Question

Sin 3x cos X = cos 3x sin x-1

Answer 1

Sin 3x cos X = cos 3x sin x-1

Answer 2

To prove the trigonometric identity sin(3x)cos(x) = cos(3x)sin(x) - 1, we need to work on the right-hand side (RHS) and try to manipulate it into the same form as the left-hand side (LHS). However, I suspect there might be a typo in the identity you provided, as it's quite uncommon for a trigonometric identity to involve a constant term like '-1'.

Let's try to prove a more standard trigonometric identity:

sin(3x)cos(x) = cos(3x)sin(x).

To prove this identity, we'll use the following trigonometric identities:

1. sin(2θ) = 2sin(θ)cos(θ)
2. cos(2θ) = cos^2(θ) - sin^2(θ) = 1 - 2sin^2(θ) = 2cos^2(θ) - 1

Proof:

Start with the right-hand side (RHS):

cos(3x)sin(x)

Now, using the double-angle formula for cosine:

cos(3x) = cos(2x + x) = cos(2x)cos(x) - sin(2x)sin(x)

We know from identity 1 that sin(2x) = 2sin(x)cos(x), so:

cos(3x) = cos(2x)cos(x) - 2sin(x)cos(x)sin(x)

Now, use identity 2 to express cos(2x) in terms of cos(x):

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

Substitute this back into the previous equation:

cos(3x) = (2cos^2(x) - 1)cos(x) - 2sin(x)cos(x)sin(x)

Simplify:

cos(3x) = 2cos^3(x) - cos(x) - 2sin(x)cos(x)sin(x)

Now, move to the left-hand side (LHS):

sin(3x)cos(x)

Using the triple-angle formula for sine:

sin(3x) = 3sin(x) - 4sin^3(x)

Now, substitute this back into the LHS equation:

sin(3x)cos(x) = (3sin(x) - 4sin^3(x))cos(x)

Expand:

sin(3x)cos(x) = 3sin(x)cos(x) - 4sin^3(x)cos(x)

Now, we can use the double-angle formula for sine:

sin^2(x) = 1 - cos^2(x)

Solve for sin(x):

sin(x) = sqrt(1 - cos^2(x))

Substitute this back into the LHS equation:

sin(3x)cos(x) = 3sin(x)cos(x) - 4(1 - cos^2(x))cos(x)

Simplify:

sin(3x)cos(x) = 3sin(x)cos(x) - 4cos(x) + 4cos^3(x)

Now, combine like terms:

sin(3x)cos(x) = 3sin(x)cos(x) + 4cos^3(x) - 4cos(x)

Now, notice that we have a common term '3sin(x)cos(x)' on both sides of the equation. We can move it to one side:

sin(3x)cos(x) - 3sin(x)cos(x) = 4cos^3(x) - 4cos(x)

Factor out cos(x) on the left-hand side:

cos(x)(sin(3x) - 3sin(x)) = 4cos^3(x) - 4cos(x)

Divide both sides by (sin(3x) - 3sin(x)):

cos(x) = (4cos^3(x) - 4cos(x))/(sin(3x) - 3sin(x))

Now, use the double-angle formula for cosine:

cos^2(x) = 1 - sin^2(x)

Solve for cos(x):

cos(x) = sqrt(1 - sin^2(x))

Substitute this back into the equation:

sqrt(1 - sin^2(x)) = (4(sqrt(1 - sin^2(x)))^3 - 4(sqrt(1 - sin^2(x))))/(sin(3x) - 3sin(x))

Now, square both sides to eliminate the square root:

1 - sin^2(x) = [(4(sqrt(1 - sin^2(x))))^3 - 4(sqrt(1 - sin^2(x))))^2]/(sin(3x) - 3sin(x))^2

At this point, the equation becomes a bit complicated, and it's challenging to show that both sides are equal. It's possible that there might be a mistake in the original identity you provided, or there could be a more specific restriction on the values of x for this identity to hold.

If you have the correct identity, please double-check it and let me know. If you need help with another trigonometric identity or have further questions, feel free to ask!