Online session 30
Inverse trigonometric functions continued
Case 2:
INVERSE COSINE FUNCTION OR ARCCOS FUNCTION ( y = cos -1(x) )
Consider the original cos function y = cos (ϴ), which has its domain as
R or (- ∞ ,∞ ) , the range [-1 , 1] and the co-domain R or ( - ∞, ∞ ).
The function is clearly not one-one and not onto since :
1)When cos ( ϴ1)= cos (ϴ2) , we find that ϴ1≠ϴ2 necessarily due to
the ASTC rule , For example
cos (300) = cos ( 3300) = √3/2 , { since cos ( 3600- y ) = cos(y))
So cos (ϴ) is not one –one in its original form .
2)Since – 1 ≤ cos (ϴ) ≤ 1 , all numbers > 1 and < -1 are not images and
hence cos (ϴ) fails to be onto in its original form .
For eg . cos ( ϴ) = 2 cannot be solved to find a value of ϴ.
Hence in its orginal form y = cos (ϴ) defined from R TO R , cannot
have an inverse .
In order to define the inverse cos function or y = cos-1(x) , we must
therefor define y = cos (ϴ) so that it is both one- one and onto or
bijective .
This can be done as follows :
A)Define the domain of y = cos ( ϴ) to be the closed interval [ 0 , π ] ,
thus removing the chance of duplications ( cos is positive in 1st
quadrant but negative in the 2nd quadrant).So the function is now
one –one.
B)Define the co –domain of cos (ϴ) to be [-1, 1] , in other words
ensure that the co-domain is the same as the range .We now have
the function being onto at this stage
C) Clearly we have cos (ϴ) from [ 0 , π] → [ 1, -1 ] , being both oneone as well as onto or bijective . We can now proceed to define the
inverse cos function or arccos or y = cos -1 (x).
Definition
We define the inverse cos function or arccos or y = cos -1(x) or
reverse cos as follows , given y = cos(ϴ) : [0,π] → [ 1, -1]
1)y = cos -1( x ) is defined from [1, -1 ] to [ 0,π]
2) The domain of y = cos -1 ( x) is [ 1, -1]
3) The range of the function is [ 0 , π]
4) Clearly cos-1 ( a real value from -1 to 1 ) = angle from π to 0
5)Standard values :
i) cos -1(0) = π/2 or 900 ii) cos -1(1) = 0 or 00
iii) cos -1(1/2) = 600or π/3 iv) cos-1(√3/2) = 300 or π/6
v) cos -1 ( 1/ √2 ) = π/4 or 450
6)Negative property of cos-1 (x)
cos -1 ( - x ) = π – cos -1 ( x) or 1800 – cos-1 (x) , when x ɛ [-1,1]
PROOF: Let cos-1 ( -x ) = ϴ→ 1 , x ɛ [ -1 , 1] , ϴ ɛ [ 0,π ]
We have - x = cos (ϴ) { cos inverse transferred from lhs gives cos
on rhs }
So x = - cos ( ϴ ) = cos( π-ϴ), Since { cos ( 1800- y ) = - cos y }
Now cos-1 (x) = π - ϴ , { cos transferred from rhs becomes cos
inverse on lhs }
Now from 1 , we have on replacing ϴ
cos-1 (x) = π – cos -1 ( - x ) , thus on rearranging the terms we have
cos-1 ( -x ) = π – cos -1 ( x)
Examples :
1)cos -1(-1 ) = π – cos -1(1) = π -0 = π or 180o
2) cos-1 ( -1/2 ) = π - cos -1(1/2) = π- π/3 = 2π/3
3) cos-1 ( - √3/2 ) = 5π/6 = 1500
4)cos -1 ( -1 / √2) = 3π/4 = 1350
7)Transfer property
i) If cos -1 ( A ) = B , THEN A = cos (B)
ii)If cos ( A ) = B , then we have A = cos -1(B)
8)Principal values and Principal Branch of the function
The principal values of cos -1 (x) are the angles lying in the closed
interval [0, π] and as can be seen from the graph this is the
principal value banch of the curve
9)The graph of the arccos function or inverse cos function or
reverse cos function is shown below and clearly we see the domain
and range demonstrated for the function .
10)On a calculator we have the inverse cos function given by
Shift ( cos )
Questions :
1)Given cos -1 ( x ) = 5π/6 , find the value of x
2)Is it possible to determine the value of cos-1 ( 5/4) , explain your
answer
3)Given cos -1 ( x) = 4π / 3, is it possible to determine the value of x
4)Given cos -1 (x) + cos -1 (y) = 2π , find 3x+ 2y
5) cos -1 (x) + cos -1 (y) + cos -1(z) = 3 π , find the value of x2 + y2 + z2