cos-2-76-cos-2-16-cos-76-cos-16-

Question Number 133929 by bemath last updated on 25/Feb/21

\cos^{2} 76 ° + \cos^{2} 16 ° - \cos 76 ° . \cos 16 ° = ?

Answered by liberty last updated on 25/Feb/21

\Rightarrow a^{2} + b^{2} = {(a + b)}^{2} - 2 a b

\cos^{2} 76 ° + \cos^{2} 16 ° - \cos 76 ° . \cos 16 ° =

{[\cos 76 ° + \cos 16 °]}^{2} - 3 \cos 76 ° \cos 16 °

= {[2 \cos 46 ° \cos 30 °]}^{2} - \frac{3}{2} [\cos 92 ° + \cos 60 °]

= 3 \cos^{2} 46 ° - \frac{3}{2} \cos 92 ° - \frac{3}{4}

= 3 [\frac{1}{2} + \frac{1}{2} \cos 92 °] - \frac{3}{2} \cos 92 ° - \frac{3}{4}

= \frac{3}{2} - \frac{3}{4} = \frac{3}{4}

Answered by EDWIN88 last updated on 25/Feb/21

\cos^{2} 76 ° + \cos^{2} 16 ° - \cos 76 ° . \cos 16 ° =

\frac{1}{2} [1 + \cos 152 ° + 1 + \cos 32 ° - \cos 92 ° - \cos 60 °]

= \frac{1}{2} [2 - \frac{1}{2} + \cos 152 ° + \cos 32 ° - \cos 92 °]

= \frac{1}{2} [\frac{3}{2} + 2 \cos 92 ° \cos 60 ° - \cos 92 °]

= \frac{1}{2} [\frac{3}{2} + \cos 92 ° - \cos 92 °] = \frac{3}{4}

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