# How do I solve this trigonometric function???

How do I prove that:

4sin^2 theta (don't know how to type it) +2cos^s theta = 4 - 2cos^2 theta

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• Anonymous

Come on, this is an EASY one.

Here's how you do identities in general. Look to see what the differences are in the two sides. Then, look for an identity that can fix those differences.

For example, the main difference between the left and right side is, there is sin^2 on the left, but not on the right. So somehow, the sin^2 has to disappear. Additionally, there are only cos^2's on the right, so whatever you do you your sin^2, it has to turn into some form of cos^2.

Are you GETTING this?

Start with the left side. Replace sin^2 (theta) with 1 - cos^2 (theta). Do the algebra that ensues to get the right side.

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• The equation can be written as

4sin^2 theta + 2cos^2 theta +2cos^2 theta = 4

4sin^2 theta + 4cos^2 theta = 4

4( sin^2 theta + cos^2 theta) = 4

since sin^2 theta + cos^2 theta = 1

4(1) = 4

4 = 4

Its proved

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• I never liked proofs much. Basically you start applying your trig rules and equivalencies to one side of the equation until it looks like the other. You might wish to do some simplification, if possible, before you start.

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• You know that (sin θ)^2 + (cos θ)^2 = 1 for all θ.

4(sin θ)^2 + 2(cos θ)^2

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

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

= 4 - 2(cos θ)^2.

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