Question

prove that
Sin 50° - sin 70° +cos 80°=0​

Answer 1

Question :-

• Prove that : sin 50° - sin 70° + cos 80° = 0.

Answer

Given : -

• sin 50° - sin 70° + cos 80°

To Prove :-

• sin 50° - sin 70° + cos 80° = 0

Identity Used :-

$\bf \:sinx - siny = 2cos(\dfrac{x + y}{2} )sin(\dfrac{x - y}{2} )$

$\bf \:sin( - x) = - sinx$

$\bf \:cos(90° - x) = sinx$

Solution :-

$\bf \:Consider \: sin 50° - sin 70° + cos 80°$

Using identity sinx - siny, we get

$\bf\implies \:2cos(\dfrac{50° + 70°}{2} )sin(\dfrac{50° - 70°}{2} ) + cos80°$

$\bf\implies \:2cos(\dfrac{120°}{2} )sin(\dfrac{ - 20}{2} ) + cos80°$

$\bf\implies \:2cos60°sin( - 10°) + cos80°$

$\bf\implies \: - 2 \times \dfrac{1}{2} \times sin10° + cos(90° - 10°)$

$\bf\implies \: - sin10° + sin10°$

$\bf\implies \:0$

$\large{\boxed{\boxed{\bf{Hence, Proved}}}}$

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