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Chapter 2 · Inverse Trigonometric Functions

Miscellaneous · Q13 (MCQ): sin⁡(tan⁡−1x)\sin(\tan^{-1}x) for ∣x∣<1|x|<1

Miscellaneous ExerciseThe triangle trick · Miscellaneous Exercise
The question

sin⁡(tan⁡−1x)\sin(\tan^{-1}x)⁠, ∣x∣<1|x| < 1 is equal to

  1. x1−x2\dfrac{x}{\sqrt{1-x^2}}
  2. 11−x2\dfrac{1}{\sqrt{1-x^2}}
  3. 11+x2\dfrac{1}{\sqrt{1+x^2}}
  4. x1+x2\dfrac{x}{\sqrt{1+x^2}}
Exam answer

Let θ=tan⁡−1x\theta = \tan^{-1}x⁠. Then tan⁡θ=x\tan\theta = x and θ∈(−π2,π2)\theta \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)⁠, so cos⁡θ>0\cos\theta > 0⁠.

cos⁡θ=11+tan⁡2θ=11+x2sin⁡θ=tan⁡θcos⁡θ=x1+x2\begin{aligned} \cos\theta &= \frac{1}{\sqrt{1 + \tan^2\theta}} = \frac{1}{\sqrt{1 + x^2}} \\ \sin\theta &= \tan\theta\cos\theta = \frac{x}{\sqrt{1 + x^2}} \end{aligned}

Answer: (D) x1+x2\dfrac{x}{\sqrt{1 + x^2}}

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

Give the inner angle a name: θ=tan⁡−1x\theta = \tan^{-1}x⁠. Then the question is "if tan⁡θ=x\tan\theta = x⁠, what is sin⁡θ\sin\theta⁠?" The identity sec⁡2θ=1+tan⁡2θ\sec^2\theta = 1 + \tan^2\theta gives cos⁡θ\cos\theta up to sign, the principal range of tan⁡−1\tan^{-1} fixes the sign (cos⁡θ>0\cos\theta > 0⁠), and sin⁡θ=tan⁡θcos⁡θ\sin\theta = \tan\theta\cos\theta finishes it.

Given, to find, and the plan
Given
A real number xx with ∣x∣<1|x| < 1⁠.
To find
sin⁡(tan⁡−1x)\sin(\tan^{-1}x) as an expression in xx⁠, and the matching option.
Plan
(1) Put θ=tan⁡−1x\theta = \tan^{-1}x and locate θ\theta⁠. (2) Find cos⁡θ\cos\theta with the correct sign. (3) Find sin⁡θ\sin\theta⁠. (4) Test every option.

Basics you need

1Range of tan⁻¹

θ=tan⁡−1x\theta = \tan^{-1}x means tan⁡θ=x\tan\theta = x and θ∈(−π2,π2)\theta\in\left(-\frac\pi2,\frac\pi2\right)⁠. Basics: tan, cot, sec and cosec

2Signs on (−π/2, π/2)

On this interval cos⁡θ>0\cos\theta > 0⁠, while sin⁡θ\sin\theta has the same sign as θ\theta⁠. Basics: sin and cos on the unit circle

3sec² = 1 + tan²

Divide sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1 by cos⁡2θ\cos^2\theta⁠: tan⁡2θ+1=sec⁡2θ\tan^2\theta + 1 = \sec^2\theta⁠.

4The triangle picture

For an acute angle, tan⁡θ=oppositeadjacent\tan\theta = \frac{\text{opposite}}{\text{adjacent}} and sin⁡θ=oppositehypotenuse\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}}⁠. Lesson: the triangle trick

Solution

  1. Name the inner angle

    Let θ=tan⁡−1x\theta = \tan^{-1}x⁠. Then tan⁡θ=x\tan\theta = x and θ∈(−π2,π2)\theta\in\left(-\frac\pi2,\frac\pi2\right)⁠. Because ∣x∣<1|x|<1⁠, in fact θ∈(−π4,π4)\theta \in\left(-\frac\pi4,\frac\pi4\right)⁠. We want sin⁡θ\sin\theta⁠.

    An inverse function inside a trig function is best handled by naming it. The definition of tan⁡−1\tan^{-1} then gives us both an equation (tan⁡θ=x\tan\theta = x⁠) and an interval for θ\theta⁠, and the interval decides signs.
    Definition and range of tan⁡−1\tan^{-1} (Basic 1).

  2. Find cos θ, with the right sign

    By Basic 3, sec⁡2θ=1+tan⁡2θ=1+x2\sec^2\theta = 1 + \tan^2\theta = 1 + x^2⁠, so

    cos⁡2θ=11+x2.\cos^2\theta = \frac{1}{1+x^2}.

    Since θ∈(−π2,π2)\theta\in\left(-\frac\pi2,\frac\pi2\right)⁠, cos⁡θ>0\cos\theta > 0⁠, so we take the positive root:

    cos⁡θ=11+x2.\cos\theta = \frac{1}{\sqrt{1+x^2}}.
    cos⁡2θ\cos^2\theta alone allows cos⁡θ=±11+x2\cos\theta = \pm\frac1{\sqrt{1+x^2}}⁠. The principal range of tan⁡−1\tan^{-1} rules out the minus sign. This is the step where the inverse function, not just trigonometry, matters.
    cos⁡θ>0\cos\theta>0 on (−π2,π2)\left(-\frac\pi2,\frac\pi2\right) (Basic 2).

