What you will be able to do use sin − 1 ( − x ) = − sin − 1 x \sin^{-1}(-x) = -\sin^{-1}x sin − 1 ( − x ) = − sin − 1 x and cos − 1 ( − x ) = π − cos − 1 x \cos^{-1}(-x) = \pi - \cos^{-1}x cos − 1 ( − x ) = π − cos − 1 x , and know which function follows which rule; prove and use sin − 1 x + cos − 1 x = π 2 \sin^{-1}x + \cos^{-1}x = \dfrac{\pi}{2} sin − 1 x + cos − 1 x = 2 π and its partners; change sec − 1 \sec^{-1} sec − 1 , cosec − 1 \operatorname{cosec}^{-1} cosec − 1 and cot − 1 \cot^{-1} cot − 1 into the three inverse functions that a calculator has. Negative numbers The sine is an odd function: sin ( − θ ) = − sin θ \sin(-\theta) = -\sin\theta sin ( − θ ) = − sin θ . The branch [ − π 2 , π 2 ] \left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right] [ − 2 π , 2 π ] is symmetric about 0 0 0 . So if θ = sin − 1 x \theta = \sin^{-1}x θ = sin − 1 x then − θ -\theta − θ is in the branch and has the sine − x -x − x . This proves
sin − 1 ( − x ) = − sin − 1 x , tan − 1 ( − x ) = − tan − 1 x , cosec − 1 ( − x ) = − cosec − 1 x . \displaystyle \sin^{-1}(-x) = -\sin^{-1}x, \qquad \tan^{-1}(-x) = -\tan^{-1}x, \qquad \operatorname{cosec}^{-1}(-x) = -\operatorname{cosec}^{-1}x. sin − 1 ( − x ) = − sin − 1 x , tan − 1 ( − x ) = − tan − 1 x , cosec − 1 ( − x ) = − cosec − 1 x . The cosine is even, and its branch [ 0 , π ] [0, \pi] [ 0 , π ] is not symmetric about 0 0 0 . Let θ = cos − 1 x \theta = \cos^{-1}x θ = cos − 1 x . Then cos ( π − θ ) = − cos θ = − x \cos(\pi - \theta) = -\cos\theta = -x cos ( π − θ ) = − cos θ = − x , and π − θ \pi - \theta π − θ is in [ 0 , π ] [0, \pi] [ 0 , π ] . So
cos − 1 ( − x ) = π − cos − 1 x , cot − 1 ( − x ) = π − cot − 1 x , sec − 1 ( − x ) = π − sec − 1 x . \displaystyle \cos^{-1}(-x) = \pi - \cos^{-1}x, \qquad \cot^{-1}(-x) = \pi - \cot^{-1}x, \qquad \sec^{-1}(-x) = \pi - \sec^{-1}x. cos − 1 ( − x ) = π − cos − 1 x , cot − 1 ( − x ) = π − cot − 1 x , sec − 1 ( − x ) = π − sec − 1 x . You saw the first rule on the graph: a half turn about the point ( 0 , π 2 ) \left(0, \dfrac{\pi}{2}\right) ( 0 , 2 π ) leaves the graph of cos − 1 \cos^{-1} cos − 1 unchanged.
Partners that add up to a right angle For every x x x in [ − 1 , 1 ] [-1, 1] [ − 1 , 1 ] ,
sin − 1 x + cos − 1 x = π 2 . \displaystyle \sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}. sin − 1 x + cos − 1 x = 2 π . Proof. Let θ = sin − 1 x \theta = \sin^{-1}x θ = sin − 1 x , so θ \theta θ is in [ − π 2 , π 2 ] \left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right] [ − 2 π , 2 π ] . Then π 2 − θ \dfrac{\pi}{2} - \theta 2 π − θ is in [ 0 , π ] [0, \pi] [ 0 , π ] , and cos ( π 2 − θ ) = sin θ = x \cos\left(\dfrac{\pi}{2} - \theta\right) = \sin\theta = x cos ( 2 π − θ ) = sin θ = x . An angle in [ 0 , π ] [0, \pi] [ 0 , π ] whose cosine is x x x is cos − 1 x \cos^{-1}x cos − 1 x . So cos − 1 x = π 2 − θ \cos^{-1}x = \dfrac{\pi}{2} - \theta cos − 1 x = 2 π − θ .
