Question

Find the value.
Sin100.sin120.sin140.sin160

Answer 1

Answer:

the value of sin100.sin160.sin140.sin200 is o.o7311

Answer 2

$\textbf{Given:}$

$sin\,100^{\circ}\;sin\,120^{\circ}\;sin\,140^{\circ}\;sin\,160^{\circ}$

$\bf\,sin\,100^{\circ}=sin(90^{\circ}+10^{\circ})=cos\,10^{\circ}$

$\bf\,sin\,120^{\circ}=sin(90^{\circ}+30^{\circ})=cos\,30^{\circ}=\frac{\sqrt3}{2}$

$\bf\,sin\,140^{\circ}=sin(90^{\circ}+50^{\circ})=cos\,50^{\circ}$

$\bf\,sin\,160^{\circ}=sin(90^{\circ}+70^{\circ})=cos\,70^{\circ}$

$\text{Now,}$

$sin\,100^{\circ}\;sin\,120^{\circ}\;sin\,140^{\circ}\;sin\,160^{\circ}$

$=cos\,10^{\circ}(\frac{\sqrt3}{2})cos\,50^{\circ}\;cos\,70^{\circ}$

$=\frac{\sqrt3}{2}[cos\,50^{\circ}cos\,10^{\circ}\;cos\,70^{\circ}]$

$=\frac{\sqrt3}{2}[cos\,(60^{\circ}-10^{\circ})cos\,10^{\circ}\;cos\,(60^{\circ}+10^{\circ})]$

$\text{We know that,}$

$\boxed{\bf\,cos(60-A)\;cosA\;cos(60+A)=\frac{1}{4}cos\,3A}}$

$=\frac{\sqrt3}{2}[\frac{1}{4}cos3(10^{\circ})]$

$=\frac{\sqrt3}{2}[\frac{1}{4}cos30^{\circ}]$

$=\frac{\sqrt3}{2}[\frac{1}{4}\frac{\sqrt3}{2}]$

$=\frac{3}{16}$

$\therefore\bf\,sin\,100^{\circ}\;sin\,120^{\circ}\;sin\,140^{\circ}\;sin\,160^{\circ}=\frac{3}{16}$

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