Question #69254

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Answer 1 · 2018-02-19T15:56:18

$(2+sqrt(2))/4$

We have:

$sin^2(67.5^@)$

Using the identity, $sin(x)=cos(90^@-x)$

$:.sin(67.5^@)=cos(90^@-67.5^@)$

$=cos(22.5^@)$

We got: $cos(22.5^@)=cos(45^@/2)$

Using the half-angle identity, $cos(x/2)=sqrt((cos(x)+1)/2)$

$:.cos(45^@/2)=sqrt((cos(45^@)+1)/2)$

We know that $cos(45^@)=sqrt(2)/2$

$<=>cos(45^@/2)=sqrt((sqrt(2)/2+1)/2)$

$=sqrt((2+sqrt(2))/4)$

$=sqrt(2+sqrt(2))/sqrt(4)$

$=sqrt(2+sqrt(2))/2$

$:.cos(22.5^@)=sqrt(2+sqrt(2))/2$

$<=>sin(67.5^@)=sqrt(2+sqrt(2))/2$

$:.sin^2(67.5^@)=(sqrt(2+sqrt(2))/2)^2$

$=(2+sqrt(2))/4$