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Question
Prove that: \[\cos 4x = 1 - 8 \cos^2 x + 8 \cos^4 x\]
Solution 1
\[LHS = \cos4x\]
\[ = \cos\left( 2 \times 2x \right)\]
\[ = 2 \cos^2 \times 2x - 1 \left[ \because \cos2\theta = 2 \cos^2 \theta - 1 \right]\]
\[ = 2 \left( 2 \cos^2 x - 1 \right)^2 - 1 \left[ \because \cos^2 2\theta = \left( 2 \cos^2 \theta - 1 \right)^2 \right]\]
\[= 2\left( 4 \cos^4 x - 4 \cos^2 x + 1 \right) - 1\]
\[ = 8 \cos^4 x - 8 \cos^2 x + 1\]
\[ = 1 - 8 \cos^2 x + 8 \cos^4 x = RHS\]
\[Hence proved .\]
Solution 2
\[LHS = \cos4x\]
\[ = \cos\left( 2 \times 2x \right)\]
\[ = 2 \cos^2 \times 2x - 1 \left[ \because \cos2\theta = 2 \cos^2 \theta - 1 \right]\]
\[ = 2 \left( 2 \cos^2 x - 1 \right)^2 - 1 \left[ \because \cos^2 2\theta = \left( 2 \cos^2 \theta - 1 \right)^2 \right]\]
\[= 2\left( 4 \cos^4 x - 4 \cos^2 x + 1 \right) - 1\]
\[ = 8 \cos^4 x - 8 \cos^2 x + 1\]
\[ = 1 - 8 \cos^2 x + 8 \cos^4 x = RHS\]
\[\text{ Hence proved } .\]
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