Inverse Trigonometric Functions

Table of domain and range of inverse trigonometric function

Inverse-forward identities are

Forward-inverse identities are

Relation between inverse functions:

  Example 1.   Evaluate sin   2

 -- "the angle whose sine

  ."

-1  is  2

  Solution.   2 

 is the sine of what angle?

2 =

sin

π

4  

That is,

Sin-1

2  = 

π

4.

The range of y = arcsin x=sin-1x

  But 

π

4

 is not the only angle whose sine is  2 

. 2 

 is the sine of every 1st and

  2nd quadrant angle whose corresponding acute angle is 

π

4 

si 3  =  .

n π 4

2 

sin ( 

π

4

 + 2π

) =  2 

.

And so on.For the function  y = sin-1x to be

single-valued, then, we must restrict the values of y.  How will we do that?  We will restrict them to those angles that have the smallest absolute value.

In that same way, we will restrict the range of each inverse trigonometric function. (Topic 3 of Precalculus.)

π4

 is the angle of smallest absolute value whose sine

2 

.

is 

  Example 2.   Evaluate sin-1 (−

)   2

Solution.  Angles whose sines are negative fall in the 3rd and 4th quadrants.  The angle of smallest absolute value is in the 4th

  quadrant.  It is the angle −

π

4.

sin-

1(−   2

) = −

π

4.

For angles whose sine is negative, we always choose a 4th quadrant angle.  In fact,

sin-1

(−x) = − sin-1x

sin-  = − sin-1½

1(−½)

   = −π

6.

*

To see that sin-1 (−x) = − sin-1x, look here:

 = θ.

 

 = −θ.

 

Here, then, is the range of the function  y = sin-1x

To restrict the range of arcsin x is equivalent to restricting the domain of sin x to those same values. This will be the case with all the restricted ranges that follow.

.)

Problem 2.   Evaluate the following in radians.a)  sin-1 0 = 0.  

b)  sin-1 1 = π/2.  

c)  sin-1 (−1) = −π/2.  

  

 π/3.

   −π/3.

   −π/6.

The range of y = tan-1 xSimilarly, we must restrict the range of  y = tan-1 x.  Like y = sin-1

x,   y =  tan-1 x has its smallest absolute values in the 1st and 4th quadrants.

For angles whose tangent is postive, we choose a 1st quadrant angle. For angles whose tangent is negative, we choose a 4th quadrant angle. Like sin-1 (−x),

tan-1 (−x) = −tan -1x.

 = −θ.

 

 = θ.

 

 

Problem 3.   Evaluate the following.

  a)  tan-1 1 = 

π    b)  tan-1 (−1) = 

−π

4 4

  c)  tan−1

 = π

3 

  d)  tan−1(−

) = −

π

3

  e)  tan-1 0 = 

0  f)   = −π

6

The range of y = cos-1 xThe values of  y = Cos-1 x  will have their smallest absolute values when y falls within the 1st or 2nd quadrants.

Example 3.   Evaluatea)   cos-1 ½

  Solution.   The radian angle whose cosine is ½ is  

π

3

 (60°).

b)   cos-1 (−½)Solution.  An angle x whose cosine is negative falls in the 2nd quadrant.

And the cosine of a 2nd quadrant angle is the negative of the cosine of its supplement. This implies:

An angle θ whose cosine is −x  is the supplement

of the angle whose cosine is x.cos-1 (−x) = π – cos-1 x.

Therefore,cos-1 =π – cos-1

(−½) ½

  =

  =2 π  3

Problem 4.   Evaluate the following.

  a)  cos-1 1 = 

0   b)  cos-1 (−1) = 

π

  c) cos−1  2

 = π

4 

  d) cos−1(−  2

) = 

3π 4

  e)  cos-1 0 = 

π

2    f)   =

5π 6

The range of y = sec-1 x

In calculus, sin−1x,  tan−1x,  and cos−1x  are the most important inverse trigonometric functions.  Nevertheless, here are the ranges that make the rest single-valued.

The Inverse Sine Function (sin-1 x)

We define the inverse sine function as

y = sin-1 x for

where y is the angle whose sine is x. This means that

x = sin y

The graph of y = sin-1 x

Let's see the graph of y = sin x first and then derive the curve of y = sin-1 x.

As we did previously , if we reflect the indicated portion of y = sin x through the line y = x, we obtain the graph of y = sin-1 x:

Once again, what you see is what you get. The graph does not extend beyond the indicated boundaries of x and y.

The domain (the possible x-values) of sin-1 x is

-1 ≤ x ≤ 1

The range (of y-values for the graph) for sin-

1 x is

-π/2 ≤ sin-1 x ≤ π/2

The Inverse Tangent Function (tan-1x)

As a reminder, here is the graph of y = tan x, that we met before in Graphs of tan x , cot x , sec x and csc x.

