Question

The value of ${\sin }^{2}5°+{\sin }^{2}10°+{\sin }^{2}15°+......+{\sin }^{2}85°+{\sin }^{2}90°$ is equal to

• A. $7$
• B. $8$
• C. $9$
• D. $9\frac{1}{2}$

Answer 1

The correct option is D

$9\frac{1}{2}$

Explanation for the correct option:

The given trigonometric expression: ${\sin }^{2}5°+{\sin }^{2}10°+{\sin }^{2}15°+......+{\sin }^{2}85°+{\sin }^{2}90°$.

$$\begin{gathered} {\sin }^{2}5°+{\sin }^{2}10°+{\sin }^{2}15°+......+{\sin }^{2}85°+{\sin }^{2}90° \\ =\left({\sin }^{2}5°+{\sin }^{2}85°\right)+\left({\sin }^{2}10°+{\sin }^{2}80°\right)+\left({\sin }^{2}15°+{\sin }^{2}75°\right)+....+{\sin }^{2}45°+{\sin }^{2}90° \\ =\left[{\sin }^{2}5°+{\sin }^2\left(90°-5°\right)\right]+\left[{\sin }^{2}10°+{\sin }^2\left(90°-10°\right)\right]+\left({\sin }^{2}15°+{\sin }^2\left(90°-15°\right)\right)+....+{\left(\frac{1}{\sqrt{2}}\right)}^2+{\left(1\right)}^2 \\ =\left[{\sin }^2\left(5°\right)+{\cos }^2\left(5°\right)\right]+\left[{\sin }^2\left(10°\right)+{\cos }^2\left(10°\right)\right]+\left[{\sin }^2\left(15°\right)+{\cos }^2\left(15°\right)\right]+....+{\left(\frac{1}{\sqrt{2}}\right)}^2+{\left(1\right)}^2\left[\because \sin \left(90°-\theta \right)=\cos \left(\theta \right)\right] \\ =\left(1+1+1+1+1+1+1+1\right)+\frac{1}{2}+1\left[\because {\sin }^2\left(\theta \right)+{\cos }^2\left(\theta \right)=1\right] \\ =9+\frac{1}{2} \\ =9\frac{1}{2} \end{gathered}$$

Therefore, the value of ${\sin }^{2}5°+{\sin }^{2}10°+{\sin }^{2}15°+......+{\sin }^{2}85°+{\sin }^{2}90°$ is equal to $9\frac{1}{2}$.

Hence, the correct option is (D).