A trigonometric block sum

Given ∑r=1ncos⁡2rθ=sin⁡(2n+1)θ−sin⁡θ2sin⁡θ\sum_{r=1}^n\cos2r\theta=\frac{\sin(2n+1)\theta-\sin\theta}{2\sin\theta}, where 0<θ<π0<\theta<\pi, show that

sin⁡210θ+sin⁡211θ+⋯+sin⁡220θ=k−sin⁡41θ−sin⁡19θ4sin⁡θ,\sin^2 10\theta+\sin^2 11\theta+\cdots+\sin^2 20\theta=k-\frac{\sin41\theta-\sin19\theta}{4\sin\theta},

where kk is a constant to be determined. marks

[modified 2024/RI/Prelim/I/5]

Solution

Rewrite the terms

Double-angle identity

Given ∑r=1ncos⁡2rθ=sin⁡(2n+1)θ−sin⁡θ2sin⁡θ\sum_{r=1}^n\cos2r\theta=\frac{\sin(2n+1)\theta-\sin\theta}{2\sin\theta} for 0<θ<π0<\theta<\pi, rewrite sin⁡210θ+sin⁡211θ+⋯+sin⁡220θ\sin^210\theta+\sin^211\theta+\cdots+\sin^220\theta using cosines.

Hint

  1. Use sin⁡2x=1−cos⁡2x2\sin^2x=\frac{1-\cos2x}2 on each term.

  2. Count the terms in the block.

Answer · Rewrite the terms

∑r=1020sin⁡2rθ=112−12∑r=1020cos⁡2rθ\sum_{r=10}^{20}\sin^2r\theta=\dfrac{11}2-\dfrac12\sum_{r=10}^{20}\cos2r\theta.

Apply the given formula

Block as a difference

Show that the sum equals k−sin⁡41θ−sin⁡19θ4sin⁡θk-\frac{\sin41\theta-\sin19\theta}{4\sin\theta}, and find kk.

Explore this idea

Hint

  1. Write the cosine block as ∑r=120−∑r=19\sum_{r=1}^{20}-\sum_{r=1}^{9}.

  2. Substitute the given formula for each.

Answer · Apply the given formula

k=11/2k=11/2; sum 11/2−[sin⁡(41θ)−sin⁡(19θ)]/(4sin⁡θ)11/2-[\sin(41\theta)-\sin(19\theta)]/(4\sin\theta).

0<θ<π0<\theta<\pi.

Explore the idea

Squared sines for θ = 0.35 radians

Each bar is sin⁡2(rθ)\sin^2(r\theta). The closed form uses the same pattern as the question: half the number of terms, minus a ratio of sines.

Limits of the sumr = 10 to 20
Try an example
Σ sin²(0.35r), r = 10 to 20
5.049
Number of terms
11
(n)/2 − (sin(2b+1)θ − sin(2a−1)θ)/(4 sin θ)
5.049
matches the direct total
Key idea

The constant k=112k=\frac{11}2 is half the number of terms. The trigonometric ratio is a correction that wobbles around zero.

Ideas to try
  1. Keep the block at 1010 to 2020. The direct total and the formula agree even though they look completely different.
  2. Change the block and note how the constant part is always half the number of terms.
Takeaway

Turn powers into multiples

A squared sine becomes a constant plus a cosine of a double angle, which sums with the given formula.