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Question

Show that :
$$\cos^{6}A-4\cos^{4}A+8\cos^{2}A=4$$


Solution

If $$\sin A + {\sin ^2}A + {\sin ^3}A = 1$$ then$$,$$
$$\sin A + {\sin ^3}A = 1 - {\sin ^2}A$$ $$ = \cos^2 A$$
$$\sin A\left( {2 - {{\cos }^2}A} \right)$$
$${\sin ^2}A\left( {4 + {{\cos }^4}A - 4{{\cos }^2}A} \right) = {\cos ^4}A$$
$$\left( {1 - {{\cos }^2}A} \right)\left( {4 - 4{{\cos }^4}A - 4{{\cos }^2}A} \right) = {\cos ^4}A$$
$$4 - 4{\cos ^2}A + {\cos ^2}A - 4{\cos ^2}A - {\cos ^6}A + 4{\cos ^4}A = {\cos ^4}A$$
$$4 - {\cos ^6}A + 4{\cos ^4}A - 8{\cos ^2}A = 0$$
$${\cos ^6}A - 4{\cos ^4}A + 8{\cos ^2}A = 4$$   Proved$$.$$

Mathematics

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