Question

cosα+cosβ=a, sinα+sinβ=b and α-β = 2θ. If $\frac{cos3\theta }{cos\theta }$= K, when a=$\frac{1}{3}$,b=$\frac{1}{2}$Find the value of

-36K.

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Answer 1

We want to find $\frac{cos3\theta }{cos\theta }$ . We will try to simplify this because $cos3\theta$ can be expressed in terms cos $\theta$

$\Rightarrow$ $\frac{cos3\theta }{cos\theta }=\frac{4co{s}^3\theta -3cos\theta }{cos\theta }=4co{s}^2\theta -3$

We are given $2\theta$ = $\alpha -\beta$ . So it is better to convert $co{s}^2\theta$ in terms of cos2$\theta$ .

$\Rightarrow$ K = 4$co{s}^2\theta -3$

= $2(1+cos2\theta )-3$

We are given $2\theta =\alpha -\beta$ . So it is better to convert $co{s}^2\theta$ in terms of cos2$\theta$

$\Rightarrow$ $K=4co{s}^2\theta -3$

= 2(1 + cos2\theta) - 3\)

K = 2cos2$\theta$ - 1

Now, we have to find cos2$\theta$ or cos$(\alpha -\beta )$

If we get expressions like

$a=cos\alpha +cos\beta$

and $b=sin\alpha +sin\beta$, one usual step is squaring and adding both. In our case we get one advantage also.

The left over terms will be $cos\alpha cos\beta andsin\alpha sin\beta$. This will help us in finding cos $(\alpha -\beta )$ .

$\Rightarrow$ $a^2+b^2$ = 2 + 2 $(cos\alpha cos\beta +sin\alpha sin\beta )$

= 2 + 2 {cos ($\alpha -\beta$)}

$\Rightarrow$ 2 cos (\alpha - \beta) = $a^2+b^2-2$

$\Rightarrow$ k+1 = $a^2+b^2-2$

$\Rightarrow$ k = ${\left(\frac{1}{3}\right)}^2+{\left(\frac{1}{2}\right)}^2-3=\frac{-95}{36}$
$\Rightarrow$ - 36K = 95
Key steps / concepts: (1) simplifying $\frac{cos3\theta }{cos\theta }$
(2) Squaring and adding terms.