Trig identities help?

Question

I can't seem to work through these proofs. Please help.

1) sin2x = 2cotxsin^2x

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

3)cos^4-sin^4x= cos2x

4)cos^2 2x+4sin^2xcos^2x = 1

Answer 1

1) recall that cot(x) = cos(x)/sin(x) and sin(2x) = 2sin(x)cos(x)

sin(2x) = 2cot(x)sin^2(x)
2sin(x)cos(x) = 2cos(x)/sin(x) sin^2(x)
2sin(x)cos(x) = 2cos(x)sin(x)

Done.

2) I think you wrote the question wrong. Just remember that

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

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

Edit: You wrote this question twice and on the other one you said you meant sin^2(2x)/sin^2(x)

As before, sin(2x) = 2sin(x)cos(x) so

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

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

And this matches what I wrote above for the LHS

Done.

3) Use the fact that a^2 - b^2 = (a-b)(a+b)

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

cos^2(x) + sin^2(x) = 1 which leaves

[cos^2(x) - sin^2(x)] = cos(2x)

And this is a common identity.

Done.

4) Use the identity I just referred to.

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

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

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

And since cos^2(x) + sin^2(x) = 1

1 = 1

Done.

Answer 2

1) 2*(cot x)*(sin x)^2 = 2*[cos x/sin x]*(sin x)^2 =
2*(cos x)*(sin x) = sin 2x .

2) You're kidding. That's enough.

Answer 3

1) sin(2x) = 2sin(x)cos(x) = 2sin^2(x)cos(x)/sin(x)
= 2sin^2(x)cot(x)

2) false. LHS = 1

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

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

Answer 4

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