Question

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 over 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

Answer 1

`\huge\boxed{\sin\theta=-(\sqrt2)/(2);\ \tan\theta=-1}`

Explanation:

We have:

`\\cos\theta=(\sqrt2)/(2),\ (3\pi)/(2)<\theta<2\pi`

For sine use:

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

Substitute:

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

θ in IV quadrant, therefore sine is negative.

`\sin\theta=-(\sqrt2)/(2)`

For tangent use:

`\tan x=(\sin x)/(\cos x)`

Substitute:

`\tan\theta=(-(\sqrt2)/(2))/((\sqrt2)/(2))=-(\sqrt2)/(2)\cdot(2)/(\sqrt2)=-1`