Question #242ce

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Answer 1 · 2017-11-23T19:13:47

#=sqrt2/2#

We know:

#sin(a+b)=sinacosb+cosasinb#

In this case, #a=35# and #b=100#, therefore:

#sin35cos100+cos35sin100=sin(35+100)=sin135=sqrt2/2#

Answer 2 · 2017-11-23T19:17:25

# 1/sqrt2, or, sqrt2/2.#

Recall that, #sinxcosy+cosxsiny=sin(x+y).#

#:. sin35^@cos100^@+cos35^@sin100^@=sin(35+100)^@,#

#=sin135^@,#

#=sin(180^@-45^@),#

#=sin45^@,#

#=1/sqrt2, or, sqrt2/2.#