Question

Solve the trigonometric equation sin(t) = 2sin(t/2)cos(t/2) repeatedly.

a) t = 0
b) t = π
c) t = π/2
d) t = π/4

Answer 1

The trigonometric equation sin(t) = 2sin(t/2)cos(t/2) is true for t = 0 and t = π. It fails for t = π/2 and t = π/4. The correct options in the final answer are a) t = 0 and b) t = π.

Step-by-step explanation:

The student is asking to solve the trigonometric equation sin(t) = 2sin(t/2)cos(t/2).

This equation can be recognized as an application of the double-angle formula for sine, where sin(2x) = 2sin(x)cos(x).

We'll evaluate the equation at different values of 't' to determine if the equation holds true.

1. t = 0: sin(0) = 0 and 2sin(0/2)cos(0/2) = 0, so the equation holds true.
2. t = π: sin(π) = 0 and 2sin(π/2)cos(π/2) = 0, so the equation holds true.
3. t = π/2: sin(π/2) = 1 and 2sin(π/4)cos(π/4) is not equal to 1, so the equation does not hold true.
4. t = π/4: sin(π/4) = √2/2 and 2sin(π/8)cos(π/8) is not equal to √2/2, so the equation does not hold true.

In conclusion, the equation holds true for t = 0 and t = π, but it does not hold true for t = π/2 and t = π/4. The correct options in the final answer are a) t = 0 and b) t = π.