Question

question 10. (5 marks) verify the identity: (2sinx - 2cosx)^2 + 4sin2x = 4

Explanation:

Step1: Expand the square term

First, expand \((2\sin x - 2\cos x)^2\) using the formula \((a - b)^2=a^2 - 2ab + b^2\). Here, \(a = 2\sin x\) and \(b = 2\cos x\), so:

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Step2: Simplify the expanded expression

Now, rewrite the left - hand side of the identity: \((2\sin x - 2\cos x)^2+4\sin2x=4\sin^{2}x - 8\sin x\cos x + 4\cos^{2}x+4\sin2x\)
We know that \(\sin2x = 2\sin x\cos x\), so substitute \(4\sin2x = 4\times(2\sin x\cos x)=8\sin x\cos x\) into the above expression:

$$
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Step3: Use the Pythagorean identity

We know the Pythagorean identity \(\sin^{2}x+\cos^{2}x = 1\). Factor out the 4 from the first two terms:

$$
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Which is equal to the right - hand side of the given identity.

Answer:

The identity \((2\sin x - 2\cos x)^2 + 4\sin2x=4\) is verified.