If a=sin ⁻¹(sin (5)) and b=cos ⁻¹(cos (5)), then a²+b² is equal to

Solution

Maths sir · 4 steps

5 radians is not in either principal range, so shift it by 2π until it is.

1. Locate 5 radians. Use π ≈ 3.14. Then 3π/2 ≈ 4.71 and 2π ≈ 6.28, so 3π/2 < 5 < 2π. That puts 5 in the fourth quadrant, and it lies outside both [-π/2, π/2] and [0, π].
2. Find a. sin has period 2π, so sin 5 = sin(5 - 2π). Since 5 - 2π ≈ -1.28 lies in [-π/2, π/2], we get a = 5 - 2π.
3. Find b. cos is even with period 2π, so cos 5 = cos(2π - 5). Since 2π - 5 ≈ 1.28 lies in [0, π], we get b = 2π - 5.

   a² + b² = 2(2π - 5)² = 2(4π² - 20π + 25)

4. So. a² + b² = 8π² - 40π + 50, which is option (c).

The idea to keep. sin⁻¹(sin x) = x and cos⁻¹(cos x) = x only inside the principal ranges. Otherwise, add or subtract multiples of 2π, and use symmetry, until the angle lands in the range.

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