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Geometry Level 2

Evaluate 4 s i n 1 0 s i n 5 0 s i n 7 0 4sin10^{\circ}sin50^{\circ}sin70^{\circ}


The answer is 0.5.

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3 solutions

Ankush Sharma
May 13, 2014

First multiply cos50 to numerator n denominator [4sin10xsin70xsin50xcos50]/cos50 =[2sin10xsin(90-20)xsin100]/cos(90-40) =[2sin10xcos20xsin(90+10)]/sin40 =[2sin10xcos20xcos10]/sin40 =[sin20xcos20]/sin40 =[2sin20xcos20]/2sin40 =sin40/2sin40 =1/2

LaTeX \LaTeX version: 4 sin 1 0 sin 5 0 sin 7 0 = 4 sin 1 0 sin 7 0 sin 5 0 cos 5 0 cos 5 0 = 2 sin 1 0 sin ( 9 0 2 0 ) sin 10 0 cos ( 9 0 4 0 ) = 2 sin 1 0 cos 2 0 sin ( 9 0 + 1 0 ) sin 4 0 = 2 sin 1 0 cos 2 0 cos 1 0 sin 4 0 = sin 2 0 cos 2 0 sin 4 0 = 2 sin 2 0 cos 2 0 2 sin 4 0 = sin 4 0 2 sin 4 0 = 1 2 \begin{array}{rcl} 4\sin10^\circ\sin50^\circ\sin70^\circ&=&\frac{4\sin10^\circ\sin70^\circ\sin50^\circ\cos50^\circ}{\cos50^\circ}\\ &=&\frac{2\sin10^\circ\sin(90^\circ-20^\circ)\sin100^\circ}{\cos(90^\circ-40^\circ)}\\ &=&\frac{2\sin10^\circ\cos20^\circ\sin(90^\circ+10^\circ)}{\sin40^\circ}\\ &=&\frac{2\sin10^\circ\cos20^\circ\cos10^\circ}{\sin40^\circ}\\ &=&\frac{\sin20^\circ\cos20^\circ}{\sin40^\circ}\\ &=&\frac{2\sin20^\circ\cos20^\circ}{2\sin40^\circ}\\ &=&\frac{\sin40^\circ}{2\sin40^\circ}\\ &=&\frac12 \end{array}

Kenny Lau - 6 years, 11 months ago

Its kind of a to be known thing that 4.sinx.sin(60-x).sin(60+x) = sin3x

Tanya Gupta - 7 years ago

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yes I solved it using this fact

Sathvik Acharya - 4 years, 3 months ago
Bernardo Sulzbach
Jun 20, 2014

Alternatively, just use the sum identities to derive the product identities (or just use the product identities directly) and reduce the problem to

1 2 sin 7 0 + sin 11 0 = 1 2 \frac{1}{2}-\sin{70^{\circ{}}}+\sin{110^{\circ{}}}=\frac{1}{2}

Wendy Natalia
Jun 10, 2014

firstly : 4(sin10)(sin50)(sin70),then you'll get the answer 0.5..since it want 3 decimal point just add 2 zero behind the 5 and it will be 0.500

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