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Question: If \(\cos\theta = \frac{1}{2},\cos\theta = - \frac{3}{5} \Rightarrow \theta = \frac{\pi}{3},\pi - \c...

If cos⁡θ=12,cos⁡θ=−35⇒θ=π3,π−cos⁡−1(35)\cos\theta = \frac{1}{2},\cos\theta = - \frac{3}{5} \Rightarrow \theta = \frac{\pi}{3},\pi - \cos^{- 1}\left( \frac{3}{5} \right), then cos⁡θ=−12\cos\theta = \frac{- 1}{2}

A

0o<θ<360o0^{o} < \theta < 360^{o}

B

cos⁡60o=12\cos 60^{o} = \frac{1}{2}

C

cos⁡(180o−60o)\cos(180^{o} - 60^{o})

D

=−cos⁡60o=−12= - \cos 60^{o} = - \frac{1}{2}

Answer

cos⁡60o=12\cos 60^{o} = \frac{1}{2}

Explanation

Solution

2sin⁡x+2cos⁡x>21−(1/2)2^{\sin x} + 2^{\cos x} > 2^{1 - (1/\sqrt{2})}

Then x=5π4x = \frac{5\pi}{4}

(1+tan⁡θ)(1+tan⁡φ)=2⇒tan⁡θ+tan⁡φ1−tan⁡θtan⁡φ=1⇒(1 + \tan\theta)(1 + \tan\varphi) = 2 \Rightarrow \frac{\tan\theta + \tan\varphi}{1 - \tan\theta\tan\varphi} = 1 \Rightarrow.