Answer :
The value of sin²(20°) + sin²(21°) + sin²(22°) + ... + sin²(89°) + sin²(90°) is B .1
We are asked to find the value of the sum of the squares of sine functions from sin²(20°) to sin²(90°).
- The expression is:
sin²(20°) + sin²(21°) + sin²(22°) + ... + sin²(89°) + sin²(90°)
Step 1: Identify the pattern
- We know that sin²(θ) and cos²(θ) are related by the identity:
sin²(θ) = cos²(90° - θ)
- This allows us to pair terms in the summation:
sin²(20°) + sin²(70°)
sin²(21°) + sin²(69°)
sin²(22°) + sin²(68°)
- For each pair, we have:
sin²(θ) + sin²(90° - θ) = sin²(θ) + cos²(θ) = 1
Step 2: Group and pair terms
- Let’s pair terms as follows:
(sin²(20°) + sin²(70°)) + (sin²(21°) + sin²(69°)) + ... + sin²(90°)
Each of these pairs adds up to 1. Since we are summing from 20° to 90°, there are 35 such pairs (from 20° to 69°), plus the single term sin²(90°), which is equal to 1.
- Step 3: Sum the terms
The 35 pairs contribute 35 × 1 = 35 to the total.
The single sin²(90°) adds 1.
- So, the total sum is:
35 + 1 = 36
Step 4: Verify the number of terms:
- There are terms from sin²(20°) to sin²(90°), so the number of terms is 90° 20° + 1 = 71.
Thus, the total number of terms is consistent with our grouping of 35 pairs and the single term.
However, we need to adjust this by removing [tex]\( \sin^2 90^\circ \)[/tex] which equals 1.