Question

compute cos 195° from the functions of 150° and 45°​

Answer 1

Step-by-step explanation:

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Answer 2

$\red\bigstar \sf \: Question : -$

compute cos 195° from the functions of 150° and 45°

$\red \bigstar \sf \: solution : -$

$\implies\sf \: cos \: 195 \degree \\ \\ \implies\sf \: cos(150 \degree \: + 45 \degree ) \\ \\\implies\sf \: cos \: 150 \degree \: cos \: 45 \degree \: - sin150 \degree \: sin \: 45 \degree \:$

$\sf \: (cos(a + b) = cosa - cos \: b - sin \: a \: - sin \: b )$

$\implies\sf \: cos150 \degree \: \times \frac{1}{ \sqrt{2}} - sin150 \degree \: \times \frac{1}{ \sqrt{2 }} \\ \\ \implies\sf \: \frac{1}{ \sqrt{2}} cos \: (180 \degree - 30 \degree) - \frac{1}{ \sqrt{2} } sin (180 \degree - 30 \degree) \\ \\ \implies\sf \: \frac{1}{ \sqrt{2}} ( - cos30 \degree) - \frac{1}{ \sqrt{2} } sin30 \degree \:$

$\sf \: cos(180 \degree - 0) = - cos0$

$\implies\sf \: sin \: (180 \degree - 0) = sin0 \\ \\ \implies\sf \: \frac{ - 1}{ \sqrt{2}} \times \frac{ \sqrt{3} }{2} - \frac{1}{ \sqrt{2} } \times \frac{1}{2} \\ \\ \implies\sf \: \frac{ - ( \sqrt{3 } + 1)}{2 \sqrt{2} } \\ \\ \implies\sf \: ( \frac{ \sqrt{6 + 2} }{4})$

hence proved

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