Cos (67°) sin (68°) + cos (23°) sin (22°) equal ?

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Answer 1 · 2017-09-24T04:14:17

$1/sqrt2, or, sqrt2/2.$

Recall that, $costheta=sin(90^@-theta).$

$:. cos67^@=sin(90^@-67^@)=sin23^@..............(1).$

Also, $because, sintheta=cos(90^@-theta), :. sin22^@=cos68^@....(2).$

Using $(1), and, (2),$ the Given Expression becomes,

$sin23^@sin68^@+cos23^@cos68^@.$

But, $sinAsinB+cosAcosB=cos(B-A),$

$"The Exp.="cos(68^@-23^@),$

$=cos45^@,$

$=1/sqrt2, or,$

$=sqrt2/2.$