Question

Use the value given and the trig identities to find the indicated function.

cosθ = (sqrt13)/7
sinθ = (?)

I don't get this. Is it 6/7?

Answer 1

Try:
\[\sin^2 \theta + \cos^2 \theta = 1 \\ \sin^2 \theta + (\frac{\sqrt{13}}{7})^2 = 1\]

Answer 2

That's what I did. At the end there would be sin^2θ = 36/49, right?

Answer 3

Yea:
\[\sin^2 \theta =1- (\frac{\sqrt{13}}{7})^2 =\frac{49}{49} - \frac{13}{49}=\frac{36}{49}\]

Answer 4

then sinθ = 6/7 when i take the square root of both sides?

Answer 5

Sorry I got hung up for a second yes that should be the result

Answer 6

Thanks!

Answer 7

Your Welcome! :D