Question

Prove that
Sin80 -cos110 = root3sin50

Answer 1

$\mathbf{\sin80^\circ-\sin20^\circ=\sqrt{3}\sin50^\circ}$

Hence proved

Step-by-step explanation:

To prove:  $\sin80^\circ-\cos110^\circ=\sqrt{3}\sin50^\circ$

Proof:

Taking left side

$\Rightarrow \sin80^\circ-\cos110^\circ$

$\Rightarrow \sin80^\circ-\cos(90^\circ+20^\circ)$      $\cos(90+\theta)=-\sin\theta$

$\Rightarrow \sin80^\circ+\sin20^\circ$        $\sin A+\sin B=2\sin(\frac{A+B}{2})\cos(\frac{A-B}{2})$

$\Rightarrow 2\sin(\frac{80^\circ+20^\circ}{2})\cos(\frac{80^\circ-20}{2})$

$\Rightarrow 2\sin50^\circ\cos30^\circ$

$\Rightarrow 2\sin50^\circ\cdot \dfrac{\sqrt{3}}{2}$

$\Rightarrow \sqrt{3}\sin50^\circ$

LHS = RHS

Hence proved

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Answer 2

$\sin80 - \cos110=\sqrt{3}\sin50$ proved

Step-by-step explanation:

Consider the provided information.

$\sin80 - \cos110=\sqrt{3}\sin50$

Consider the LHS.

$\sin80 - \cos(90+20)$

$\sin80 +\sin20$ (∴Cos 90+θ = -Sin θ)

Use the identity: $\sin A +\sin B = 2\sin \frac{(A+B)}{2}\cos\frac{(A-B)}{2}$

$2\sin\frac{80+20}{2}\cos\frac{80-20}{2}$

$2\sin50\cos30$

$2\sin50\frac{\sqrt{3}}{2}$

$\sqrt{3}\sin50$

Hence, LHS=RHS

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