Question

I am bit confused about inverse trig functions. Can you manipulate them like this:

e.g. arccos (0.6-sin x)=arcsin (cosx-0.2)

Is it the same as if I write

sin (0.6-sin x)=cos (cosx-0.2)?

Answer 1

first, why do you think that you can?

Answer 2

I don't know, because I thought it could be like other reciprocal functions?

Answer 3

you ok

Answer 4

You can't simply.However, to get rid of either the arccos or arcsin take cos or sin to both sides:
arccos (0.6-sin x)=arcsin (cosx-0.2)

cos(arccos(0.6-sinx)) = cos((arcsin(cosx-0.2))

0.6 - sinx = cos(arcsin(cosx-0.2))

As we know : Cos^2x + Sin^2x = 1

cosx = sqrt(1-sin^2x)

0.6 - sinx = sqrt(1-sin^2(arcsin(cosx-0.2))

0.6 - sinx = sqrt(1-(cosx-0.2)^2)

(0.6-sinx)^2 = (1- (cosx-0.2)^2)

Answer 5

Can you take a snapshot of the original problem? I don't get the question.

Answer 6

@Loser66, He thinks that he can flip the the arcsin with cos, and arccos with sin by taking cos, sin to each site separately without involving the other side.

Answer 7

@TrojanPoem, thanks!! very clear steps. i understand it now!

Answer 8

You're welcome.