Question

The value of $\sin^{2}1^{\circ}+\sin^{2}2^{\circ}+\sin^{2}3^{\circ}+\dots+\sin^{2}88^{\circ}+\sin^{2}89^{\circ}$ is equal to

• A. $\frac{45}{2}$
• B. $\frac{49}{2}$
• C. $\frac{89}{2}$
• D. $45$
• E. $89$

Hint:

Trigonometry Tip: In a series from $1^\circ$ to $89^\circ$, there are always $44$ pairs that sum to $1$, plus the middle $45^\circ$ term which is $1/2$. The sum is always $44.5$ or $89/2$.

Answer 1

The Correct Option is C

Concept:
This problem relies on the complementary angle identity $\sin(90^\circ - \theta) = \cos\theta$ and the Pythagorean identity $\sin^2\theta + \cos^2\theta = 1$. By pairing terms from the beginning and the end of the series, we can sum them easily.

Step 1: Group complementary angles.
The series contains 89 terms. We can pair the first term with the last, the second with the second-to-last, and so on: $$(\sin^2 1^\circ + \sin^2 89^\circ) + (\sin^2 2^\circ + \sin^2 88^\circ) + \dots + (\sin^2 44^\circ + \sin^2 46^\circ) + \sin^2 45^\circ$$

Step 2: Apply the complementary identity.
Since $\sin(90^\circ - \theta) = \cos\theta$, we can rewrite the second term in each pair. For example, $\sin^2 89^\circ = \sin^2(90^\circ - 1^\circ) = \cos^2 1^\circ$. Substituting this into our pairs gives: $$(\sin^2 1^\circ + \cos^2 1^\circ) + (\sin^2 2^\circ + \cos^2 2^\circ) + \dots$$

Step 3: Apply the Pythagorean identity.
Using $\sin^2\theta + \cos^2\theta = 1$, each of these paired groups evaluates exactly to $1$. Since we paired numbers from 1 to 44 with numbers from 89 down to 46, we have exactly 44 pairs. $$1 + 1 + 1 \dots \text{ (44 times)} = 44$$

Step 4: Evaluate the unpaired middle term.
The only term left without a pair is the middle term, $\sin^2 45^\circ$. We know that $\sin 45^\circ = \frac{1}{\sqrt{2}}$. Squaring this gives: $$\sin^2 45^\circ = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2}$$

Step 5: Calculate the final total sum.
Add the sum of the pairs to the value of the middle term: $$\text{Total Sum} = 44 + \frac{1}{2}$$ $$\text{Total Sum} = \frac{88}{2} + \frac{1}{2} = \frac{89}{2}$$ Hence the correct answer is (C) $\frac{89{2}$}.