18cos18(1-2sin^2 18)#

#cos18=4sin18cos18(1-2sin^2 18)#

Divide both sides by #cos18#

#1=4sin18(1-2sin^2 18)#

#8sin^3 18-4sin18+1=0#

Factorise

#(2sin18-1)(4sin^2 18+ 2sin18-1)=0#

Therefore,

#sin18=1/2# which is #O/#

The solution to the quadratic equation is

#sin18=(-2+-sqrt(2^2-4*4*(-1)))/(2*4)#

#=(-2+-sqrt20)/8#

#=(-2+-2sqrt5)/8#

#=(-1+-sqrt5)/4#

Discard #sin18=(-1-sqrt5)/4# as this is #<0#

We keep only #sin18=(-1+sqrt5)/4#

So,

#cos72=(-1+sqrt5)/4#

#sin72=sqrt(1-cos^2 72)#

#=sqrt(1-((-1+sqrt5)/4)^2)#

#=sqrt((16-(1+5-2sqrt5)))/4#

#=sqrt(10+2sqrt5)/4#

Related questions
Impact of this question
958 views around the world
You can reuse this answer
Creative Commons License