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.
since <span>the slope-intercept form of a linear equation is </span>
, where 
is the slope and 
is the Y-intercept value.