Prove that Sin 10 degree sin 30 degree sin 50 degree sin 70 degree = 1/16?

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Prove that Sin 10 degree sin 30 degree sin 50 degree sin 70 degree = 1/16?

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Answer 1 · 2018-06-02T13:33:55

$L H S = sin 10 sin 30 sin 50 sin 70$ $LHS=sin10sin30sin50sin70$

$= cos (90 - 10) sin 30 cos (90 - 50) cos (90 - 70)$ $=cos(90-10)sin30cos(90-50)cos(90-70)$

$= cos (80) \cdot \frac{1}{2} \cdot cos (40) cos (20)$ $=cos(80)*1/2*cos(40)cos(20)$
$= \frac{1}{4 sin 20} cos (80) cos (40) \cdot 2 sin 20 cos (20)$ $=1/(4 sin20)cos(80)cos(40)*2sin20cos(20)$

$= \frac{1}{8 sin 20} cos (80) \cdot 2 cos (40) sin (40)$ $=1/(8 sin20)cos(80)*2cos(40)sin(40)$

$= \frac{1}{16 sin 20} \cdot 2 cos (80) sin (80)$ $=1/(16 sin20)*2cos(80)sin(80)$

$= \frac{1}{16 sin 20} \cdot sin (160)$ $=1/(16 sin20)*sin(160)$

$= \frac{1}{16 sin 20} \cdot sin (180 - 20)$ $=1/(16 sin20)*sin(180-20)$

$= \frac{1}{16 sin 20} \cdot sin (20) = \frac{1}{16} = R H S$ $=1/(16 sin20)*sin(20)=1/16=RHS$

Answer 2 · 2018-06-02T14:50:13

As, $sin x sin (60 + x) sin (60 - x) = \frac{1}{4} sin 3 x$ $sinxsin(60+x)sin(60-x)=1/4sin3x$

$\to sin 10 sin 30 sin 50 sin 70$ $rarrsin10sin30sin50sin70$

$= \frac{1}{2} [sin 10 sin (60 + 10) sin (60 - 10)]$ $=1/2[sin10sin(60+10)sin(60-10)]$

$= \frac{1}{2} [\frac{1}{4} sin (3 \times 10)] = \frac{1}{8} sin 30 = \frac{1}{8} \cdot \frac{1}{2} = \frac{1}{16}$ $=1/2[1/4sin(3xx10)]=1/8sin30=1/8*1/2=1/16$

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Prove that Sin 10 degree sin 30 degree sin 50 degree sin 70 degree = 1/16?

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