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Basic identities

A small set of identities lets you rewrite almost any inverse trigonometric expression. This subtopic catalogues the foundational identities , those that do not require addition formulas. They follow from the principal range conventions and from basic trigonometric identities like sin2θ+cos2θ=1\sin^2 \theta + \cos^2 \theta = 1.

Complementary identities

The most important identity in the chapter:

sin1x+cos1x=π2,x[1,1].\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}, \quad x \in [-1, 1].

Proof. Let θ=sin1x\theta = \sin^{-1} x, so sinθ=x\sin \theta = x and θ[π/2,π/2]\theta \in [-\pi/2, \pi/2]. Then cos(π/2θ)=sinθ=x\cos(\pi/2 - \theta) = \sin \theta = x, with π/2θ[0,π]\pi/2 - \theta \in [0, \pi]. Hence cos1x=π/2θ\cos^{-1} x = \pi/2 - \theta, and addition gives π/2\pi/2. \blacksquare

By the same argument:

tan1x+cot1x=π2,xR.\tan^{-1} x + \cot^{-1} x = \frac{\pi}{2}, \quad x \in \mathbb{R}.

sec1x+csc1x=π2,x1.\sec^{-1} x + \csc^{-1} x = \frac{\pi}{2}, \quad |x| \ge 1.

These three identities are used in nearly every problem of the chapter.

Reciprocal identities

For x1|x| \ge 1 we have csc1x=sin1(1/x)\csc^{-1} x = \sin^{-1}(1/x) and sec1x=cos1(1/x)\sec^{-1} x = \cos^{-1}(1/x).

Reason. If θ=csc1x\theta = \csc^{-1} x then cscθ=x\csc \theta = x, so sinθ=1/x\sin \theta = 1/x. Since θ[π/2,π/2]{0}\theta \in [-\pi/2, \pi/2] \setminus \{0\}, we have θ=sin1(1/x)\theta = \sin^{-1}(1/x).

For x>0x > 0 we also have cot1x=tan1(1/x)\cot^{-1} x = \tan^{-1}(1/x), but this fails for x<0x < 0 , there one must write cot1x=π+tan1(1/x)\cot^{-1} x = \pi + \tan^{-1}(1/x).

Negative argument identities

IdentityDomain
sin1(x)=sin1x\sin^{-1}(-x) = -\sin^{-1} xx[1,1]x \in [-1, 1]
cos1(x)=πcos1x\cos^{-1}(-x) = \pi - \cos^{-1} xx[1,1]x \in [-1, 1]
tan1(x)=tan1x\tan^{-1}(-x) = -\tan^{-1} xxRx \in \mathbb{R}
cot1(x)=πcot1x\cot^{-1}(-x) = \pi - \cot^{-1} xxRx \in \mathbb{R}
sec1(x)=πsec1x\sec^{-1}(-x) = \pi - \sec^{-1} x$
csc1(x)=csc1x\csc^{-1}(-x) = -\csc^{-1} x$

Cross identities

Often we need to convert one inverse function into another. For x[0,1]x \in [0, 1]:

sin1x=cos11x2=tan1x1x2.\sin^{-1} x = \cos^{-1} \sqrt{1 - x^2} = \tan^{-1} \frac{x}{\sqrt{1 - x^2}}.

Reason. If sinθ=x\sin \theta = x with θ[0,π/2]\theta \in [0, \pi/2], then cosθ=1x2\cos \theta = \sqrt{1 - x^2}, hence cos1(1x2)=θ\cos^{-1}(\sqrt{1 - x^2}) = \theta. And tanθ=x/1x2\tan \theta = x/\sqrt{1 - x^2}.

For x0x \ge 0, tan1x=sin1(x/1+x2)=cos1(1/1+x2)\tan^{-1} x = \sin^{-1}(x/\sqrt{1 + x^2}) = \cos^{-1}(1/\sqrt{1 + x^2}).

These conversions are essential for combining inverse trig terms into a single one.

Worked examples

Example 1. Simplify sin1x+cos1x\sin^{-1} x + \cos^{-1} x for x=0.3x = 0.3.

By the identity, the sum is π/2\pi/2, regardless of the value of xx (as long as x1|x| \le 1).

Example 2. Simplify sin1(35)+cos1(35)\sin^{-1}(\tfrac{3}{5}) + \cos^{-1}(\tfrac{3}{5}).

Same identity: answer π/2\pi/2.

