Prove that:
Question:

Prove that:

(i) $\sin \left(\frac{\pi}{3}-x\right) \cos \left(\frac{\pi}{6}+x\right)+\cos \left(\frac{\pi}{3}-x\right) \sin \left(\frac{\pi}{6}+x\right)=1$

(ii) $\sin \left(\frac{4 \pi}{9}+7\right) \cos \left(\frac{\pi}{9}+7\right)-\cos \left(\frac{4 \pi}{9}+7\right) \sin \left(\frac{\pi}{9}+7\right)=\frac{\sqrt{3}}{2}$

 

(iii) $\sin \left(\frac{3 \pi}{8}-5\right) \cos \left(\frac{\pi}{8}+5\right)+\cos \left(\frac{3 \pi}{8}-5\right) \sin \left(\frac{\pi}{8}+5\right)=1$

Solution:

(i) $\frac{\pi}{3}=60^{\circ}, \frac{\pi}{6}=30^{\circ}$

LHS $=\sin \left(60^{\circ}-x\right) \cos \left(30^{\circ}+x\right)+\cos \left(60^{\circ}-x\right) \sin \left(30^{\circ}+x\right)$

$=\sin \left[\left(60^{\circ}-x\right)+\left(30^{\circ}+x\right)\right]$

(U $\operatorname{sing}$ the formula $\sin A \cos B+\cos A \sin B=\sin (A+B)$   and taking $A=60^{\circ}-x$ and $\left.B=30^{\circ}+x\right)$

$=\sin 90^{\circ}$

$=1$

= RHS

Hence proved.

(ii) $\sin \left(\frac{4 \pi}{9}+7\right) \cos \left(\frac{\pi}{9}+7\right)-\cos \left(\frac{4 \pi}{9}+7\right) \sin \left(\frac{\pi}{9}+7\right)$

$=\sin \left[\left(\frac{4 \pi}{9}+7\right)-\left(\frac{\pi}{9}+7\right)\right] \quad[\sin A \cos B-\cos A \sin B=\sin (A-B)]$

$=\sin \frac{3 \pi}{9}$

$=\sin \frac{\pi}{3}$

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

(iii) $\sin \left(\frac{3 \pi}{8}-5\right) \cos \left(\frac{\pi}{8}+5\right)+\cos \left(\frac{3 \pi}{8}-5\right) \sin \left(\frac{\pi}{8}+5\right)$

$=\sin \left[\left(\frac{3 \pi}{8}-5\right)+\left(\frac{\pi}{8}+5\right)\right] \quad[\sin A \cos B+\cos A \sin B=\sin (A+B)]$

$=\sin \frac{4 \pi}{8}$

$=\sin \frac{\pi}{2}$

$=1$

 

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