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Exercise 3.1 · Q5

Q.If cos⁡4αcos⁡2β+sin⁡4αsin⁡2β=1\dfrac{\cos^4\alpha}{\cos^2\beta}+\dfrac{\sin^4\alpha}{\sin^2\beta}=1, prove that

(i) sin⁡4α+sin⁡4β=2sin⁡2αsin⁡2β\sin^4\alpha+\sin^4\beta=2\sin^2\alpha\sin^2\beta
(ii) cos⁡4βcos⁡2α+sin⁡4βsin⁡2α=1\dfrac{\cos^4\beta}{\cos^2\alpha}+\dfrac{\sin^4\beta}{\sin^2\alpha}=1.
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Substituting x=sin⁡2α, y=sin⁡2βx=\sin^2\alpha,\ y=\sin^2\beta turns the given condition into a perfect square (x−y)2=0(x-y)^2=0, forcing x=yx=y; both required identities then reduce to trivial statements once sin⁡2α=sin⁡2β\sin^2\alpha=\sin^2\beta is known.

Step 1. Set up substitutions. Let x=sin⁡2αx=\sin^2\alpha (so cos⁡2α=1−x\cos^2\alpha=1-x) and y=sin⁡2βy=\sin^2\beta (so cos⁡2β=1−y\cos^2\beta=1-y). The given condition cos⁡4αcos⁡2β+sin⁡4αsin⁡2β=1\dfrac{\cos^4\alpha}{\cos^2\beta}+\dfrac{\sin^4\alpha}{\sin^2\beta}=1 becomes (1−x)21−y+x2y=1\dfrac{(1-x)^2}{1-y}+\dfrac{x^2}{y}=1.

Step 2. Clear denominators. Multiplying through by y(1−y)y(1-y):

(1−x)2y+x2(1−y)=y(1−y).(1-x)^2y+x^2(1-y)=y(1-y).

Expand the left side: y(1−2x+x2)+x2−x2y=y−2xy+x2y+x2−x2y=y−2xy+x2y(1-2x+x^2)+x^2-x^2y=y-2xy+x^2y+x^2-x^2y=y-2xy+x^2.

So the equation becomes y−2xy+x2=y−y2y-2xy+x^2=y-y^2.

Step 3. Simplify. Cancel yy from both sides: −2xy+x2=−y2⇒x2−2xy+y2=0⇒(x−y)2=0⇒x=y-2xy+x^2=-y^2 \Rightarrow x^2-2xy+y^2=0 \Rightarrow (x-y)^2=0 \Rightarrow x=y.

So the given condition forces sin⁡2α=sin⁡2β\sin^2\alpha=\sin^2\beta — and therefore also cos⁡2α=1−x=1−y=cos⁡2β\cos^2\alpha=1-x=1-y=\cos^2\beta.

Step 4. Prove (i). With sin⁡2α=sin⁡2β=x\sin^2\alpha=\sin^2\beta=x: sin⁡4α+sin⁡4β=x2+x2=2x2\sin^4\alpha+\sin^4\beta=x^2+x^2=2x^2, and 2sin⁡2αsin⁡2β=2(x)(x)=2x22\sin^2\alpha\sin^2\beta=2(x)(x)=2x^2. These are equal, proving (i). …

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