Question

GEOMETRIA- EQUAÇÕES E RELAÇÕES FUNDAMETAIS

Answer 1

I have a Portuguese Water Dog who's translating for me.

a.

$\sin x + \cos x = -1$

Usually I'll say the linear combination of a sine and a cosine of the same angle is a phase shift and a dilation etc. but here we can take a shortcut. The only time we'll get -1 as the sum is when one of the terms is -1, then the other term will be zero because of $\cos^2 x + \sin ^2 x = 1$.

$\sin x = -1 \textrm{ or } \cos x = -1$

Answer: x = 3π/2 or x=π

b.

$\cos x = \frac 1 2$

That's the biggest cliche of trig 30/60/90 so x=plus or minus 60 degrees. We're told we're in the fourth quadrant, so

$\sin x = - \dfrac{\sqrt{3}}{2}$

Answer: -√3/2

c.

This time we're in the third quadrant, negative sine

$\cos^2 x + \sin ^2 x = 1$

$\cos x = - \sqrt{ 1- \sin ^2 x} = -\sqrt{1 - (-3/4)^2} = -\sqrt{1-9/16} = -\sqrt{7/16} = -\frac 1 4 \sqrt 7$

Answer: -(1/4)√7

d.

Second quadrant positive sine, negative tangent

$\sin x = \sqrt{1 - \cos^2x} = \sqrt{ 1 - (-3 \sqrt{2}/5)^2} = \sqrt{1 - 18/25}=\frac 1 5 \sqrt{7}$

$\tan x = \dfrac{\sin x}{\cos x} = \dfrac{\sqrt{7}/5 }{-3 \sqrt{2} / 5} = -\dfrac{\sqrt{14}}{6}$

Answer: sin x = √7/5, tan x = -√14/6

e.

We're given the tangent and the sine

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

$\cos x = \dfrac{\sin x }{\tan x } = \dfrac{\sqrt{5}/5}{-1/2} = - \dfrac 2 5 \sqrt{5}$

Answer: (-2/5)√5