Question

prove that cos50+sin20 =cos10​

Answer 1

To Prove :

$\sf : \; \implies cos50^{\circ} + sin20^{\circ} = cos10^{\circ}$

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❍ Required proof for the same is ::

$\sf : \; \implies cos50^{\circ} + sin20^{\circ} = cos10^{\circ}$

~Using Identity : sin( x )= cos( 90 - x )

$\sf : \; \implies cos50^{\circ} + cos(90-20)^{\circ} = cos10^{\circ}$

$\sf : \; \implies cos50^{\circ} + cos70^{\circ} = cos10^{\circ}$

~Using Identity : cos( a ) + cos( b )= 2cos[( a + b )/2]cos[( a - b )/2]

$\sf : \; \implies 2cos \bigg( \dfrac{70+50}{2} \bigg)cos \bigg( \dfrac{70-50}{2} \bigg) = cos10^{\circ}$

$\sf : \; \implies 2cos \bigg( \dfrac{120}{2} \bigg)cos \bigg( \dfrac{20}{2} \bigg) = cos10^{\circ}$

~ As we know that : cos60° = 1/2

$\sf : \; \implies 2cos( 60 )^{\circ}cos(10)^{\circ} = cos10^{\circ}$

$\sf : \; \implies 2 \times \dfrac{1}{2} \times cos(10)^{\circ} = cos10^{\circ}$

$\sf : \; \implies 1 \times cos10^{\circ} = cos10^{\circ}$

$\boxed{\bf{ \bigstar \;\; cos10^{\circ} = cos10^{\circ}}}$

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LHS = RHS

Hence Proved!