First slide

Multiple and sub- multiple Angles

Question

If $\tan θ = \frac{a}{b},$ then $b \cos 2 θ + a \sin 2 θ$ is equal to

Moderate

A

a
B

b
C

$\frac{a}{b}$
D

$\frac{b}{a}$

Solution

write $\cos 2 θ, \sin 2 θ$ in terms of $\tan θ$
$\cos 2 θ = \frac{1 - \tan^{2} θ}{1 + \tan^{2} θ} \sin θ = \frac{2 \tan θ}{1 + \tan^{2} θ} b \cos 2 θ + a \sin 2 θ = b (\frac{1 - \frac{a^{2}}{b^{2}}}{1 + \frac{a^{2}}{b^{2}}}) + a (\frac{2 \frac{a}{b}}{1 + \frac{a^{2}}{b^{2}}}) = b (\frac{b^{2} - a^{2}}{b^{2} + a^{2}}) + \frac{2 a^{2} b}{a^{2} + b^{2}} = b (\frac{b^{2} + a^{2}}{b^{2} + a^{2}}) = b$

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