Reflecting this portion of the graph in the line y = x, we obtain the graph of y = tan-1 x:

This time the graph does extend beyond what you see, in both the negative and positive directions of x.

The domain (the possible x-values) of tan-1 x is

All values of x

The range (of y-values for the graph) for arctan x is

-π/2 < tan-1 x < π/2

The graph of the inverse of cosine x is found by reflecting the graph of cos x through the line y = x.

We now reflect every point on this portion of the cos x curve through the line y = x.

The result is the graph y =cos-1 x:

That's it for the graph - it does not extend beyond what you see here. (If it did, there would be multiple values of y for each value of x and then we would no longer have a function.)

The domain (the possible x-values) of cos-1

x is

-1 ≤ x ≤ 1

The range (of y-values for the graph) for cos-1 x is

0 ≤ cos-1 x ≤ π

The Inverse Tangent Function (arctan)

As a reminder, here is the graph of y = tan x, that we met before in Graphs of tan, cot, sec and csc.

Reflecting this portion of the graph in the line y = x, we obtain the graph of y = tan-1 x :

This time the graph does extend beyond what you see, in both the negative and positive directions of x.

The domain (the possible x-values) of tan-1 x is

All values of x

The range (of y-values for the graph) for tan-1 x is

-π/2 < tan-1 x < π/2

The Inverse Tangent Function

Let's begin by thinking about the graph of

. If we want to solve , we may draw

a horizontal line units above the axis and choose one of the points which lies on the intersection of the graph and the horizontal line.

From this demonstration, you can see that, as you vary the horizontal line, there are always lots of solutions. However, there is

always a unique solution between and . For this reason, we have a well-defined function if define the inverse tangent function by saying

if is the value between and

such that .

This is similar to the square root function: there are two values which satisfy but

we agree, by convention, that the square root of 4 is the positive value.

Here are some famous values of the inverse tangent function:

-1

0 0

1

In fact, we can sketch the graph of the inverse tangent as below

Other inverse trigonometric functions

In the same way, we can build up the inverse sine and cosine functions:

if is the value between and

such that .

We can understand the graph of the inverse sine function

Also,

if is the value between and

such that .

Notice that the domain of both of these functions is restricted: if , there is no angle such that . This means that

we require in both of these functions.

You should verify for yourself the following relationships:

Let's first recall the graph of y = cos x , so we can see where the graph of y = cos-1 x comes from.

We now choose the portion of this graph from x = 0 to x = π.

The result is the graph y = cos-1 x:

That's it for the graph - it does not extend beyond what you see here. (If it did, there would be multiple values of y for each value of x and then we would no longer have a function.)

The domain (the possible x-values) of cos-1 x is

-1 ≤ x ≤ 1

The range (of y-values for the graph) for cos-1 x is

0 ≤ cos-1 x ≤ π

The Inverse Secant Function (sec-1x)

The graph of y = sec x, that we met before in Graphs of tan x , cot x , sec x and cscx:

The graph of y = sec-1x:

The domain of sec-1x is

All values of x, except -1 < x < 1

The range of sec-1x is

0 ≤ arcsec x ≤ π, sec-1x ≠ π/2

The Inverse Cosecant Function (arccsc)

The graph of y = csc x, that we met before in Graphs of tan x , cot x , sec x and cscx:

The graph of y = csc-1 x:

The graph extends in the negative and positive x-directions.

The domain of csc-1 x is

All values of x, except -1 < x < 1

The range of csc-1 x is

-π/2 ≤ csc-1 x ≤ π/2, csc-1 x ≠ 0

The Inverse Cotangent Function (cot-

1)

The graph of y = cot x, that we met before in Graphs of tan x , cot x , sec x and cscx is as follows:

 

Taking the highlighted portion as above, and reflecting it in the line y = x, we have the graph of y = cot-1 x:

The graph extends in the negative and positive x-directions.

So the domain of cot-1 x is:

All values of x

The range of cot-1 x is

−π/2 < cot-1 x ≤ π/2

For suitable values of x and y

sin-1 x + sin-1 y= sin-1 (x√1-y2+ y√1-x2)

sin-1 x - sin-1 y=sin-1 (x√1-y2- y√1-x2)

cos-1 x + cos-1 y= cos-1 (xy- √1-x2√1-y2)

cos-1 x - cos-1 y= cos-1 (xy+√1-x2√1-y2)

tan-1 x + tan-1y = tan−1 X+Y1−XY         ; xy<1

tan-1 x – tan-1 y= tan−1 X−Y1+XY ; xy>-1

2tan-1 x= tan−1 2 x1−x ² = sin−1 2 x

1+ x ² = cos−1 1−x ²1+x ²

Trigonometry examples

Example 1:

Solve the following equation:

Suggested answer:

Example 2:

Prove the following equation:

Suggested answer:

The angles in theoretical work will be in radian measure. Thus if

  we are given a radian angle, 

π

6

 for example, then we can evaluate a

function of it.

