## a1234 one year ago 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?

1. anonymous

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

2. a1234

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

3. anonymous

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

4. a1234

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

5. anonymous

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

6. a1234

Thanks!

7. anonymous