TS ICET 23 May 2018 Shift 2 Paper

Question : 119 of 200

Marks: +1, -0

sin 420° cos 390° + cos(-300°) sin(330°) =

Solution:

\sin 420^{∘}=\sin (360^{∘}+60)

=\sin 60^{∘}=√{3}/2

\cos 390^{∘}=\cos (360^{∘}+30^{∘})

=\cos 30^{∘}=√{3}/2

\cos (-300^{∘}) =\cos 300^{∘}

=\cos (360^{∘}-60^{∘})

=\cos 60^{∘}=1/2

\sin 330^{∘}=\sin (360^{∘}-30^{∘})

=-\sin 30^{∘}=-1/2

\GE=\;{√{3}}/{2} × \;{√{3}}/{2}+\;{1}/{2} ×(-\;{1}/{2})=\;{3-1}/{4}=\;{1}/{2}

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