Question

Answer check please. Trig.

Answer 1

Use fundamental Identities to write \[\cos \theta \] in terms of \[\cot \theta \]



I have \[\cos \theta\ = cot \theta\ / \sqrt{2}\]

Been having trouble with identities.

Answer 2

From the Pythagorean theorem identities right?

or am i looking in the wrong direction?

Answer 3

Actually, there's a much simpler way:

\(\cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)}\)

Isolate \(\cos(\theta)\)

Answer 4

That would make it,

\[\cos \theta = \cot \theta \sin \theta\]

the answers in my book make it only in terms of the second term which is what confuses me. I don't know if this completes the problem, or if they want it only in terms of \[\cot \theta \]

Answer 5

Show me an example of a similar problem where the answer is in the book.

Answer 6

show \[\cot \theta \]
in terms of \[\sin \theta\]

\[\cot \theta = (1- \sin^2 \theta) / \sin \theta\]

Answer 7

I see

Answer 8

I'm assuming they want us to manipulate identities to only have it in terms of what they want. Like sin(x) in the previous example.

Answer 9

Hang on a sec

Answer 10

Can it be in terms of cot(theta) and sin(theta) ?

Answer 11

I don't have the actual answer for that problem, so I don't know.

Answer 12

I have

\[\cot \theta = \cos \theta / \sin \theta \] isolate cos

\[\cos \theta = \sin \theta \times \cot \theta \]

Pythagorean theorem for sin

\[\cot \theta \times 1 - \cos ^2 \theta = \cos^2 \theta\]

add cos to both sides

\[\cot \theta \times 1 = 2\cos ^2 \theta \]

divide by two and sqrt ( i assume we can eliminate 1 since cot x 1 = cot)

\[\sqrt{\cot \theta / 2} = \cos \theta\]

However, I don't know if my math is definite, or even correct unless I can check it.

Answer 13

I'm not really sure about your steps, but I figured it out

Answer 14

Ok. Am I correct?
If not, please provide a hint.

Answer 15

I got something different that I am sure is correct, but I'll check to see if they are equivalent

Answer 16

Would you mind providing a source, or example on how I can double check answers.

Or is it a matter of using a triangle with known values and testing the formulas against each other?

Answer 17

I'm just going to post what I did

Answer 18

\(\cot(x)) = \frac{cos(x)}{sin(x)}\)
\(co