Can someone tell me how to prove that, sin 78° + cos 132° = (√5 - 1)/4, in the easiest method?

Question

12243 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-02-20T17:27:04

#LHS=sin 78° + cos 132°#

#=sin (90^@-12°) + cos 132°#

#=cos12° + cos 132°#

#=2cos72°cos60°#

#=2cos(90^@-18°)*1/2#

#=sin18°#

# = (√5 - 1)/4# [Please see below]

Let #x=18^@#

#=>5x=90^@#

#=>cos3x=cos(90^@-2x)#

#=>4cos^3x-3cosx=sin2x#

#=>4cos^3x-3cosx=2sinxcosx#

As #cosx!=0#
We get

#=>4cos^2x-3=2sinx#

#=>4-4sin^2x-3=2sinx#

#=>4sin^2x+2sinx-1=0#

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

#=>sinx=(-2+sqrt20)/(2*4)#

#=>sinx=(-2+2sqrt5)/(2*4)#

#=>sinx=(sqrt5-1)/4#

#=>sin18^@=(sqrt5-1)/4#