Question

(cos4a-cos2a)/sin3a help me

Answer 1

(cos 4α - cos 2α)/sin 3α = -2 * sin 3α * sin α / sin 3α = -2 sin α

Answer 2

We can start by using the identity:

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

Substituting x = 2a, we get:

cos(4a) = 2cos^2(2a) - 1

Now, using the same identity with x = a, we get:

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

Substituting these expressions back into the original expression, we get:

(cos4a - cos2a)/sin3a = [(2cos^2(2a) - 1) - (2cos^2(a) - 1)]/sin3a

Simplifying the numerator, we get:

(cos4a - cos2a)/sin3a = (2cos^2(2a) - 2cos^2(a))/sin3a

Now, using the identity:

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

Substituting x = a, we get:

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

So we have:

(cos4a - cos2a)/sin3a = (2cos^2(2a) - 2cos^2(a))/(3sin(a) - 4sin^3(a))

Next, using the identity:

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

Substituting x = a, we get:

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

Substituting this expression back into the numerator, we get:

(cos4a - cos2a)/sin3a = [2(1 - 2sin^2(2a)) - 2(1 - 2sin^2(a))]/(3sin(a) - 4sin^3(a))

Simplifying the numerator, we get:

(cos4a - cos2a)/sin3a = (4sin^2(a) - 8sin^2(2a))/(3sin(a) - 4sin^3(a))

Finally, using the identity:

sin(2x) = 2sin(x)cos(x)

Substituting x = a, we get:

sin(2a) = 2sin(a)cos(a)

Substituting this expression back into the numerator, we get:

(cos4a - cos2a)/sin3a = (4sin^2(a) - 8sin(2a)cos(2a))/(3sin(a) - 4sin^3(a))

Simplifying the numerator further, we get:

(cos4a - cos2a)/sin3a = (4sin^2(a) - 8sin(2a)(1 - 2sin^2(a)))/(3sin(a) - 4sin^3(a))

(cos4a - cos2a)/sin3a = (4sin^2(a) - 8sin(2a) + 16sin^3(a))/(3sin(a) - 4sin^3(a))

Therefore, the final expression is:

(cos4a - cos2a)/sin3a = (4sin^2(a) - 8sin(2a) + 16sin^3(a))/(3sin(a) - 4sin^3(a))