Use compound angle formulae to express Second harmonic i2= sin2t and the the third harmonic i3=sin3t in terms of the first harmonic sint

sin2t = 2 sint cost

sin3t = sint cos2t + cost sin2t
(cos2t = 1 - 2sin^2 t)

Thanks :)

Whats the workings out to get to these answers ? Jw ta

To express the second harmonic i2 = sin(2t) in terms of the first harmonic sin(t), we can use the compound angle formula for sine:

sin(A + B) = sin(A)cos(B) + cos(A)sin(B).

Let's rewrite sin(2t) as sin(t + t) and use the compound angle formula:

sin(2t) = sin(t + t) = sin(t)cos(t) + cos(t)sin(t).

Using the identity cos(t) = cos(-t) and sin(t) = -sin(-t), we can rewrite the equation as:

sin(2t) = 2sin(t)cos(t).

Thus, the second harmonic i2 = sin(2t) can be expressed in terms of the first harmonic sin(t) as:

i2 = 2sin(t)cos(t).

Now, let's move on to expressing the third harmonic i3 = sin(3t) in terms of the first harmonic sin(t). Again, we'll utilize the compound angle formula for sine:

sin(A + B + C) = sin(A)cos(B)cos(C) + cos(A)sin(B)cos(C) + cos(A)cos(B)sin(C) + sin(A)sin(B)sin(C).

Let's rewrite sin(3t) as sin(t + t + t) and use the compound angle formula:

sin(3t) = sin(t + t + t)
= sin(t)cos(2t)cos(t) + cos(t)sin(2t)cos(t) + cos(t)cos(2t)sin(t) + sin(t)sin(2t)sin(t).

By substituting sin(2t) = 2sin(t)cos(t) from the previous step, the equation simplifies to:

sin(3t) = sin(t)cos(2t)cos(t) + cos(t)(2sin(t)cos(t))cos(t) + cos(t)cos(2t)sin(t) + sin(t)(2sin(t)cos(t))sin(t).

Further simplifying the equation, we have:

sin(3t) = sin(t)cos^2(t) + 2cos^2(t)sin(t) + cos(t)sin^2(t) + 2sin^2(t)sin(t).

Rearranging like terms, we get:

sin(3t) = 3sin(t)cos^2(t) + 3sin^2(t)cos(t).

Thus, the third harmonic i3 = sin(3t) can be expressed in terms of the first harmonic sin(t) as:

i3 = 3sin(t)cos^2(t) + 3sin^2(t)cos(t).