Question

Tan(x-1) ( sin2x-2cos2x) = 2(1-2sinxcosx)

Answer 1

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We can simplify the expression using trigonometric identities.

First, we can use the double angle formula for sine to write sin(2x) = 2sin(x)cos(x).

Next, we can use the double angle formula for cosine to write cos(2x) = cos^2(x) - sin^2(x) = 1 - 2sin^2(x). Rearranging this equation gives 2sin^2(x) - 2cos^2(x) = -cos(2x) + 1.

Substituting these identities into the original expression gives:

tan(x-1) ( sin2x-2cos2x) = tan(x-1) [2sin(x)cos(x) - 2(1 - 2sin^2(x))]

= 2tan(x-1)sin(x)cos(x) - 2tan(x-1) + 4tan(x-1)sin^2(x)

We can use the identity tan(x) = sin(x)/cos(x) to simplify this expression further:

2tan(x-1)sin(x)cos(x) - 2tan(x-1) + 4tan(x-1)sin^2(x)

= 2sin(x)cos(x)/(cos(x-1)) - 2sin(x)/(cos(x-1)) + 4sin^2(x)/(cos(x-1))

Multiplying both sides of the equation by cos(x-1) gives:

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

Expanding the left-hand side of the equation gives:

2sin(x)cos(x) - 2sin(x) + 4sin^2(x)cos(x) - 4sin^2(x) = 2 - 4sin(x)cos(x)

Simplifying this equation gives:

4sin^2(x) - 2sin(x) - 2 = 0

This is a quadratic equation in sin(x), which can be solved using the quadratic formula.