Problems – Solve Inverse Trigonometric Functions Problems:

Problem 1:

Prove that tan−1 x + tan−1 2 x1−x ² = tan−1 3 x−x ³

1−3 x ² , x<1

√ 3

Solution:

Let x = tan θ , then θ = tan-1 x. we have

You will take R.H.S to prove the given expression

R.H.S =     tan−1 3 x−x ³1−3 x ²

           = tan-1 (tan 3θ)            = 3 θ   = 3 tan-1 x

           = tan-1 x + 2 tan-1 x

           =   tan−1 x + tan−1 2 x1−x ²          = L.H.S

Hence, the given expression will be proved.

OR

We can take L.H.S. = tan−1 x + tan−1 2 x1−x ² = tan−1(

x+ 2x1−x ²

1−[ 2 x2

1−x 2])

By using

tan-1 x + tan-1y = tan−1 X+Y1−XY         =R.H.S.

Example problem 2:

Solve tan-1 2x + tan-1 3x = π/4

Solution:

Given: tan-1 2x + tan-1 3x = π/4

Or

 tan-1 ((2x + 3x)/(1 - 2x . 3x)) = π/4

tan-1 ((5x)/(1 - 6x2)) = π/4

∴ (5x)/(1 - 6x2) = tan π/4 = 1

or

6x2 + 5x – 1 = 0

That means, (6x – 1)(x + 1) = 0

Which gives

             x = 1/6 or x = -1

since x = -1 does not satisfy the equation ,the equation of the L.H.S is negative, so  x = 1/6 is the only solution of the given equation.

Practice Question – Solve Inverse Trigonometric Functions Problems:

Practice problem 1: Prove the given expresssion 3sin-1 x = sin-1 (3x – 4x3 ), x ∈ (-1/2,1/2 )

Practice problem 2: Prove the given expression 3 cos-1 x = cos-1 (4x3 – 3x), x ∈ (1/2, 1)

Practice problem 3: Find the value of  tan-1 [ 2 cos(2 sin-1

(½) )] [answer is п/4]

ASSIGNMENT:

Question.1 Evaluate: (i) sin-1(sin10) (ii) cos-1 (cos10) (iii) tan-1(tan(-6))

Answer: (i) if –π/2 ≤x ≤π/2, then sin-1(sinx)=x but x= 10 radians does not lie between –π/2 and π/2

3π – 10 lies between –π/2 and π/2 ∴ sin-

1(sin(3π-10)) = 3π-10. Similarly for (ii) cos-1 (cos10) = cos-1 (cos(4π-10)) = 4π-10. [10 radians does not lie between 0 and π. ∴ 0≤4π-10≤π]

For (iii) tan-1(tan(-6)) = tan-1(tan(2π-6)) = 2π-6 . { -6 radians does not lie in [ –π/2 , π/2]}

Question.2 If x = cos-1(cos4) and y = sin-1(sin3), then which holds? (give reason)

(i) x=y=1 (ii) x+y+1=0 (iii) x+2y=2 (iv) tan(x+y) = -tan7.

Question.3 if cos−1 xa + cos−1 y

b = θ, then prove that x ²a ² -

2 xyab cosθ

+ y ²a ² = sin2θ

[Hint: cos−1 xa + cos−1 y

b = cos−1 ¿ - √1− x ²a ² √1− y ²

b ² ] = θ ⇨ ¿cosθ)2 = ¿)2

Simplify it]

Question.4 (i) sin-1x + sin-1y + sin-1z = π, then prove that

X4+y4+z4+4x2y2z2 = 2(x2y2+y2z2+z2x2)

(ii) If tan−1 x + tan−1 y + tan−1 z = π/2 ; prove that xy+yz+xz = 1.

(iii) If tan−1 x + tan−1 y + tan−1 z = π , prove that x+y+z = xyz.

[Hint: for (i) sin-1x + sin-1y = π - sin-1z ⇨ cos(sin-1x + sin-1y) =cos( π - sin-1z)

Use cos(A-B) = cosAcosB – sinAsinB and cos(π – 𝛂)= -cos𝛂

It becomes √(1−x ²)(1− y ²) - xy = - √1−z ² and simply it.

