Question

Prove that sin 40° + sin 20° - cos 10º = 0​

Answer 1

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consider LHS

\[\sin40^\circ+\sin20^\circ\]

\[=2\sin\left(\frac{40^\circ+20^\circ}{2}

\right)\cos\left(\frac{40^\circ-20^\circ}{2}

\right)\left\{\because\sinA\sinB=2\sin\left(|frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right\right\3\]\[=2\sin30^\circ\cos10^\circ\]\}=2\times\frac{1}{2}\cis10^\circ\]\=\cos\left(10^\circ\right)\]

Hence,LHS=RHS

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Answer 2

$\large\underline{\sf{Solution-}}$

Consider,

$\rm :\longmapsto\:sin40\degree + sin20\degree - cos10\degree$

$\rm \: = (\:sin40\degree + sin20\degree) - cos10\degree$

We know,

$\boxed{ \tt{ \: sinx + siny = 2sin\bigg[\dfrac{x + y}{2} \bigg]cos\bigg[\dfrac{x - y}{2} \bigg]}}$

Using this, we get

$\rm \: = \:2sin\bigg[\dfrac{40\degree + 20\degree }{2} \bigg]cos\bigg[\dfrac{40\degree - 20\degree }{2} \bigg] - cos10\degree$

$\rm \: = \:2sin\bigg[\dfrac{60\degree }{2} \bigg]cos\bigg[\dfrac{20\degree}{2} \bigg] - cos10\degree$

$\rm \: = \:2sin\bigg[30\degree \bigg]cos\bigg[10\degree \bigg] - cos10\degree$

$\rm \: = \:2 \times \dfrac{1}{2} \times cos10\degree - cos10\degree$

$\rm \: = \: cos10\degree - cos10\degree$

$\rm \: = \:0$

Hence,

$\rm :\longmapsto\:\boxed{ \tt{ \: sin40\degree + sin20\degree - cos10\degree = 0 \: \: }}$

Additional Information :-

$\boxed{ \tt{ \: sinx - siny = 2cos\bigg[\dfrac{x + y}{2} \bigg]sin\bigg[\dfrac{x - y}{2} \bigg]}}$

$\boxed{ \tt{ \: cosx + cosy = 2cos\bigg[\dfrac{x + y}{2} \bigg]cos\bigg[\dfrac{x - y}{2} \bigg]}}$

$\boxed{ \tt{ \: cosx - cosy \: = - \: 2sin\bigg[\dfrac{x + y}{2} \bigg]sin\bigg[\dfrac{x - y}{2} \bigg]}}$

$\boxed{ \tt{ \: 2cosxcosy = cos(x + y) + cos(x - y) \: }}$

$\boxed{ \tt{ \: 2sinxsiny = cos(x - y) - cos(x + y) \: }}$

$\boxed{ \tt{ \: 2sinxcosy = sin(x + y) + sin(x - y) \: }}$