Prove that: cos 'theta' = 4 cos³ 'theta'/3 - 3 cos'theta'/3 | Sub Multiple Angles | SciPiPupil

Questions: Prove the following simple identity: $cos \theta = 4 cos^3 \frac{ \theta}{3} - 3 cos \frac{ \theta}{3}$

Solution:

To prove: $cos \theta = 4 cos^3 \frac{ \theta}{3} - 3 cos \frac{ \theta}{3}$

Taking LHS

$cos \theta$

$= cos \left ( \frac{ \theta}{3} + \frac{2 \theta}{3} \right )$

$= cos \frac{ \theta}{3} × cos \frac{ 2\theta}{3} - sin \frac{ \theta}{3} × sin \frac{ 2\theta}{3}$

$= cos \frac{ 2\theta}{3} \left ( 2cos^2 \theta -1 \right )   - sin \frac{ \theta}{3}  × 2 sin \frac{\theta}{3} cos \frac{\theta}{3}$

$= 2cos^3 \frac{\theta}{3} - cos \frac{\theta}{3} - 2 sin^2 \frac{\theta}{3} cos \frac{ \theta}{3}$

$= 2cos^3 \frac{\theta}{3} - cos \frac{\theta}{3} - 2 cos \frac{\theta}{3} \left ( 1 - cos^2 \frac{ \theta}{3} \right )$

$= 2 cos ^3 \frac{\theta}{3} - cos \frac{\theta}{3} - \left ( 2 cos \frac{\theta}{3} - 2 cos^3 \frac{\theta}{3} \right )$

$= 2 cos ^3 \frac{\theta}{3} - cos \frac{\theta}{3} - 2 cos \frac{\theta}{3} + 2 cos^3 \frac{\theta}{3}$

$= 4 cos^3 \frac{ \theta}{3} - 3 cos \frac{ \theta}{3}$

RHS #proved

Some SUB Multiple Angles Formulae

$sin A = 2 sin\frac{A}{2}cos\frac{A}{2}$

$cos A = cos^2 \frac{A}{2} - sin^2 \frac{A}{2}$

$tan A = \dfrac{2 tan \frac{A}{2} }{1 - tan^2 \frac{A}{2}}$

$cot A = \dfrac{cot^2 \frac{A}{2} -1}{2 cot \frac{A}{2}}$

$sin A = 3 sin \frac{A}{3} - 4sin^3 \frac{A}{3}$

$cos A = 4cos^3 \frac{A}{3} - 3 cos \frac{A}{3}$

$tan A = \dfrac{3tan \frac{A}{3} - tan^3 \frac{A}{3}}{ 1 - 3tan^2 \frac{A}{3}}$

$cot A = \dfrac{3cot \frac{A}{3} - cot^3 \frac{A}{3}}{ 1 - 3cot^2 \frac{A}{3}}$

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