Question

(a + b)^2 = 4 ab, sin^2 theta is possible only when

Answer 1

By Pythagoras

4absin2(θ)=4ab(1−cos2(θ))4absin2⁡(θ)=4ab(1−cos2⁡(θ))

Now subtract 4ab4ab on both sides of your equation to see

−4abcos2(θ)=(a+b)2−4ab=(a−b)2≥0−4abcos2⁡(θ)=(a+b)2−4ab=(a−b)2≥0

Therefore we must have either a=ba=b or a,ba,b must have different signs. But in the latter case the original equation implies a=−ba=−b. If (a,b)≠(0,0)(a,b)≠(0,0), then this also determines the angle θθ.

So the only solutions are

⎧⎩⎨⎪⎪a=b,θ=(2k+1)π2,k∈Za=−b,θ=kπ,k∈Za=b=0,θarbitrary

Answer 2

Step-by-step explanation:

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