How do you use the half angle formula to find the value of #sin^2(112.5)#?

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Answer 1 · 2018-04-30T18:59:35

#sin^2(112.5)^circ=(2-sqrt2)/4#

Take,

#T=sin^2(112.5)^circ#

We know that,

#sin^2theta=(1-cos2theta)/2,where,theta=112.5#

#:.T=(1-cos2(112.5)^circ)/2#

#=1/2(1-cos225^circ)#

#=1/2[1-cos(180^circ+45^circ)]#

#=1/2[1-cos(pi+45^circ)]#

#=1/2[1+cos45^circ]...to[as, cos(pi+theta)=-costheta]#

#=1/2[1-1/sqrt2]#

#=1/2[1-sqrt2/2]#

#=1/2[(2-sqrt2)/2]#

#:.X=(2-sqrt2)/4#