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.#