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Question

If $$\sin A+\sin^2 A+\sin^3 A=1$$, then find the value of $$\cos^6 A-4\cos ^4A+8\cos ^2A$$.


Solution

$$\sin A + \sin^2A + \sin^3A = 1$$
$$\sin A + \sin^3A = 1 - \sin^2A = \cos^2A$$
$$\sin A(1+\sin^2A) = \cos^2A$$
$$\sin A(2 -\cos^2A) = \cos^2A$$ (since, $$\sin^2A=1-\cos^2A$$)

Squaring both sides,
$$\sin^2A(4-4\cos^2A +\cos^4A) = \cos^4A$$
$$(1-\cos^2A)(4-4\cos^2A +\cos^4A) = \cos^4A$$
$$4-4\cos^2A +\cos^4A-4\cos^2A+4\cos^4A-\cos^6A = \cos^4A$$
$$4 -\cos^6A +4\cos^4A -8\cos^2A = 0$$
$$\cos^6A - 4 \cos^4A + 8\cos^2A = 4$$

Maths

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