In the product sin 1 0 sin 3 0 - sin89 0 - 1 2 n then find the value of n-

Given ( sin 1 ^ 0 ) ( sin 3 ^ 0 ) ......( sin89 ^ 0 ) = 1 2 ^ n ⇒ ( sin1 ^ 0 ) ( sin2 ^ 0 ) ( sin3 ^ 0 ) ( sin4 ^ 0 ) ( sin6 ^ 0 ) ( sin7 ^ 0 ) .... ...

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Byju's Answer

Standard XII

Mathematics

Principal Solution of Trigonometric Equation

In the produc...

Question

In the product $(sin 1^{0}) (sin 3^{0}) . . . . . . ({sin 89}^{0}) = \frac{1}{2^{n}}$ then find the value of n.

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Solution

Given-
$(sin 1^{0}) (sin 3^{0}) . . . . . . ({sin 89}^{0}) = \frac{1}{2^{n}}$
$\Rightarrow \frac{({sin 1}^{0}) ({sin 2}^{0}) ({sin 3}^{0}) ({sin 4}^{0}) ({sin 6}^{0}) ({sin 7}^{0}) . . . . . . ({sin 88}^{0}) ({sin 89}^{0})}{({sin 2}^{0}) ({sin 4}^{0}) ({sin 6}^{0}) ({sin 8}^{0}) . . . . . . ({sin 88}^{0})} = \frac{1}{2^{n}}$
$\Rightarrow \frac{({sin 1}^{0} . {sin 89}^{0}) ({sin 2}^{0} . {sin 88}^{0}) ({sin 3}^{0} . {sin 87}^{0}) . . . . . . ({sin 44}^{0} . {sin 46}^{0}) {sin 45}^{0}}{{sin 2}^{0} {sin 4}^{0} {sin 6}^{0} . . . . . . {sin 88}^{0}} = \frac{1}{2^{n}}$
$\Rightarrow \frac{(sin 1. cos 1) (sin 2. cos 2) (sin 3. cos 3) (sin 4. cos 4) . . . . . . (sin 44. cos 44) {sin 45}^{0}}{{sin 2}^{0} {sin 4}^{0} {sin 6}^{0} . . . . . . {sin 88}^{0}} = \frac{1}{2^{n}}$
${∵ sin θ = cos (90 - θ)}$
$\Rightarrow \frac{(2 sin 1. cos 1) (2 sin 2. cos 2) (2 cos 3. sin 3) . . . . . . (2 sin 44. cos 44) {sin 45}^{0}}{2^{44} . sin 2. sin 4. sin 6...... {sin 88}^{0}} = \frac{1}{2^{n}}$
$\Rightarrow \frac{(sin 2 sin 4 sin 6 sin 8...... sin 88) {sin 45}^{0}}{2^{44} (sin 2^{0} {sin 4}^{0} {sin 6}^{0} {sin 8}^{0} . . . . . . {sin 88}^{0})} = \frac{1}{2^{n}}$
$\Rightarrow \frac{{sin 45}^{0}}{2^{44}} = \frac{1}{\sqrt{2} . 2^{44}} = \frac{1}{2^{89 / 2}} = \frac{1}{2^{n}}$
$\Rightarrow n = 89 / 2$

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In the product $(sin 1^{0}) (sin 3^{0}) . . . . . . ({sin 89}^{0}) = \frac{1}{2^{n}}$ then find the value of n.