Question

find the exact value of cos[arcsin(-3/5) + arccos (-4/5)]

Answer 1

\[\Large\bf\sf \arcsin\left(\frac{-3}{5}\right)=\theta, \qquad\qquad \arccos\left(\frac{-4}{5}\right)=\phi\]
\[\Large\bf\sf \sin \theta = \frac{-3}{5},\qquad\qquad \left(\frac{opposite}{hypotenuse}\right)\]\[\Large\bf\sf \cos \phi =\frac{-4}{5},\qquad\qquad \left(\frac{adjacent}{hypotenuse}\right)\]

Answer 2

Pythagorean Theorem to find the missing sides.

Answer 3

\[\Large\bf\sf \cos\left[\arcsin\left(\frac{-3}{5}\right)+\arccos\left(\frac{-4}{5}\right)\right]\]Can be written as,\[\Large\bf\sf \cos\left[\theta+\phi\right]\]Using our Cosine Angle Sum Identity:\[\Large\bf\sf \cos\left[\theta+\phi\right]=\cos \theta \cos \phi - \sin \theta \sin \phi\]

Answer 4

From there we could use our triangles to fill in the pieces.

Answer 5

This problem is kinda difficult.
We can't use the same triangle for both inverse functions because:
arcsine is only negative in the `fourth quadrant`,
while arccosine is only negative in `quadrant two`.

So they can't refer to the same angle unfortunately. :(
Maybe there is some trick that allows you to do it.. but I dunno.

Answer 6

so i know how to find the rest of the triangles but where would i use them?

Answer 7

So if you're able to find the missing sides.
Then plug the relationships into the last messy looking setup.\[\Large\bf\sf \color{orangered}{\cos \theta }\cos \phi - \sin \theta \sin \phi\quad=\quad \color{orangered}{\frac{adj}{hyp}}\cos \phi - \sin \theta \sin \phi\]This first one would use the triangle with the theta angle.
You would plug in all 4 of the relationships like this.
Understand what I mean..? :O kinda?

Answer 8

(adj/hyp 1st triangle)(adj/hyp 2st triangle)-(opp/hyp 1st triangle)(opp/hyp 2st triangle)

Answer 9

???

Answer 10

Yes, the relationships look correct.
In case you're confused by what I'm saying,
here is how the orange one would simplify:\[\Large\bf\sf \color{orangered}{\cos \theta }\cos \phi - \sin \theta \sin \phi\quad=\quad \color{orangered}{\frac{adj}{hyp}}\cos \phi - \sin \theta \sin \phi\]\[\Large\bf\sf =\quad \color{orangered}{\frac{4}{5}}\cos \phi - \sin \theta \sin \phi\]

Answer 11

(4/5)(-4/5)-(-3/5)(3/5) ????

Answer 12

Mmmm yah I think that looks right! :)

Answer 13

thx so much!

Answer 14

what about... evaluate exactly the cos(2x) and sin (x/w) if tan x = -8/15

Answer 15

ik cos = 15/17 and sin = -8/17

Answer 16

but what formula would i use????

Answer 17

\[\Large\bf\sf \cos(2x)\quad=\quad \cos^2x-\sin^2x\]

Answer 18

For the sine one, I'm not sure.
I don't know what they mean by w. Hmmmmm

Answer 19

It's some arbitrary angle cut.
So ummm.. Hmm I'm trying to think of how we deal with that.

Answer 20

x/2*

Answer 21

sorry

Answer 22

Oh lol

Answer 23

haha

Answer 24

Sine Half-Angle Identity:
\[\Large\bf\sf \sin\left(\frac{x}{2}\right)\quad=\quad \pm\sqrt{\frac{1-cosx}{2}}\]Try to remember some of these buster! :O

Answer 25

haha

Answer 26

so just plug in sin and cos where they are in the equations correct?

Answer 27

Mmmmmmmmmmmmmm.... yes.
That plus/minus might cause a problem on the sine though. Hmmm.

Answer 28

thx for all your help. Now time for bed.