A right triangle with a small square at the right angle. The hypotenuse is labelled 1, the vertical side is labelled x, and the horizontal side is labelled the square root of 1 minus x squared. The angle at the lower left corner is marked in orange as theta = sin inverse x. The angle at the top corner is marked in green as cos inverse x = pi over 2 minus theta. The title says that the two acute angles add up to pi over 2. The same idea gives two more:
tan − 1 x + cot − 1 x = π 2 ( every real x ) , sec − 1 x + cosec − 1 x = π 2 ( ∣ x ∣ ≥ 1 ) . \displaystyle \tan^{-1}x + \cot^{-1}x = \frac{\pi}{2} \ \ (\text{every real } x), \qquad \sec^{-1}x + \operatorname{cosec}^{-1}x = \frac{\pi}{2} \ \ (|x| \ge 1). tan − 1 x + cot − 1 x = 2 π ( every real x ) , sec − 1 x + cosec − 1 x = 2 π ( ∣ x ∣ ≥ 1 ) .
Reciprocal rules For ∣ x ∣ ≥ 1 |x| \ge 1 ∣ x ∣ ≥ 1 the number 1 x \dfrac{1}{x} x 1 is in [ − 1 , 1 ] [-1, 1] [ − 1 , 1 ] , and we saw that
cosec − 1 x = sin − 1 1 x , sec − 1 x = cos − 1 1 x . \displaystyle \operatorname{cosec}^{-1}x = \sin^{-1}\frac{1}{x}, \qquad \sec^{-1}x = \cos^{-1}\frac{1}{x}. cosec − 1 x = sin − 1 x 1 , sec − 1 x = cos − 1 x 1 . The tangent needs a little care, because the branches of tan − 1 \tan^{-1} tan − 1 and cot − 1 \cot^{-1} cot − 1 are different:
tan − 1 1 x = { cot − 1 x if x > 0 , cot − 1 x − π if x < 0. \displaystyle \tan^{-1}\frac{1}{x} = \begin{cases} \cot^{-1}x & \text{if } x > 0, \\ \cot^{-1}x - \pi & \text{if } x < 0. \end{cases} tan − 1 x 1 = { cot − 1 x cot − 1 x − π if x > 0 , if x < 0. For x = − 1 x = -1 x = − 1 you can check this: tan − 1 ( − 1 ) = − π 4 \tan^{-1}(-1) = -\dfrac{\pi}{4} tan − 1 ( − 1 ) = − 4 π , while cot − 1 ( − 1 ) = 3 π 4 \cot^{-1}(-1) = \dfrac{3\pi}{4} cot − 1 ( − 1 ) = 4 3 π , and 3 π 4 − π = − π 4 \dfrac{3\pi}{4} - \pi = -\dfrac{\pi}{4} 4 3 π − π = − 4 π .
A worked example Find (a) cos − 1 ( − 3 5 ) + cos − 1 3 5 \cos^{-1}\left(-\dfrac{3}{5}\right) + \cos^{-1}\dfrac{3}{5} cos − 1 ( − 5 3 ) + cos − 1 5 3 , (b) tan − 1 ( − 2 ) + cot − 1 ( − 2 ) \tan^{-1}(-2) + \cot^{-1}(-2) tan − 1 ( − 2 ) + cot − 1 ( − 2 ) , (c) sin − 1 1 3 − cos − 1 ( − 1 3 ) \sin^{-1}\dfrac{1}{3} - \cos^{-1}\left(-\dfrac{1}{3}\right) sin − 1 3 1 − cos − 1 ( − 3 1 ) .
(a) cos − 1 ( − 3 5 ) = π − cos − 1 3 5 \cos^{-1}\left(-\dfrac{3}{5}\right) = \pi - \cos^{-1}\dfrac{3}{5} cos − 1 ( − 5 3 ) = π − cos − 1 5 3 . So the sum is π \pi π , whatever the value of cos − 1 3 5 \cos^{-1}\dfrac{3}{5} cos − 1 5 3 . (b) The two are partners, and the rule holds for every real number. The sum is π 2 \dfrac{\pi}{2} 2 π . (c) Write cos − 1 ( − 1 3 ) = π − cos − 1 1 3 = π − ( π 2 − sin − 1 1 3 ) = π 2 + sin − 1 1 3 \cos^{-1}\left(-\dfrac{1}{3}\right) = \pi - \cos^{-1}\dfrac{1}{3} = \pi - \left(\dfrac{\pi}{2} - \sin^{-1}\dfrac{1}{3}\right) = \dfrac{\pi}{2} + \sin^{-1}\dfrac{1}{3} cos − 1 ( − 3 1 ) = π − cos − 1 3 1 = π − ( 2 π − sin − 1 3 1 ) = 2 π + sin − 1 3 1 . So the difference is − π 2 -\dfrac{\pi}{2} − 2 π . Practise with new numbers
Try it in Python The program checks the reflection rule and the partner rules at several numbers, including negative ones.
