Question

prove the following
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Answer 1

We have two methods to do the above question.

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M E T H O D 1 :

$\sf \implies \: sin( \: 112 \degree) + cos( \: 74 \degree) - sin( \: 68 \degree) + cos( \: 106 \degree)$

$\sf \implies \: sin( \: 90 \degree + 22 \degree) + cos ( \: 90 \degree - 16 \degree) - sin( \: 90 \degree - 22 \degree) + cos( \: 90 \degree + 16 \degree)$

$\sf \implies \: cos( \: 22 \degree) + sin( \: 16 \degree ) - cos( \: 22 \degree) - sin( \: 16 \degree)$

$\sf \implies \: cos( \: 22 \degree) - cos( \: 22 \degree) + sin( \: 16 \degree) - sin( \: 16 \degree)$

$\sf\implies~0+0$

$\sf \implies \: 0$

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N O T E :

$\sf \implies \: sin( \: 90 \degree + 22 \degree) = cos( \: 22 \degree)$

• II quadrant is positive for sin

$\sf \implies \: cos( \: 90 \degree- 16 \degree) = sin( \: 16 \degree)$

• I quadrant is positive for cos

$\sf \implies \: - \: sin( \: 90\degree - 22 \degree) = - \: cos( \: 22 \degree)$

• I quadrant is positive for sin

$\sf \implies \: cos( \: 90 \degree + 16 \degree) = - \: sin( \: 16 \degree)$

• II quadrant is negative for cos

M E T H O D 2 :

$\sf \implies \: sin( \: 112 \degree) + cos( \: 74 \degree) - sin( \: 68 \degree) + cos( \: 106 \degree)$

$\sf \implies \: cos( \: 106 \degree) + cos( \: 74 \degree) + sin( \: 112 \degree) - sin( \: 68 \degree)$

$\sf \implies \: 2 \: cos \bigg( \: \dfrac{106 \degree + 74 \degree}{2} \bigg) \: cos \bigg( \: \dfrac{106 \degree - 74 \degree}{2} \bigg ) + 2 \: sin \bigg( \: \dfrac{112 \degree - 68 \degree}{2} \bigg) \: cos \bigg( \: \dfrac{112 \degree + 68 \degree}{2} \bigg)$

$\sf \implies \: 2 \: cos \bigg( \: \dfrac{180 \degree}{2} \bigg) \: cos \bigg( \: \dfrac{32 \degree}{2} \bigg) + sin \bigg( \: \dfrac{44 \degree}{2} \bigg) \: cos \bigg( \: \dfrac{180 \degree}{2} \bigg)$

$\sf \implies \: 2 \: cos ( \: 90 \degree) \: cos( \: 16 \degree) + 2 \: sin( \: 22 \degree) \: cos( \: 90 \degree)$

$\sf\implies~ 2~(~0~)~cos(~16\degree)+2~sin(~22\degree)~(~0~)$

$\sf\implies~0+0$

$\sf \implies \: 0$

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N O T E :

$\sf \implies \: cos( \: 90 \degree) = 0$

$\sf\implies~ sinC-sinD=2~sin\bigg(~\dfrac{C-D}{2}~\bigg)~ cos\bigg(~\dfrac{C+D}{2}~\bigg)$

$\sf\implies~cosC+cosD=2~cos\bigg(~\dfrac{C+D}{2}~\bigg)~cos\bigg(~\dfrac{C-D}{2}~\bigg)$