Question

Sin t sin 3t sin 5t = 1/4(-sin t + sin 3t +sin 7t - sin 9t).​

Answer 1

Recall that

cos(a + b) = cos(a) cos(b) - sin(a) sin(b)

cos(a - b) = cos(a) cos(b) + sin(a) sin(b)

and by subtracting the first equation from the second, we get

cos(a - b) - cos(a + b) = 2 sin(a) sin(b)

So, we can write

sin(t) sin(3t) = 1/2 (cos(2t) - cos(4t))

and expanding the left side in the original equation gives

sin(t) sin(3t) sin(5t) = 1/2 (cos(2t) - cos(4t)) sin(5t)

… = 1/2 cos(2t) sin(5t) - 1/2 cos(4t) sin(5t)

Similarly, recall that

sin(a + b) = sin(a) cos(b) + cos(a) sin(b)

sin(a - b) = sin(a) cos(b) - cos(a) sin(b)

===> sin(a + b) + sin(a - b) = 2 sin(a) cos(b)

Then

cos(2t) sin(5t) = 1/2 (sin(7t) + sin(3t))

and

cos(4t) sin(5t) = 1/2 (sin(9t) + sin(t))

So we have

sin(t) sin(3t) sin(5t) = 1/2 cos(2t) sin(5t) - 1/2 cos(4t) sin(5t)

… = 1/2 (1/2 (sin(7t) + sin(3t))) - 1/2 (1/2 (sin(9t) + sin(t)))

… = 1/4 sin(7t) + 1/4 sin(3t) - 1/4 sin(9t) - 1/4 sin(t)

… = 1/4 (-sin(t) + sin(3t) + sin(7t) - sin(9t))

as required.