If 'theta' = 60°, verify that: cos 3'theta' = 4cos³'theta' - 3cos'theta' | Trigonometry

Question: If $\theta = 60°$, verify that: $cos3\theta = 4cos^3 \theta - 3cos\theta$.

Solution:

Given: $\theta = 60°$

To prove: $cos3\theta = 4cos^3 \theta - 3cos\theta$

We know,

$cos 60° = \dfrac{1}{2}$

$cos(90° + \theta) = - sin \theta$

$sin 90° = 1$

Taking LHS

$= cos 3\theta$

$= cos(3\theta)$

$= cos(3×60°)$

$= cos(180°)$

$= cos(90°+90°)$

$= - sin90°$

$= - 1$ [or you can use calculator]

Taking RHS

$= 4cos^3 \theta - 3cos\theta$

$= 4(cos \theta)^3 - 3cos\theta$

$= 4(cos 60°)^3 - 3cos60°$

$= 4 \left ( \dfrac{1}{2} \right )^3 - 3× \dfrac{1}{2}$

$= 4× \dfrac{1}{8} - \dfrac{3}{2}$

$= \dfrac{1}{2} - \dfrac{3}{2}$

$= \dfrac{1 -3}{2}$

$= \dfrac{-2}{2}$

$= -1$

Since, LHS = RHS

$cos3\theta = 4cos^3 \theta - 3cos\theta$ #proved

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