Suppose you wish to express sin(3t) in terms of sint and cost. Apply the sum formula to sin(3t) = sin(t+2t) to obtain an expression that contains sin(2t)=sin(t+t) and cos(2t)=cos(t+t). Apply the sum formulas to those two expressions. Enter the resulting expression for sin(3t) here:____

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

= [2sint*cos^2t + sint(cos^2t - sin^2t)
= 3sint*cos^2t -sin^3t
You can get rid of the cosine function.
= 3sint(1 - sin^2t)- sin^3t
= 3sint -4 sin^3t

To express sin(3t) in terms of sin(t) and cos(t), we can use the sum formula for sine:

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

Applying this formula to sin(3t) = sin(t + 2t), we can express sin(3t) as:

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

Now, we need to express sin(2t) and cos(2t) in terms of sin(t) and cos(t) using the sum formulas for sine and cosine:

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

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

For sin(2t) = sin(t + t):

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

And for cos(2t) = cos(t + t):

cos(2t) = cos(t)cos(t) - sin(t)sin(t)

Simplifying sin(2t) and cos(2t):

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

cos(2t) = 2cos^2(t) - 1

Now, substituting these expressions into the previous expression for sin(3t), we get:

sin(3t) = sin(t)cos(2t) + cos(t)sin(2t)
= sin(t)(2sin(t)cos(t)) + cos(t)(2cos^2(t) - 1)
= 2sin^2(t)cos(t) + 2cos(t)cos^2(t) - cos(t)
= 2sin^2(t)cos(t) + 2cos^3(t) - cos(t)

Thus, the resulting expression for sin(3t) in terms of sin(t) and cos(t) is:

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

To express sin(3t) in terms of sin(t) and cos(t), we can use the sum formula for sine:

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

We are given sin(3t) = sin(t+2t), which we can rewrite as sin(t+2t) = sin(t+ t + t).

Applying the sum formula, we can express sin(t+2t) as:

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

Now, we need to represent sin(2t) and cos(2t) in terms of sin(t) and cos(t) using the sum formulas.

The sum formulas for sine and cosine are:

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

For sin(2t), we can let A = t and B = t, so sin(2t) = sin(t + t) = sin(t)cos(t) + cos(t)sin(t) = 2sin(t)cos(t)

And for cos(2t), we can also let A = t and B = t, so cos(2t) = cos(t + t) = cos(t)cos(t) - sin(t)sin(t) = cos^2(t) - sin^2(t)

Now, substituting sin(2t) and cos(2t) back into our expression for sin(t+2t), we get:

sin(t+2t) = sin(t)cos(2t) + cos(t)sin(2t)
= sin(t)(2sin(t)cos(t)) + cos(t)(cos^2(t) - sin^2(t))

Expanding and simplifying:

sin(t+2t) = 2sin^2(t)cos(t) + cos(t)cos^2(t) - cos(t)sin^2(t)
= 2sin^2(t)cos(t) + cos(t)(1 - sin^2(t)) - cos(t)sin^2(t)
= 2sin^2(t)cos(t) + cos(t) - cos(t)sin^2(t) - cos(t)sin^2(t)
= 2sin^2(t)cos(t) + cos(t) - 2cos(t)sin^2(t)

Therefore, the resulting expression for sin(3t) is:

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