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Trigo problem :

Find the value of $\sin25^° \sin35^° \sin85^°$.

My approach :

Using $2\sin A\sin B = \cos(A-B) -\cos(A+B)$

$$ \begin{align} & \phantom{={}}[\cos10^{°} -\cos60^°] \sin85^° \\ & = \frac{1}{2}[2\cos10^°\sin85^° -2\cos60^° \sin85^°] \\ & = \frac{1}{2}[2\cos10^°\sin85^° -2 \frac{1}{2} \sin85^°] \\ & = \frac{1}{2}[2\cos10^°\sin85^° -\sin85^°] \\ & = \frac{1}{2} [\sin 95^° + \sin 75^° -\sin85^°] \end{align} $$

After this I am unable to solve further.... please guide... thanks.

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Note that $\sin(95^\circ)=\sin(85^\circ)$.

We can get an explicit formula for $\sin(75^\circ)$ in various ways, probably most simply by writing it as $\sin(30^\circ+45^\circ)$ and using the Addition Law.

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  • $\begingroup$ oh! got it thanks.. I skipped that cancellation of sin $95^{\circ} $ and $sin85^{\circ}$ .. thanks once againthanks.. $\endgroup$ – sachin Sep 30 '13 at 3:41
  • $\begingroup$ You are welcome. A surprising result! $\endgroup$ – André Nicolas Sep 30 '13 at 3:43
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\begin{align} & \phantom{={}} \sin25^{°}\sin35^{°}\sin85^{°} \\ & =\sin85^{°}\sin35^{°}\sin25^{°} \\ & =\frac{1}{2}(2\sin85^{°}\sin35^{°})\sin25^{°} \\ & =\frac{1}{2}[ \cos50^{°} -\cos120^{°}] \sin25^{°} \\ & = \frac{1}{4}[ 2\cos50^{°}\sin25^{°} +2\cos60^{°} \sin25^{°}] \\ & = \frac{1}{4}[ 2\cos50^{°}\sin25^{°} +2 \frac{1}{2} \sin25^{°}] \\ & = \frac{1}{4}[ 2\cos50^{°}\sin25^{°} +\sin25^{°}] \\ & = \frac{1}{4} [ \sin75^{°} -\sin 25^{°} +\sin25^{°}] \\ & =\frac{1}{4} \sin75^{°} \end{align}

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