Question

If cos Θ = negative 4 over 7, what are the values of sin Θ and tan Θ? (2 points)

Answer 1

I accept that θ in second quadrant, therefore sinθ > 0, cosθ < 0 and tanθ < 0.

$\cos\theta=-\dfrac{4}{7}\\\\We\ know:\sin^2x+\cos^2x=1\to\sin^2x=1-\cos^2x\to\sin x=\sqrt{1-\cos^2x}\\\\subtitute:\\\\\sin x=\sqrt{1-\left(-\dfrac{4}{7}\right)^2}=\sqrt{1-\dfrac{16}{49}}=\sqrt{\dfrac{33}{49}}=\dfrac{\sqrt{33}}{7}$

$We\ know:\tan x=\dfrac{\sin x}{\cos x}\\\\subtitute:\\\\\tan x=\dfrac{\sqrt{33}}{7}:\left(-\dfrac{4}{7}\right)=-\dfrac{\sqrt{33}}{7}\cdot\dfrac{7}{4}=-\dfrac{\sqrt{33}}{4}$