Question

# If $$\cos \theta + \cos 7\theta+\cos 3\theta+\cos 5\theta=0$$, then $$\theta=$$

A
(2n+1)π2;nZ
B
(2n+1)π4;nZ
C
(2n+1)π8;nZ
D
(2n+1)π16;nZ

Solution

## The correct options are A $$\left( 2n + 1 \right) \cfrac{\pi}{2}; n \in Z$$ B $$\left( 2n + 1 \right) \cfrac{\pi}{4}; n \in Z$$ C $$\left( 2n + 1 \right) \cfrac{\pi}{8}; n \in Z$$Given that:-$$\cos{\theta} + \cos{7 \theta} + \cos{3 \theta} + \cos{5 \theta} = 0$$To find:- $$\theta = ?$$$$\cos{\theta} + \cos{7 \theta} + \cos{3 \theta} + \cos{5 \theta} = 0$$$$\left( \cos{\theta} + \cos{7 \theta} \right) + \left( \cos{3 \theta} + \cos{5 \theta} \right) = 0$$$$2 \cos{\left( \cfrac{\theta + 7 \theta}{2} \right)} \cos{\left( \cfrac{\theta - 7 \theta}{2} \right)} + 2 \cos{\left( \cfrac{3 \theta + 5 \theta}{2} \right)} \cos{\left( \cfrac{3 \theta - 5 \theta}{2} \right)} = 0 \; \left[ \because \cos{A} + \cos{B} = 2 \cos{\cfrac{A + B}{2}} \cos{\cfrac{A - B}{2}} \right]$$$$2 \left( \cos{4 \theta} \cos{3 \theta} + \cos{4 \theta} \cos{ \theta} \right) = 0 \; \left[ \because \cos{\left( - \theta \right)} = \cos{\theta} \right]$$$$\Rightarrow \; 2 \cos{4 \theta} \left( \cos{3 \theta} + \cos{\theta} \right) = 0$$$$\Rightarrow \; 2 \cos{4 \theta} \left( 2 \cos{2 \theta} \cos{\theta} \right) = 0$$$$\Rightarrow \; 4 \cos{\theta} \cos{2 \theta} \cos{4 \theta} = 0$$$$\Rightarrow \; \cos{\theta} = 0 \text{ or } \cos{2 \theta} = 0 \text{ or } \cos{4 \theta} = 0$$Case I:-For $$\cos{\theta} = 0$$$$\Rightarrow \; \theta = \left( 2n + 1 \right) \cfrac{\pi}{2}; n \in Z ..... \left( 1 \right)$$Case II:-For $$\cos{2 \theta} = 0$$$$\Rightarrow \; 2 \theta = \left( 2n + 1 \right) \cfrac{\pi}{2}$$$$\Rightarrow \; \theta = \left( 2n + 1 \right) \cfrac{\pi}{4}; n \in Z ..... \left( 2 \right)$$Case III:-For $$\cos{4 \theta} = 0$$$$\Rightarrow \; 4 \theta = \left( 2n + 1 \right) \cfrac{\pi}{2}$$$$\Rightarrow \; \theta = \left( 2n + 1 \right) \cfrac{\pi}{8}; n \in Z ..... \left( 3 \right)$$Hence equation $$\left( 1 \right), \left( 2 \right) \& \left( 3 \right)$$ are the solution of $$\cos{\theta} + \cos{7 \theta} + \cos{3 \theta} + \cos{5 \theta} = 0$$.Mathematics

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