Prove that: 4 A (60^ - A ) (60^ + A )= 3 A. Hence deduce that: 20^ 40^ 60^ 80^ =3 16

l L H S=4 A (60^ -A ) (60^ +A ) =2 A [2 (60^ -A ) (60^ +A ) ] =2 A [ (60^ - A )- (60^ + A ) - (60^ - A )+ (60^ + A ) ] [ 2 A B = ( A - B )- ( A + B )] =2 A [ (-2 A)- 120^ ] =2 A [ 2 A- 120^ ][ (- )= ] =2 A 2 A-2 A 120^ =[ ( A +2 A)+ ( A -2 A)]-2 A (-1 2 ) l [ 2 A B= (A+B)+ (A-B) and 120^ =-1 2 ] = 3 A+ (-A)+ A = 3 A- A+ A= 3 A= RHS [ (- )=- ] LHS = RHS Hence proved. Now, 4 A (60^ -A ) (60^ +A )= 3 A On putting A =20^ , we get 4 20^ (60^ -20^ ) (60^ +20^ )= 3 (20^ ) l 4 20^ 40^ 80^ = 60^ = 3 2 20^ 40^ 80^ = 3 8 20^ 40^ 3 2 80^ = 3 8 3 2 [multiplying both sides by 3 2 ] 20^ 40^ 60^ 80^ =3 16 [ 3 2 = 60^ ]

Prove that: 4 A (60^ - A ) (60^ + A )= 3 A. Hence deduce that: 20…

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Question

Prove that: $4 \sin A \sin \left(60^{\circ}- A \right) \sin \left(60^{\circ}+ A \right)=\sin 3 A$.
Hence deduce that: $\sin 20^{\circ} \times \sin 40^{\circ} \times \sin 60^{\circ} \times \sin 80^{\circ}=\frac{3}{16}$

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Answer

$\begin{array}{l}L H S=4 \sin A \times \sin \left(60^{\circ}-A\right) \times \sin \left(60^{\circ}+A\right) \\ =2 \sin A\left[2 \sin \left(60^{\circ}-A\right) \sin \left(60^{\circ}+A\right)\right] \\ =2 \sin A\left[\cos \left\{\left(60^{\circ}- A \right)-\left(60^{\circ}+ A \right)\right\}-\cos \left\{\left(60^{\circ}- A \right)+\left(60^{\circ}+ A \right)\right\}\right] \\ {[\because 2 \sin A \times \sin B =\cos ( A - B )-\cos ( A + B )]} \\ =2 \sin A\left[\cos (-2 A)-\cos 120^{\circ}\right] \\ =2 \sin A\left[\cos 2 A-\cos 120^{\circ}\right][\because \cos (-\theta)=\cos \theta] \\ =2 \sin A \times \cos 2 A-2 \sin A \times \cos 120^{\circ} \\ =[\sin ( A +2 A)+\sin ( A -2 A)]-2 \sin A\left(-\frac{1}{2}\right)\end{array}$
$\begin{array}{l}{\left[\because 2 \sin A \times \cos B=\sin (A+B)+\sin (A-B) \text { and } \cos 120^{\circ}=-\frac{1}{2}\right]} \\ =\sin 3 A+\sin (-A)+\sin A \\ =\sin 3 A-\sin A+\sin A=\sin 3 A=\text { RHS }[\because \sin (-\theta)=-\sin \theta] \\ \therefore \text { LHS }=\text { RHS }\end{array}$
Hence proved.
Now, $4 \sin A \sin \left(60^{\circ}-A\right) \times \sin \left(60^{\circ}+A\right)=\sin 3 A$
On putting $A =20^{\circ}$, we get
$4 \sin 20^{\circ} \times \sin \left(60^{\circ}-20^{\circ}\right) \sin \left(60^{\circ}+20^{\circ}\right)=\sin 3 \times\left(20^{\circ}\right)$
$\begin{array}{l}\Rightarrow 4 \sin 20^{\circ} \times \sin 40^{\circ} \times \sin 80^{\circ}=\sin 60^{\circ}=\frac{\sqrt{3}}{2} \\ \Rightarrow \sin 20^{\circ} \times \sin 40^{\circ} \times \sin 80^{\circ}=\frac{\sqrt{3}}{8} \\ \Rightarrow \sin 20^{\circ} \times \sin 40^{\circ} \times \frac{\sqrt{3}}{2} \times \sin 80^{\circ}=\frac{\sqrt{3}}{8} \times \frac{\sqrt{3}}{2}\end{array}$
[multiplying both sides by $\frac{\sqrt{3}}{2}$ ]
$\therefore \sin 20^{\circ} \times \sin 40^{\circ} \times \sin 60^{\circ} \times \sin 80^{\circ}=\frac{3}{16}\left[\because \frac{\sqrt{3}}{2}=\sin 60^{\circ}\right]$

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