Question

Show that \cos(3\alpha ) = 4\cos^3\alpha − 3 \cos\alpha.

Collected in the board: Trigonometry

Steven Zheng posted 2 months ago


Answer

Use the sum of two angles for the cosine functions, the double angle formulas for both the sine and cosine functions and Pythagorean Identities.


\cos(3\alpha )

= \cos(\alpha+2\alpha )

= \cos \alpha\cos 2\alpha -\sin \alpha \sin 2\alpha

= \cos \alpha(2\cos^2\alpha -1)-\sin \alpha\cdotp 2\sin\alpha \cos\alpha

=2\cos^3\alpha -\cos\alpha -2\sin^2\alpha \cos\alpha

=2\cos^3\alpha -\cos\alpha -2(1-\cos^2\alpha ) \cos\alpha

= 4\cos^3\alpha − 3 \cos\alpha

Steven Zheng posted 2 months ago

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