First slide

Multiple and sub- multiple Angles

Question

$2 \cos θ + \sin θ = 1 then 7 \cos θ + 6 \sin θ =$

Difficult

A

1 or 2
B

2 or 3
C

2 or 4
D

2 or 6

Solution

We have $2 \cos θ + \sin θ = 1$
$\begin{array}{l} \Rightarrow 2 (\frac{1 - \tan^{2} \frac{θ}{2}}{1 + \tan^{2} \frac{θ}{2}}) + \frac{2 \tan \frac{θ}{2}}{1 + \tan^{2} \frac{θ}{2}} = 1 (∵ \cos θ = \frac{1 - \tan^{2} \frac{θ}{2}}{1 + \tan^{2} \frac{θ}{2}} a n d \sin θ = \frac{2 \tan \frac{θ}{2}}{1 + \tan^{2} \frac{θ}{2}}) \\ \Rightarrow 2 - 2 \tan^{2} \frac{θ}{2} + 2 \tan \frac{θ}{2} = 1 + \tan^{2} \frac{θ}{2} \\ \Rightarrow 3 \tan^{2} \frac{θ}{2} - 2 \tan \frac{θ}{2} - 1 = 0 \\ \Rightarrow \tan \frac{θ}{2} = 1, \tan \frac{θ}{2} = - \frac{1}{3} \\ \Rightarrow θ = \frac{π}{2}, \tan \frac{θ}{2} = \frac{- 1}{3} \\ F o r θ = \frac{π}{2}, w e h a v e 7 \cos θ + 6 \sin θ = 6 \\ F o r \tan \frac{θ}{2} = \frac{- 1}{3}, w e h a v e 7 \cos θ + 6 \sin θ = 7 (\frac{1 - \tan^{2} \frac{θ}{2}}{1 + \tan^{2} \frac{θ}{2}}) + 6 (\frac{2 \tan \frac{θ}{2}}{1 + \tan^{2} \frac{θ}{2}}) \end{array}$
$= \frac{7 (1 - \frac{1}{9}) + 12 (- \frac{1}{3})}{1 + \frac{1}{9}} = 2$

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