Question

how do i get only 1 trig function in: 2cos^-1(y)-sin^-1(y)=0?

Answer 1

These are the inverse functions?
Or did you mean to write \(\Large\rm \frac{1}{\cos y}\)?

Answer 2

thse are inverse

Answer 3

Ok ok I think I got it.
Boy you've got some tricky questions tonight :D lol

Answer 4

Let this inverse function be equal to some arbitrary angle,
\[\Large\rm \arccos(y)=\theta\]We'll write it in terms of cosine,\[\Large\rm \cos(\theta)=y\]Recall that sine and cosine are `co-functions`,
therefore,\[\Large\rm \cos(\theta)=\sin\left(\frac{\pi}{2}-\theta\right)=y\]

Answer 5

Write it back in terms of the inverse function,\[\Large\rm \arcsin(y)=\frac{\pi}{2}-\theta\]Solving for the angle theta,\[\Large\rm \theta=\frac{\pi}{2}-\arcsin(y)\]And relating this to what we stated earlier,\[\Large\rm \arccos(y)=\theta=\frac{\pi}{2}-\arcsin(y)\]

Answer 6

So that's how we'll replace our arccos(y) with something involving arcsin(y).
Any confusion there? :o

Answer 7

it seems so simple now. i couldn't figure it out before, but this make sense

Answer 8

thanks alot zepdrix

Answer 9

np \c:/
hmm neat problem, i'd never really thought about switching inverses like that before

Answer 10

brilliant!

Answer 11

its just the 1st part of the problem, cuz its for finding area between curves

Answer 12

Ooo interesting +_+