Example 3. Compute tan1(13)+cot1(13)\tan^{-1}(\tfrac{1}{3}) + \cot^{-1}(\tfrac{1}{3}).

Answer π/2\pi/2.

Example 4. Show sin1(sin2π3)=π/3\sin^{-1}(\sin \tfrac{2\pi}{3}) = \pi/3.

sin(2π/3)=sin(π2π/3)=sin(π/3)=3/2\sin(2\pi/3) = \sin(\pi - 2\pi/3) = \sin(\pi/3) = \sqrt 3/2. The principal value answer is sin1(3/2)=π/3\sin^{-1}(\sqrt 3/2) = \pi/3.

Example 5. Express sin1(35)\sin^{-1}(\tfrac{3}{5}) in terms of cos1\cos^{-1}.

If sinθ=3/5\sin \theta = 3/5 with θ[0,π/2]\theta \in [0, \pi/2], then cosθ=4/5\cos \theta = 4/5. So sin1(3/5)=cos1(4/5)\sin^{-1}(3/5) = \cos^{-1}(4/5).

Example 6. Show cos1(1/2)=πcos1(1/2)=ππ/3=2π/3\cos^{-1}(-1/2) = \pi - \cos^{-1}(1/2) = \pi - \pi/3 = 2\pi/3.

Direct: cos(2π/3)=1/2\cos(2\pi/3) = -1/2 and 2π/3[0,π]2\pi/3 \in [0, \pi]. So the answer is 2π/32\pi/3.

Try it yourself

  1. Evaluate sin1(2/2)+cos1(2/2)\sin^{-1}(\sqrt 2/2) + \cos^{-1}(\sqrt 2/2).
  2. Evaluate tan1(3)+cot1(3)\tan^{-1}(\sqrt 3) + \cot^{-1}(\sqrt 3).
  3. Show sec1(2)+csc1(2)=π/2\sec^{-1}(2) + \csc^{-1}(2) = \pi/2.
  4. Express sin1(4/5)\sin^{-1}(4/5) as cos1\cos^{-1} of something.
  5. Express sin1(5/13)\sin^{-1}(5/13) as tan1\tan^{-1} of something.
  6. Simplify sin1(1/2)+cos1(1/2)\sin^{-1}(-1/2) + \cos^{-1}(-1/2).
  7. Evaluate tan1(1)+cot1(1)\tan^{-1}(-1) + \cot^{-1}(-1) , careful with branches.
  8. Show cos1(x)=πcos1x\cos^{-1}(-x) = \pi - \cos^{-1} x by direct verification at x=1/2x = 1/2.
  9. Show sin1(3/5)+cos1(3/5)=π/22sin1(3/5)\sin^{-1}(-3/5) + \cos^{-1}(3/5) = \pi/2 - 2 \sin^{-1}(3/5).
  10. Find the value of sin1(1)+cos1(1)\sin^{-1}(1) + \cos^{-1}(1).
  11. Evaluate tan1(2)+cot1(2)\tan^{-1}(2) + \cot^{-1}(2).
  12. Verify the identity sec1x=cos1(1/x)\sec^{-1} x = \cos^{-1}(1/x) at x=2x = 2.
  13. For x>0x > 0 show cot1x=tan1(1/x)\cot^{-1} x = \tan^{-1}(1/x).
  14. For x<0x < 0 find a correct identity replacing cot1x=tan1(1/x)\cot^{-1} x = \tan^{-1}(1/x).

Pitfalls / Tricks

  • The identity sin1x+cos1x=π/2\sin^{-1} x + \cos^{-1} x = \pi/2 holds for all x[1,1]x \in [-1, 1], including the boundary values.
  • cot1x=tan1(1/x)\cot^{-1} x = \tan^{-1}(1/x) only for x>0x > 0; for x<0x < 0 add π\pi.
  • Always restrict xx to the natural domain before applying any identity. Asking for sin1(2)\sin^{-1}(2) is meaningless in real analysis.
  • For symbolic manipulations, sin1(sinθ)=θ\sin^{-1}(\sin \theta) = \theta only when θ[π/2,π/2]\theta \in [-\pi/2, \pi/2].
  • The identity cos1(x)=πcos1x\cos^{-1}(-x) = \pi - \cos^{-1} x is asymmetric (compare to sin\sin): the cosine inverse is not an odd function.

Next we tackle the additive identities for sums of two inverse trig terms.

Practice quiz

Quick check on this topic.

Quiz
Quick check : Basic identities
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