First slide

Multiple and sub- multiple Angles

Question

If $\cos α + \cos β = a, \sin α + \sin β = b$ and $α - β = 2 θ,$ then $\frac{\cos 3 θ}{\cos θ} =$

Moderate

A

$a^{2} + b^{2} - 2$
B

$a^{2} + b^{2} - 3$
C

$3 - a^{2} - b^{2}$
D

$(a^{2} + b^{2}) / 4$

Solution

From the given relations we have
$\begin{array}{l} a^{2} + b^{2} = \cos^{2} α + \cos^{2} β + 2 \cos α \cos β + \sin^{2} α + \\ \sin^{2} β + 2 \sin α \sin β \\ = 2 + 2 \cos (α - β) \\ = 2 + 2 \cos 2 θ = 4 \cos^{2} θ \end{array}$
Now $\frac{\cos 3 θ}{\cos θ} = \frac{4 \cos^{3} θ - 3 \cos θ}{\cos θ}$
$= 4 \cos^{2} θ - 3 = a^{2} + b^{2} - 3$ .

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