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The value of $\sin ^2 5^{\circ} + \sin ^2 10^{\circ} + \sin ^2 15^{\circ} + \dots + \sin^2 90^{\circ}$ is

1. $8$
2. $9$
3. $9.5$
4. None of these

recategorized | 18 views

Answer = $C.$
We will make the pairs of $\sin^2\theta + \cos^2\theta$, where $\theta$ is ame for both $\sin$ and $\cos$

$$\sin^25^0 + \sin^285^0 = \sin^25^0 + \sin^2(90^0-5^0) = \sin^25^0 + \cos^25^0 =1$$

(Using $\color{blue}{\sin^2\theta + \cos^2\theta = 1}$)

$$\sin^2{85}^0 = \cos^25^0 (\color{blue}{\because \sin\theta = \cos({90^0-\theta})})$$

similarly,

$$\sin^210 \;and\; \sin^280$$

$$\vdots$$

$$\sin^245^0 = \bigg(\frac{1}{\sqrt2}\bigg)^2 = \frac{1}{2} = 0.5$$

$$\vdots$$

$$\sin^240^0 \;and\; \sin^250$$

$$\vdots$$

$$\sin^290^0 = (1)^2 = 1$$

So, this way $8$ pairs will be formed, leaving $\sin^245^0$ and $\sin^290^0$

But we know that:

$$\sin^245^0 = \bigg(\frac{1}{\sqrt2}\bigg)^2 = \frac{1}{2} = 0.5$$

$$\sin^290^0 = 1^2 = 1$$

So, answer $= 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 +0.5 = 8\times1 +0.5 =9.5$
by Boss (13.4k points)
edited by
+1
Edit:- answer is 9.5, not 8.5
0
Yes!

Thanks.