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- anonymous

PreCalc & Trig Help, Picture Below!

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- anonymous

PreCalc & Trig Help, Picture Below!

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- anonymous

- 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

- anonymous

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.

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- ajprincess

What is the value of hypotenuse?

- anonymous

(√13)? I hope..

- ajprincess

yup right:)

- ajprincess

nw what is \(\cos\theta\)

- anonymous

3/√13

- ajprincess

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

- anonymous

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

- anonymous

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?

- ajprincess

yup:)

- anonymous

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

- anonymous

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

- 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

- 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:)

- anonymous

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

- ajprincess

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

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