Question #e1294

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Answer 1 · 2017-12-29T23:46:03

#sinOcosO=0#

.

#sinO +cosO=1#

We square both sides:

#(sinO+cosO)^2=1#

#sin^2O+cos^2O+2sixOcosO=1#

But we know that for any angle #alpha# we have:

#sin^2alpha +cos^2alpha=1#

We substitute:

#1+2sinOcosO=1#

#2sinOcosO=0#

#sinOcosO=0#