Question

(sinx+cosx)^2-1-sin4x/(cos4x-cos2x)

Answer 1

(sin²a+2sinxcosx+cos²x-sin²x-cos²x-2sin2xcos2x)/(cos4x-cos2x)=0

Answer 2

To simplify the expression (sinx+cosx)^2-1-sin4x/(cos4x-cos2x), let's break it down step by step:

Step 1: Expand (sinx+cosx)^2 (sinx+cosx)^2 = sin^2(x) + 2sinxcosx + cos^2(x) = 1 - cos^2(x) + 2sinxcosx + cos^2(x) = 1 + sin(2x)

Step 2: Simplify sin4x/(cos4x-cos2x) To simplify this expression, we need to use trigonometric identities. Let's express sin4x and cos4x in terms of sin2x and cos2x:

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

Now, substitute these values into the expression:

sin4x/(cos4x-cos2x) = 2sin(2x)cos(2x) / (cos^2(2x) - sin^2(2x) - cos(2x)) = 2sin(2x)cos(2x) / (cos^2(2x) - sin^2(2x) - cos^2(2x) + 1) = 2sin(2x)cos(2x) / (1 - 2cos^2(2x))

Step 3: Substitute the results back into the original expression:

(1 + sin(2x)) - 1 - [2sin(2x)cos(2x) / (1 - 2cos^2(2x))]

Now, we can simplify further by combining like terms:

1 - 1 = 0

The simplified expression is: sin(2x) - [2sin(2x)cos(2x) / (1 - 2cos^2(2x))]