## anonymous one year ago sin theta= 3/5 cos beta = 2sqrt6/5 Find sin(theta+beta). Is the answer 6sqrt6+4/25?

1. shamim

R u able to find out Cos theta=?

2. anonymous

Yes 4/5

3. UsukiDoll

$\sin (\theta) = \frac{3}{5}, \cos(\beta) = 2\sqrt{\frac{6}{5}}$ ? I'm confused on what cosine beta should be

4. shamim

Sin beta=?

5. anonymous

1/5

6. UsukiDoll

oh yeah we need sin beta and cosine beta otherwise the angle sum identity formula isn't going to work

7. UsukiDoll

$\sin(\theta + \beta) = \sin(\theta)\cos(\beta)+\cos(\theta)\sin(\beta)$

8. UsukiDoll

so far we have sin (theta) = 3/5 |dw:1433043667731:dw| so cos (theta) = 4/5

9. anonymous

Yep got that

10. UsukiDoll

ummm what was the cosine beta?

11. UsukiDoll

is it $\cos(\beta) = 2 \frac{\sqrt{6}}{5} ?$ is the square root inside the 6/5? that part is throwing me off

12. UsukiDoll

$\sin(\theta + \beta) = \frac{3}{5}\cos(\beta)+\frac{4}{5}\sin(\beta)$ I don't understand what cosine beta is... but I do know that I need another triangle so I can grab sin beta and we add this all together

13. UsukiDoll

is the 6/5 fraction inside the whole square root or is it just square root 6 / 5 ?

14. UsukiDoll

$\cos(\beta) = 2 \frac{\sqrt{6}}{5} ?$ or $\cos(\beta) = 2 \sqrt{\frac{6}{5}}$

15. UsukiDoll

I need clarification on that part... otherwise I might end up with the wrong answer.

16. UsukiDoll

|dw:1433044545197:dw|

17. UsukiDoll

$\sin(\theta + \beta) = \frac{3}{5}\frac{2\sqrt{6}}{5}+\frac{4}{5} (\frac{1}{5})$

18. UsukiDoll

$\frac{6\sqrt{6}}{25}+\frac{4}{25}$

19. UsukiDoll

I relooked at cos beta and realized that 5 was the longest side of the triangle and that 2 sqrt{6} was the adjcent part... needed to find the opposite part of the triangle so I could find sin beta.

20. UsukiDoll