Run it. Which two columns are equal in each row, and what is the first sum?
1. Add -1 and 1 to the first list. Do the rules still hold at the ends of the domain?
2. Add 0 to the second list. What is cot − 1 0 \cot^{-1}0 cot − 1 0 ?
Common mistakes Using sin − 1 ( − x ) = π − sin − 1 x \sin^{-1}(-x) = \pi - \sin^{-1}x sin − 1 ( − x ) = π − sin − 1 x . That rule belongs to the cosine. The inverse sine is odd. Using cos − 1 ( − x ) = − cos − 1 x \cos^{-1}(-x) = -\cos^{-1}x cos − 1 ( − x ) = − cos − 1 x . That would be negative, and the range of cos − 1 \cos^{-1} cos − 1 is [ 0 , π ] [0, \pi] [ 0 , π ] . Writing tan − 1 1 x = cot − 1 x \tan^{-1}\dfrac{1}{x} = \cot^{-1}x tan − 1 x 1 = cot − 1 x for every x x x . It holds for x > 0 x > 0 x > 0 only. Using the partner rule outside the domain. sin − 1 x + cos − 1 x = π 2 \sin^{-1}x + \cos^{-1}x = \dfrac{\pi}{2} sin − 1 x + cos − 1 x = 2 π needs ∣ x ∣ ≤ 1 |x| \le 1 ∣ x ∣ ≤ 1 . Remember these sin − 1 \sin^{-1} sin − 1 , tan − 1 \tan^{-1} tan − 1 and cosec − 1 \operatorname{cosec}^{-1} cosec − 1 are odd. For cos − 1 \cos^{-1} cos − 1 , cot − 1 \cot^{-1} cot − 1 and sec − 1 \sec^{-1} sec − 1 , f ( − x ) = π − f ( x ) f(-x) = \pi - f(x) f ( − x ) = π − f ( x ) . sin − 1 x + cos − 1 x = π 2 \sin^{-1}x + \cos^{-1}x = \dfrac{\pi}{2} sin − 1 x + cos − 1 x = 2 π , tan − 1 x + cot − 1 x = π 2 \tan^{-1}x + \cot^{-1}x = \dfrac{\pi}{2} tan − 1 x + cot − 1 x = 2 π , sec − 1 x + cosec − 1 x = π 2 \sec^{-1}x + \operatorname{cosec}^{-1}x = \dfrac{\pi}{2} sec − 1 x + cosec − 1 x = 2 π . sec − 1 x = cos − 1 1 x \sec^{-1}x = \cos^{-1}\dfrac{1}{x} sec − 1 x = cos − 1 x 1 and cosec − 1 x = sin − 1 1 x \operatorname{cosec}^{-1}x = \sin^{-1}\dfrac{1}{x} cosec − 1 x = sin − 1 x 1 for ∣ x ∣ ≥ 1 |x| \ge 1 ∣ x ∣ ≥ 1 . tan − 1 1 x = cot − 1 x \tan^{-1}\dfrac{1}{x} = \cot^{-1}x tan − 1 x 1 = cot − 1 x for x > 0 x > 0 x > 0 , and cot − 1 x − π \cot^{-1}x - \pi cot − 1 x − π for x < 0 x < 0 x < 0 . Check yourself Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer .
Write sin − 1 ( − 1 2 ) \sin^{-1}\left(-\dfrac{1}{2}\right) sin − 1 ( − 2 1 ) in terms of sin − 1 1 2 \sin^{-1}\dfrac{1}{2} sin − 1 2 1 , and find its value. Find cos − 1 ( − 1 3 ) + cos − 1 1 3 \cos^{-1}\left(-\dfrac{1}{3}\right) + \cos^{-1}\dfrac{1}{3} cos − 1 ( − 3 1 ) + cos − 1 3 1 . If sin − 1 x = π 5 \sin^{-1}x = \dfrac{\pi}{5} sin − 1 x = 5 π , find cos − 1 x \cos^{-1}x cos − 1 x . Find tan − 1 5 + cot − 1 5 \tan^{-1}5 + \cot^{-1}5 tan − 1 5 + cot − 1 5 . Is tan − 1 1 x = cot − 1 x \tan^{-1}\dfrac{1}{x} = \cot^{-1}x tan − 1 x 1 = cot − 1 x true for x = − 1 x = -1 x = − 1 ? Check with the values. References NCERT, Mathematics Textbook for Class XII, Part I, Chapter 2: Inverse Trigonometric Functions. OpenStax, Algebra and Trigonometry 2e, Section 8.3: Inverse Trigonometric Functions. OpenStax, Calculus Volume 1, Section 1.4: Inverse Functions.