First slide

Trigonometric transformations

Question

$\cos 28^{0} + \sin 28^{0} = k^{3} then \cos 17^{0} =$

Easy

A

$\frac{k^{3}}{\sqrt{2}}$
B

$\frac{- k^{3}}{\sqrt{2}}$
C

$\sqrt{2} k^{3}$
D

$- \sqrt{2} k^{3}$

Solution

We have $\cos 28^{\circ} + \sin 28^{\circ} = k^{3}$
$\begin{array}{l} \Rightarrow \cos 28 ° + \cos 62 ° = k^{3} \\ \Rightarrow 2 \cos 45 ° \cos 17 ° = k^{3} (∵ \cos C + \cos D = 2 \cos (\frac{C + D}{2}) \cos (\frac{C - D}{2})) \\ \Rightarrow \cos 17 ° = \frac{k^{3}}{\sqrt{2}} \end{array}$

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