## Atkinsoha Group Title PreCalc & Trig Help, Picture Below! one year ago one year ago

1. Atkinsoha Group Title

2. ajprincess Group Title

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 Group Title

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 Group Title

What is the value of hypotenuse?

5. Atkinsoha Group Title

(√13)? I hope..

6. ajprincess Group Title

yup right:)

7. ajprincess Group Title

nw what is $$\cos\theta$$

8. Atkinsoha Group Title

3/√13

9. ajprincess Group Title

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

10. Atkinsoha Group Title

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

11. Atkinsoha Group Title

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 Group Title

yup:)

13. Atkinsoha Group Title

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

14. Atkinsoha Group Title

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 Group Title

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 Group Title

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 Group Title

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

18. ajprincess Group Title

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