## Atkinsoha 2 years ago PreCalc & Trig Help, Picture Below!

1. Atkinsoha

2. ajprincess

Using pythagoras' theorem u can find the length of the hypotenuse. a^2+b^2=c^2 Here a=2, b=3, c=? |dw:1362407511223:dw||dw:1362407540894:dw| Does that help? @Atkinsoha

3. Atkinsoha

I did all of that.. I'm stuck on maybe how to solve it, since I can't seem to come up with any of the answers that are given.

4. ajprincess

What is the value of hypotenuse?

5. Atkinsoha

(√13)? I hope..

6. ajprincess

yup right:)

7. ajprincess

nw what is $$\cos\theta$$

8. Atkinsoha

3/√13

9. ajprincess

yup:) what is $$\cos^2\theta$$?

10. Atkinsoha

That's were I'm lost apparently. 3^2 = 9 and the (√13)^2 = 13?

11. Atkinsoha

WAIT. Is it A? 5/13? because then it would be 2(9/13)-1 which would be (18/13)-1 and one would be 13/13 so it would be 5/13? A?

12. ajprincess

yup:)

13. Atkinsoha

HOLY BEJEBOUS! Thank you so much! I guess I just needed someone to break it down!

14. Atkinsoha

Can you help me with this one? I'm sure it's really easy, but I've been gone from school.. so haven't learned it :P

15. ajprincess

let us say arcsinx=theta nd arccosx=beta so sin theta=x nd cos beta =x sin(arcsinx+arccosx)=sin(theta+beta) =sin theta*cos beta+sin beta * cos theta

16. ajprincess

cos theta=sqrt(1-sin^2theta) sin beta=sqrt(1-cos^2 beta) plug in the values and u will get the algebraic expression:)

17. Atkinsoha

Yeahh.. I don't understand at all. haha.

18. ajprincess

Really sorry. Hav to go nw. Please post this as a new ques. Hope some will sre help u.:) Really sorry