[Hint: for (ii) tan-1 x + tan-1y = tan−1 X+Y1−XY ]

Question.5 Write the following functions in the simplest form:

(i)tan−1 ¿ ) (ii) tan−1 ¿) (iii) tan−1 √ a−xa+ x

, -a<x<a

[Hint: for (i) write cosx = cos2x/2 – sin2x/2 and 1+sinx =(cosx/2 +sinx/2)2 , then use tan(A-B),

answer is π/4 – x/2 ]

[ Hint: for (ii) write cosx = sin(π/2 – x) and sinx = cos(π/2 – x), then

use formula of 1-cos(π/2 – x)= 2sin2(π/4 – x/2) and sin(π/2 – x) = 2 sin(π/4 – x/2) cos(π/4 – x/2)

Same method can be applied for (i) part also. Answer is π/4 + x/2]

[ for (iii) put x=a cos𝛂, then answer will be ½ cos−1 xa ]

Question.6 If y = cot−1¿¿) - tan−1(√cosx ), prove that siny = tan2(x/2).

[Hint: y = π2 - 2 tan−1 ¿¿ , use formula 2tan−1 x = cos−1 ¿)]

Question.7 (i) Prove that tan−11 +tan−1 2 + tan−13 = π.

(ii) Prove that cot−1¿) +cot−1¿) + cot−1¿) = 0.

[Hint: for (i) tan−1 2 = π2 - cot−12 =

π2−tan−1 1

2 , then use formula of

tan-1 x + tan-1y = tan−1 X+Y1−XY    ]  

[Hint: write cot−1 x  = tan−1 1x ]

Question.8 Solve the following equations:

(i) sin−1 x + sin−1(1−x) = cos−1 x .

(ii) tan−1 √x (x+1) + sin−1√ x ²+ x+1 = π2 .

[Hint: write cos−1 x = π2 - sin−1 x, put sin−1 x = y]

[Hint: use tan−1 x=¿ cos−1 1

√1+x ² ]

Question.9 Using principal values, evaluate cos−1 ¿¿) + sin−1¿). [answer is π]

Question.10 Show that tan(12

sin−1 34 ) =

4−√ 73 and justify why

the other value is ignored?

[ Hint: put 12

sin−1 34 =∅ ⇨ ¾ = sin2∅ = 2tan∅/(1+tan2∅), find tan∅]

** SOME HOT QUESTIONS:

1. Which is greater tan 1 or tan-11?

2. Find the value of sin(2tan−1 23) + cos( tan−1 √ 3¿¿ .

3. Find the value of x which satisfies the equation sin−1 x + sin−1(1−x) = cos−1 x .

4. Solve the equation: sin−16 x + sin−1¿) = -π/2.

5. Show that tan−1 ¿tanx2) =

12

cos−1¿) .

6. If 𝛂 = tan−1(2 tan ² α ) - 12

sin−1 ¿¿), then find the general value of 𝛂.ANSWERS WITH HINTS:1. Since 1> π/4 ⇨ tan1> 1> tan-11.

2. sin(2tan−1 23) + cos( tan−1 √ 3¿¿= sin2x + cosy ⇨

2 tanx1+ tan ² x +

1

√1+ tan ² y = 2.( 2

3)

1+(4/9)+

12 =

3726 .

3. sin ¿ + sin−1(1−x)¿ = sin(cos−1 x) by using sin(A+B)=sinA cosB + cosA sinB

⇨ x√1−(1−x ) ² + (1-x)√1−x ² = √1−x ² ∵ sin(cos−1 x) = √1−cos ²(¿cos−1 x )¿

⇨ 2x – x2 = 1 ⇨ x = 0 or ½.

4. sin−16 x = - π2 - sin−1¿¿) ⇨ 6x = sin[-

π2 - sin−1¿¿)]

= -cos[sin−1¿¿ ] = -cos[cos−1 √1−108 x ² ] = - √1−108 x ² etc.

5. 1/2(2tan−1 ¿tanx2 )) , use formula 2tan−1 A = cos−1 1−A ²

1+ A ² and

tan2x/2 = 1−cosx1+cosx .

6. Put tan𝛂 = t and use sin2𝛂 = 2 tanα

1+ tan ² α and cos2𝛂 = 1−tan ² α1+tan ² α

then put t/3 = T,answer is 𝛂 = nπ, nπ+π/4. tan−1 2t ² - tan−1 t = ½ sin−1 6 t

9+t ² ⇨ tan−1 2 t ²−t1+2 t ³ = ½ sin−1 2 t /3

1+(t /3)² = ½ sin−1 2 T1+(T ) ²

tan−1 2 t ²−t1+2 t ³ = (2T), then tan𝛂 =½ 0 ,1.

.