2.3 Properties of Inverse Trigonometric Functions

Screen Readable NCERT Class XII Mathematics textbook for blind and low vision students prepared by Professor T K Bansal.

In this section, we shall prove some important properties of inverse trigonometric functions. It may be mentioned here that these results are valid within the principal value branches of the corresponding inverse trigonometric functions and wherever they are defined. Some results may not be valid for all values of the domains of inverse trigonometric functions. In fact, they will be valid only for some values of x for which inverse trigonometric functions are defined. We will not go into the details of these values of x in the domain as this discussion goes beyond the scope of this textbook.

Let us recall that,


\[y\ =\ \sin^{−1} x\ \implies\ x\ =\ \sin y,\]


and


\[x\ =\ \sin y,\ \implies\ y\ =\ \sin^{−1} x.\]

This is equivalent to

\[\sin (\sin^{−1} (x))\ =\ x,\ x\ ∈\ [−1,\ 1]\]


and


\[\sin^{−1} (\sin x)\ =\ x,\ x\ ∈\ [\frac{−π}{2},\ \frac{π}{2}]\]

Example 3

Show that

\[i.\ \sin^{−1}(2x\sqrt{1−x^2})\ =\ 2 \sin^{−1} x,\ \frac{−1}{√2}\ ≤\ x\ ≤\ \frac{1}{√2}\]

\[ii.\ \sin^{−1}(2x\sqrt{1−x^2})\ =\ 2 \cos^{−1}x,\ \frac{1}{√2}\ ≤\ x\ ≤\ 1\]

Solution:

(i) Let x = sin θ.


Then sin^−1 (x) = θ.


We have


sin^−1 (2x √(1 − x^2)


= sin^−1 (2 sin θ √(1 − sin^2 θ))


= sin^−1 (2sin θ cos θ)


= sin^−1 (sin 2θ) = 2 θ


= 2 sin^−1 (x)

(ii) Take x = cos θ, then proceeding as above, we get, sin^−1(2x √(1 − x^2))


= 2cos^−1 x

Example 4

Express


\[\tan^{−1} (\cos \frac{x}{1− \sin x}),\ \frac{−3π}{2}\ <\ x\ <\ \frac{π}{2}\]


in the simplest form.

Solution:

We write tan^−1 (cos x/(1 – sinx))


= tan^−1[(cos^2 (x/2) − sin^2( x/2))/(cos^2 (x/2) + sin^2 (x/2) − 2sin (x/2) cos (x/2))]


= tan^−1 [{(cos (x/2) + sin (x/2))(cos (x/2) − sin (x/2))}/{(cos (x/2) – sin(x/2))^2}]


= tan^−1 [(cos (x/2) + sin (x/2))/(cos (x/2) − sin (x/2))]


= tan^−1 [(1 + tan (x/2))/(1 − tan (x/2))]


= tan^−1[tan((π/4) + (x/2))]


= (π/4) + (x/2)

Alternatively,

tan^−1(cosx/(1 – sinx)) = tan^−1 [sin((π/2) − x)/1 − cos((π/2) − x)]


= tan^−1[sin((π − 2x)/2)/1 − cos((π − 2x)/2)]


= tan^−1 [{2sin ((π − 2x)/4) cos((π − 2x)/4)}/{2sin^2((π − 2x)/4)}]


= tan^−1 [cot((π − 2x)/4)] = tan^−1 [tan((π/2) − (π − 2x)/4)]


= tan^−1 [tan (π/4 + x/2)] = π/4 + x/2

Example 5

Write


\[\cot^{−1} \frac{1}{\sqrt{x^2−1}},\ x\ >\ 1\]


in the simplest form.

Solution:

Let x = sec θ, then √(x^2 − 1 )


= √(sec^2 θ – 1)


= tan θ.

Therefore, cot^−1 (1/√(x^2 – 1))


= cot^−1(cot θ) = θ


= sec^−1 (x), which is the simplest form.

EXERCISE 2.2

Prove the following:

Question 1


\[3 \sin^{−1} x\ =\ \sin^{−1} 3x\ −\ 4x^3, x\ ∈\ [\frac{−1}{2},\ \frac{1}{2}]\]

Question 2


\[3 \cos^{−1} x\ =\ \cos^{−1} (4x^3\ −\ 3x), x ∈ [1/2, 1]\]

Write the following functions in the simplest form:

Question 3


\[\tan^{−1} \sqrt{(1+x^2
}\ −\ \frac{1}{x},\ x ≠ 0\]

Question 4


\[\tan^{−1}( \sqrt{\frac{1−\cos x}{1+\cos x})},\ 0 < x < π\]

Question 5


\[\tan^{−1}( \frac{\cos x − \sin x}{\cos x + \sin x}),\ −π/4\ <\ x\ <\ 3π/4\]

Question 6


\[\tan^{−1} \frac{x}{\sqrt{a^2 − x^2}},\ |x| < a\]

Question 7


\[\tan^{−1}( \frac{3a^{2x} − x^3}{a^3 − 3ax^2}),\ a > 0;\ −a/√3\ <\ x\ <\ a/√3\]

Find the values of each of the following:

Question 8


\[\tan^{−1} [2 \cos (2\sin^{−1} \frac{1}{2})]\]

Question 9


\[\tan \frac{1}{2} [\sin^{−1} \frac{2x}{1+x^2}\ +\ \cos^{−1} \frac{1−y^2}{1+y^2}]\, |x|<1, y>0,\ xy<1\]

Find the values of each of the expressions in Exercises 16 to 18.

Question 10


\[\sin^{−1} (\sin \frac{2π}{3})\]

Question 11


\[\tan^{−1} (\tan \frac{3π}{4})\]

Question 12


\[\tan (\sin^{−1} \frac{3}{5}\ +\ \cot^{−1} \frac{3}{2})\]

Question 13


\[\cos^{−1} (\cos \frac{7π}{6}) =\]


(A) 7π/6


(B) 5π/6


(C) π/3


(D) π/6

Question 14


\[\sin (\frac{π}{3}\ −\ \sin^{−1}(\frac{−1}{2}))\ =\ \]


(A) 1/2


(B) 1/3


(C) 1/4


(D) 1

Question 15


\[\tan^{−1}√3\ −\ \cot^{−1} −√3\ =\]


(A) π


(B) −π/2


(C) 0


(D) 2√3