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Mathematics Question on Trigonometric Functions

If sin⁡ θ=2425\sin \, \theta = \frac{24}{25} and 0∘<θ<90∘0^{\circ} < \theta < 90^{\circ} then what is the value of sin⁡(θ2)\sin \left( \frac{\theta}{2} \right) ?

A

1225\frac{12}{25}

B

725\frac{7}{25}

C

35\frac{3}{5}

D

45\frac{4}{5}

Answer

35\frac{3}{5}

Explanation

Solution

We have, sin⁡ θ=2425,0∘<θ<90∘\sin \, \theta = \frac{24}{25} , 0^{\circ} < \theta < 90^{\circ} cos⁡2θ=1−sin⁡2θ=1−(2425)2\cos^2 \theta = 1 - \sin^2 \theta = 1 - \left( \frac{24}{25} \right)^2 Since lies in first quadrant ⇒ cos⁡θ=725\Rightarrow \, \cos \theta = \frac{7}{25} cos⁡θ=1−2sin⁡2θ2\cos \theta = 1 - 2 \sin^2 \frac{\theta}{2} 2sin⁡2θ2=1−cos⁡θ=1−725 2 \sin^2 \frac{\theta}{2} = 1 - \cos \theta = 1 - \frac{7}{25} 2sin⁡2θ2=18252 \sin^2 \frac{\theta}{2} = \frac{18}{25} sin⁡2θ2=925 ⇒ sin⁡θ2=±35\sin^2 \frac{\theta}{2} = \frac{9}{25} \, \Rightarrow \, \sin \frac{\theta}{2} = \pm \frac{3}{5} ⇒ sin⁡θ2=35\Rightarrow \, \sin \frac{\theta}{2} = \frac{3}{5} [Negative sign discarded since θ\theta is in first quadrant]