Question

cos[tan^-1(-1) + cos^-1 (-4/5)
How would I find the exact value of this one?

Will FAN and MEDAL

Answer 1

@zepdrix sorry to bother you again but can you help me again lol

Answer 2

Figure this one out yet? :)

Answer 3

Nope all I have is that tan inverse of -1 is -pi/4

Answer 4

@zepdrix

Answer 5

Ok that at least gets us started,\[\Large\rm \cos\left[-\frac{\pi}{4}+\arccos\left(-\frac{4}{5}\right)\right]\]Recall that the composition of a function and it's inverse just gives you the argument back. I'm saying this because... we're going to end up with cos(arccos(-4/5)) at some point, so we'll want to keep this in mind:\[\Large\rm \cos\left(\arccos\left(x\right)\right)=x\]

Answer 6

Our inverse function represents some angle.
So what we REALLY have, even though it might not look like it right now,
is something like this: \(\Large\rm \cos(\theta+\beta)\)

Answer 7

Remember your Cosine Angle Sum Identity?

Answer 8

\[\cos(\theta+\beta)=\cos \theta \cos \beta - \sin \theta \sin \beta\]

Answer 9

?

Answer 10

Ya, we want to use that.

Answer 11

Oh oh oh, we're going to end up with a sin(arccos) as well..
so I guess we better just do the triangle business
ok ok ok

Answer 12

So I'm saying that this arccos represents some arbitrary angle,\[\Large\rm \arccos\left(-\frac{4}{5}\right)=\theta\]We'll rewrite this relationship like this,\[\Large\rm \cos \theta=-\frac{4}{5}\]And draw a triangle to show the relationship.

Answer 13

something like that, ya?

Answer 14

Yep

Answer 15

So using this information,
\[\Large\rm \color{royalblue}{\arccos\left(-\frac{4}{5}\right)=\theta}\]
Our problem,\[\Large\rm \cos\left[-\frac{\pi}{4}+\color{royalblue}{\arccos\left(-\frac{4}{5}\right)}\right]\]really looks like this,\[\Large\rm \cos\left[-\frac{\pi}{4}+\color{royalblue}{\theta}\right]\]

Answer 16

Then we can apply our Angle Addition Formula for Cosine,\[\Large\rm =\cos\left(-\frac{\pi}{4}\right)\cos\left(\color{royalblue}{\theta}\right)-\sin\left(-\frac{\pi}{4}\right)\sin\left(\color{royalblue}{\theta}\right)\]

Answer 17

Got it

Answer 18

And then use the triangle to fill in the pieces related to theta, ya? :)

Answer 19

Yes, should would that change sin theta

Answer 20

yes, I guess if we look at our triangle, we get something like 3/5 for our sin(theta) ya?

Answer 21

Yes

Answer 22

\[\Large\rm =\cos\left(-\frac{\pi}{4}\right)\cos\left(\color{royalblue}{\theta}\right)-\sin\left(-\frac{\pi}{4}\right)\sin\left(\color{royalblue}{\theta}\right)\]
\[\Large\rm =\cos\left(-\frac{\pi}{4}\right)\left(-\frac{4}{5}\right)-\sin\left(-\frac{\pi}{4}\right)\frac{3}{5}\]

Answer 23

You'll have to simplify a little further :o
Remember how to figure these out? Unit circle might help.

Answer 24

Would the cos be 3pi/4's x

Answer 25

hmm? 0_o

Answer 26

we're taking the cosine of an angle, we don't end up with any pi business.
we end up with one of those special values, ya?

1/2, sqrt(3)/2, sqrt(2)/2, 1, 0

one of those guys, remember?