(Sin theta+ sin3theta + sin5theta)÷ (cos theta + cos3theta +cos 5theta)= tan3theta
To prove this identity, we can start by simplifying the left side of the equation.
We know that:
sin(A) + sin(B) = 2sin((A+B)/2)cos((A-B)/2)
Using this identity for the numerator, we have:
sin(theta) + sin(3theta) + sin(5theta) = 2sin(4theta)cos(theta) + 2sin(4theta)cos(2theta) = 2sin(4theta)(cos(theta) + cos(2theta))
Similarly, we know that:
cos(A) + cos(B) = 2cos((A+B)/2)cos((A-B)/2)
Using this identity for the denominator, we have:
cos(theta) + cos(3theta) + cos(5theta) = 2cos(4theta)cos(theta) + 2cos(4theta)cos(2theta) = 2cos(4theta)(cos(theta) + cos(2theta))
Therefore, the original expression simplifies to:
(2sin(4theta)(cos(theta) + cos(2theta))) / (2cos(4theta)(cos(theta) + cos(2theta))) = tan(4theta)
Therefore, (sin(theta) + sin(3theta) + sin(5theta)) / (cos(theta) + cos(3theta) + cos(5theta)) = tan(3theta)