First slide

Trigonometric equations

Question

If $\sin θ + \cos e c θ = 2$ , then $\sin^{n} θ + \cos e c^{n} θ$ is equal to

Easy

A

2
B

$2^{n}$
C

$4^{n}$
D

none of these

Solution

We can write $\sin^{2} θ + 1 = 2 \sin θ$
$\Rightarrow \sin^{2} θ - 2 \sin θ + 1 = 0 \Rightarrow {(\sin θ - 1)}^{2} = 0$
$\Rightarrow \sin θ = 1 \Rightarrow \cos e c θ = 1$
And thus $\sin^{n} θ + \cos e c^{n} θ = 2$

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