Cos inverse of cos 350 - sin inverse of sin 350. Could someone help me with this

cos inverse of cos 350 = 350
sin inverse of sin 350 = 350
350 - 350 = 0

But, my text says otherwise:
cos inverse of cos 350 = cos inverse of cos 10 = 10
sin inverse of sin 350 = sin inverse of sin (-10) = -10
Therefore, 10 - (-10) = 20



Why cant I follow my method, what's wrong?




Thanks in advance!

the domain of sin inverse functions are from -90 degrees to 90, so when given 350, you have to convert it to -10

as for cosine functions i believe the domain is 0 to 180, which again with 350 degrees, you convert to an angle inside its domain. -10 is not in the domain, but cosine is an even function (the graph is reflected about the y axis), which means cos(-x) = cosx, so the negative becomes positive, and -10 becomes 10, it becomes part of the cosine inverse domain

sine on the other hand is an odd function, so you keep it negative

did i say domain, i meant range, the domain of inverse sine and cosine functions would be -1 to 1

The first answer is a good explanation as to why the method in your textbook works. As for the reason your method doesn't work, the problem is that the "inverse" sine and cosine functions are not true inverse functions. In other words, the inverse sine of the sine of x is not necessarily equal to x, and the inverse cosine of the cosine of x is not necessarily equal to x. The x would need to be in the appropriate domain.

ASTC (all, sin, tan, cos)
sin and cos and tan are >= 0 for quadrant 0 through 90 degrees
sin is >= 0 for quadrant 90 through 180
tan is >= 0 for quadrant 180 through 270
cos is >= 0 for quadrant 270 through 360

Reason:
consider a point, coordinates (x,y)
In polar coordinates (r*cos(theta), r*sin(theta))

If the point is in quadrant 90 through 180, then x <= 0 and y >= 0

y = r*sin(theta)
so if y >= 0 then r*sin(theta) >= 0

Now r >= 0,
so sin(theta) >= 0.
This shows that in quadrant 90 through 180, sin is positive,

arccos always returns value between 0° and 180°
arcsin always returns value between -90° and 90°

So angle of 350° has same cos value as angle of 10°
and angle of 350° has same sin value as angle of -10°

arccos(cos(350)) - arcsin(sin(350))
= 10 - (-10)
= 20

-- Ματπmφm --

we know that------(I) cos ( 360 -A ) = cos A &
(II) sin (360 - A) = -sin A
also sin (- A) = -sinA
so cos350 = cos(360-10 ) = cos10
sin 350 = sin(360 -10 ) = - sin10 = sin (-10 )
so cos-1 ( cos350) = cos-1{cos10} = 10
sin-1(sin350) = sin-1{ sin(-10)}= - 10
so 10 - (-10) = 20 proved