Let S = { θ ∈ [0,2π ):tan (π cos θ ) + tan (π sin θ ) = 0}. Then Σ_(θ ∈ S) sin ²( θ + π/4 ) is equal to __________.

Solution

Maths sir · 4 steps

Turn the tan equation into sin θ + cos θ = n, and the quantity you are summing becomes n²/2.

1. Remove the tan. Write the equation as tan(π cos θ) = tan(-π sin θ). Two angles with equal tangents differ by a multiple of π, so π cos θ = nπ - π sin θ.

   sin θ + cos θ = n, n ∈ {-1, 0, 1}

2. Allowed values of n. sin θ + cos θ = √2 sin(θ + π/4) lies in [-√2, √2], so the integer n can only be -1, 0 or 1. Squaring the same identity gives sin²(θ + π/4) = n²/2.

   n	θ in [0, 2π)	sin²(θ+π/4) each	Contribution
   0	3π/4, 7π/4	0	0
   1	0, π/2	1/2	1
   -1	π, 3π/2	1/2	1

3. Check the tans. Each tan term must be defined, so cos θ and sin θ must not equal ±1/2. At all six angles, sin θ and cos θ are 0, ± 1 or ±1/√2. Every tan term is defined there and the equation holds, so all six angles are in S.
4. So. The sum is 0 + 1 + 1 = 2.

The common slip: 1. Keeping only n = 0 and n = 1 and missing sin θ + cos θ = -1 (that is, θ = π, 3π/2) gives 1, not 2.

The idea to keep. When tan A + tan B = 0, write A + B = nπ, then use the range of the expression to list the few integers n can be. Always check that each tan is defined at the solutions.

Answer from the NTA final answer key. Solution by PrepWiser.