Question

if cosgama÷a=singama÷b then cos2gama+bsin2gama​

Answer 1

$\frac{ \cos( \gamma ) }{a } = \frac{ \sin( \gamma ) }{b } \\ \cos( \gamma ) = \sin( \gamma ) \frac{a}{b} \\ a \cos(2 \gamma ) + b \sin(2 \gamma ) = a(1 - 2 { \sin}^{2} \gamma ) + b(2 \sin( \gamma ) \cos( \gamma ) \\ putting \cos( \gamma ) = \sin( \gamma ) \frac{a}{b} \\ a( \cos(2 \gamma) ) + b \sin(2 \gamma ) = a - 2a \: { \sin }^{2} \gamma + b(2 \sin( \gamma ) \times \sin( \gamma ) \frac{a}{b} \\ \: \: \: \: \: \: = a - 2 { \sin }^{2} \gamma \: a + 2 { \sin}^{2} \gamma \: a \\ \: \: \: \: \: \: \: = a$

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