Question

Explain the difference between using the trigonometric ratios (sin, cos, tan) to solve for a missing angle in a right triangle versus using the reciprocal ratios (sec, csc, cot). You must use complete sentences and any evidence needed (such as an example) to prove your point of view.

Answer 1

@whpalmer4

Answer 2

Do you know what the reciprocal ratios are?

Answer 3

I've never seen a catchy mnemonic for them, but

\[\cot x = \frac{1}{\tan x}\]\[\sec x = \frac{1}{\cos x}\]\[\csc x = \frac{1}{\sin x}\]

Answer 4

That means that \[\cot x = \frac{\text{Adjacent}}{\text{Opposite}}\]\[\sec x = \frac{\text{Hypotenuse}}{\text{Adjacent}}\]\[\csc x = \frac{\text{Hypotenuse}}{\text{Opposite}}\]

Answer 5

will this be my answer??

Answer 6

I'm just explaining the idea to you, not writing your answer.

So if we have that same triangle, and we wanted to find \(\csc x\), what would it be?

Answer 7

hypostuse/oppisote

Answer 8

okay, but what would the value be?

Answer 9

idk

Answer 10

csx

Answer 11

?

Answer 12

What is the hypotenuse in that triangle? What is the opposite side from angle \(x\)?

Answer 13

\[\csc x = \frac{\text{Hypotenuse}}{\text{Opposite}}\]

Answer 14

thhe adjacent side

Answer 15

numbers, man, numbers! Look at the triangle?