Question

Find sin θ if cot θ = –2 and cos θ < 0.
I don't understand this lesson at all.

Answer 1

can somebody pleaseee help me quickly? The population (in millions) of a certain country can be approximated by the function: P(x)=50*1.02^x where x is the number of years after 2000. Which of the following calculations will tell in what year the population can be expected to reach 100 million? a. ln(2/1.02)+2000 b. ln(2)/ln(1.02)+2000 c. ln(2/1.02) d. ln(2)/ln(1.02)

Answer 2

Remember your identity for cotangent? Converting it to sines and cosines might give us some information.\[\large \cot \theta=\frac{\cos \theta}{\sin \theta}\]Hmmmm, they told us that cos θ < 0. It's negative.
Then how can cot θ be -2? That will only be true if Sin θ is POSITIVE right? Cause then we have a negative cosine over a positive sine, giving us a negative -2 some how.

In which quadrant is Cosine negative, and Sine positive?

Answer 3

is it quadrant II?

Answer 4

Yesssss, so we're in Quadrant II, and cot θ = -2... hmmm lemme think a sec :d

Answer 5

they give me options with radicals in the numerator for sin?
i don't understand what you mean

Answer 6

Umm can you give me an idea of what options we have? Then maybe I can figure this out. Do any of them involve sqrt5?

Answer 7

yes they both do. but they have one with denominator of 5 and the other 2?

Answer 8

Ok cool I think I've got a handle on this :) Let's see if this makes sense. I'm gonna draw a picture real quick.

Answer 9

ok.....

Answer 10

So we know that our angle is somewhere in the second quadrant, so we'll draw a line where our angle is, and then drop a line down to form a triangle with the x-axis.

The reason we do this is because we can use our Trig Triangle relationships to label the sides.

Answer 11

right opp, adj, and hyp?

Answer 12

Yah! :)
So what is Cotangent in relation to the sides?

Well we remember that Tangent is OPPOSITE over ADJACENT, yes?
So Cotangent will be ADJACENT over OPPOSITE.

Answer 13

We can think of \[\large \cot \theta = -2 \quad \text{as}\quad \cot \theta = \frac{-2}{1} \quad = \frac{adjacent}{opposite}\]See how we're going to label the sides? :o