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Ksivusya [100]
2 years ago
7

If cos Θ = square root 2 over 2 and 3 pi over 2 < Θ < 2π, what are the values of sin Θ and tan Θ? sin Θ = square root 2 ov

er 2; tan Θ = −1 sin Θ = negative square root 2 over 2; tan Θ = 1 sin Θ = square root 2 over 2; tan Θ = negative square root 2 sin Θ = negative square root 2 over 2; tan Θ = −1
Mathematics
1 answer:
densk [106]2 years ago
8 0

Answer:

\huge\boxed{\sin\theta=-\dfrac{\sqrt2}{2};\ \tan\theta=-1}

Step-by-step explanation:

We have:

\\cos\theta=\dfrac{\sqrt2}{2},\ \dfrac{3\pi}{2}

For sine use:

\sin^2x+\cos^2x=1\to\sin^2x=1-\cos^2x

Substitute:

\sin^2\theta=1-\left(\dfrac{\sqrt2}{2}\right)^2\\\\\sin^2\theta=1-\dfrac{(\sqrt2)^2}{2^2}\\\\\sin^2\theta=1-\dfrac{2}{4}\\\\\sin^2\theta=\dfrac{4}{4}-\dfrac{2}{4}\\\\\sin^2\theta=\dfrac{4-2}{4}\\\\\sin^2\theta=\dfrac{2}{4}\to\sin\theta=\pm\sqrt{\dfrac{2}{4}}\\\\\sin\theta=\pm\dfrac{\sqrt2}{\sqrt4}\\\\\sin\theta=\pm\dfrac{\sqrt2}{2}

θ in IV quadrant, therefore sine is negative.

\sin\theta=-\dfrac{\sqrt2}{2}

For tangent use:

\tan x=\dfrac{\sin x}{\cos x}

Substitute:

\tan\theta=\dfrac{-\frac{\sqrt2}{2}}{\frac{\sqrt2}{2}}=-\dfrac{\sqrt2}{2}\cdot\dfrac{2}{\sqrt2}=-1

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