Inverse Trigonometric Functions — JEE Mains Mathematics

Formulas (3)

• Domain conditions on the argument · Domains
  $$\sin^{-1}u,\ \cos^{-1}u:\ |u|\le1,\qquad \sec^{-1}u,\ \csc^{-1}u:\ |u|\ge1$$
• Ranges and complementary pairs · Complementary pairs
  $$\sin^{-1}x+\cos^{-1}x=\tan^{-1}x+\cot^{-1}x=\frac{\pi}{2}$$
• Inverse of a trigonometric value outside the principal range · Back into the principal range
  $$\sin^{-1}(\sin x)=\pi-x\ \ \left(\tfrac{\pi}{2}\le x\le\tfrac{3\pi}{2}\right),\qquad \cos^{-1}(\cos x)=2\pi-x\ \ (\pi\le x\le2\pi)$$

Watch out for (3)

• Do not multiply by a denominator of unknown sign
  → Domain conditions on the argument
• Check whether each endpoint is reached
  → Ranges and complementary pairs
• sin⁻¹(sin x) is not always x
  → Inverse of a trigonometric value outside the principal range

Formulas (2)

• Reading ratios off a right triangle · Sum formula
  $$\sin(A+B)=\sin A\cos B+\cos A\sin B$$
• Double, half and triple angles of an inverse · Double angle in terms of the tangent
  $$\tan2A=\frac{2t}{1-t^2},\quad \sin2A=\frac{2t}{1+t^2},\quad \cos2A=\frac{1-t^2}{1+t^2}$$

Watch out for (2)

• A triangle has no signs
  → Reading ratios off a right triangle
• Pick the half angle that fits the range
  → Double, half and triple angles of an inverse

Formulas (3)

• Substitute and track the interval · Standard substitutions
  $$2\tan^{-1}x=\sin^{-1}\frac{2x}{1+x^2}\ \ (|x|\le1),\qquad 2\tan^{-1}x=\cos^{-1}\frac{1-x^2}{1+x^2}\ \ (x\ge0)$$
• Turning a composition into algebra · Composition to algebra
  $$\sin(\tan^{-1}x)=\frac{x}{\sqrt{1+x^2}},\qquad \cos(\sin^{-1}x)=\sqrt{1-x^2}$$
• Complementary identities with a condition · Complementary identity
  $$\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2},\quad -1\le x\le1$$

Watch out for (3)

• The interval for x decides the formula
  → Substitute and track the interval
• Squaring brings extra roots
  → Turning a composition into algebra
• The identity needs the same argument
  → Complementary identities with a condition

Formulas (2)

• The addition formula for inverse tangents · Addition formula
  $$\tan^{-1}a+\tan^{-1}b=\tan^{-1}\frac{a+b}{1-ab}\quad(ab<1)$$
• Telescoping sums of inverse tangents · Telescoping term
  $$\tan^{-1}\frac{b-a}{1+ab}=\tan^{-1}b-\tan^{-1}a$$

Watch out for (2)

• When the product exceeds 1
  → The addition formula for inverse tangents
• Keep the order of the difference
  → Telescoping sums of inverse tangents

Formulas (3)

• Equations in inverse tangents · Taking tangents
  $$\tan^{-1}a+\tan^{-1}b=c\ \Rightarrow\ \frac{a+b}{1-ab}=\tan c$$
• Equations in inverse sines and cosines · Isolate, then take the sine
  $$\sin^{-1}u=\theta\ \Rightarrow\ u=\sin\theta,\quad -\tfrac{\pi}{2}\le\theta\le\tfrac{\pi}{2}$$
• When the domain pins x down · Squeezed argument
  $$u\ge1\ \text{and}\ u\le1\ \Rightarrow\ u=1$$

Watch out for (3)

• Taking tangents loses the range
  → Equations in inverse tangents
• Squaring adds roots
  → Equations in inverse sines and cosines
• Find the domain first
  → When the domain pins x down