  3. Find sin θ

    sin⁡θ=tan⁡θ⋅cos⁡θ=x⋅11+x2=x1+x2.\sin\theta = \tan\theta\cdot\cos\theta = x\cdot\frac{1}{\sqrt{1+x^2}} = \frac{x}{\sqrt{1+x^2}}.

    The sign takes care of itself: for negative xx⁠, θ\theta is negative, sin⁡θ<0\sin\theta < 0⁠, and so is x1+x2\frac{x}{\sqrt{1+x^2}}⁠.

    θ 1 (adjacent) x (opposite) √(1 + x²) tan θ = x/1 = x sin θ = x/√(1+x²) cos θ = 1/√(1+x²) drawn for x = 3/4
    For 0<x<10 < x < 1⁠: a right triangle with adjacent side 1 and opposite side xx has tan⁡θ=x\tan\theta = x⁠, hypotenuse 1+x2\sqrt{1+x^2} by Pythagoras, so sin⁡θ=x1+x2\sin\theta = \frac{x}{\sqrt{1+x^2}}⁠. Drawn for x=34x = \frac34⁠, hypotenuse 54\frac54⁠. For negative xx use the algebra of Steps 2 and 3; the formula is the same.
    tan⁡θ=sin⁡θcos⁡θ\tan\theta = \frac{\sin\theta}{\cos\theta}⁠, so sin⁡θ=tan⁡θcos⁡θ\sin\theta = \tan\theta\cos\theta⁠. With cos⁡θ\cos\theta known exactly (including its sign), this gives sin⁡θ\sin\theta with the correct sign automatically.
    tan⁡θ=sin⁡θcos⁡θ\tan\theta = \frac{\sin\theta}{\cos\theta}⁠; triangle picture (Basic 4).

  4. Test every option

    Take x=12x = \frac12⁠: tan⁡−112=0.4636476090\tan^{-1}\frac12 = 0.4636476090 and sin⁡(0.4636476090)=0.4472135955\sin(0.4636476090) = 0.4472135955⁠.

    OptionValue at x=12x=\frac12VerdictReason
    (A) x1−x2\frac{x}{\sqrt{1-x^2}}0.5773502692wrongThis is tan⁡(sin⁡−1x)\tan(\sin^{-1}x)⁠: the roles of sin and tan are swapped.
    (B) 11−x2\frac{1}{\sqrt{1-x^2}}1.1547005384wrongIt is greater than 1, so it cannot be a sine. (It is sec⁡(sin⁡−1x)\sec(\sin^{-1}x)⁠.)
    (C) 11+x2\frac{1}{\sqrt{1+x^2}}0.8944271910wrongThis is cos⁡(tan⁡−1x)\cos(\tan^{-1}x) from Step 2, not the sine. It is also always positive, but sin⁡(tan⁡−1x)<0\sin(\tan^{-1}x) < 0 for x<0x < 0⁠.
    (D) x1+x2\frac{x}{\sqrt{1+x^2}}0.4472135955correctMatches Step 3, and the number above.
    Checking each option against the exact formula, and against a number, guards against a slip. Exactly one option survives.
Result

sin⁡(tan⁡−1x)=x1+x2\sin(\tan^{-1}x) = \dfrac{x}{\sqrt{1+x^2}}⁠. The answer is (D).

(Textbook answer: D.)

Check it with numbers

x=12x = \frac12⁠: 1+14=1.1180339887\sqrt{1+\frac14} = 1.1180339887⁠, and 0.51.1180339887=0.4472135955\frac{0.5}{1.1180339887} = 0.4472135955⁠, the same as sin⁡(tan⁡−10.5)\sin(\tan^{-1}0.5) ✓.

x=−12x = -\frac12⁠: sin⁡(tan⁡−1(−0.5))=−0.4472135955\sin(\tan^{-1}(-0.5)) = -0.4472135955⁠, which matches (D) and shows (C) has the wrong sign.

Common mistakes
  1. Answering (C): computing cos⁡θ\cos\theta and forgetting to multiply by tan⁡θ\tan\theta⁠.
  2. Drawing the triangle with hypotenuse 1 and opposite xx⁠. That describes sin⁡θ=x\sin\theta = x⁠, the setting of sin⁡−1x\sin^{-1}x⁠, and leads to (A) or (B).
  3. Writing cos⁡θ=±11+x2\cos\theta = \pm\frac1{\sqrt{1+x^2}} and not choosing the sign. The principal range of tan⁡−1\tan^{-1} settles it.
What to take away
  1. To evaluate trig(inverse trig), name the inner angle, use one Pythagorean identity, and fix the sign from the principal range.
  2. Remark: the condition ∣x∣<1|x| < 1 is not needed. sin⁡(tan⁡−1x)=x1+x2\sin(\tan^{-1}x) = \frac{x}{\sqrt{1+x^2}} holds for every real xx⁠, since Steps 2 and 3 only used θ∈(−π2,π2)\theta\in\left(-\frac\pi2,\frac\pi2\right)⁠.

Every fact used, and where it comes from

FactUsed inSource
tan⁡−1\tan^{-1}⁠: range (−π2,π2)\left(-\frac\pi2,\frac\pi2\right)⁠; y=tan⁡−1x⇒tan⁡y=xy=\tan^{-1}x \Rightarrow \tan y = xSteps 1, 2Section 2.2 (Lesson: principal values)
1+tan⁡2θ=sec⁡2θ1+\tan^2\theta = \sec^2\theta⁠; cos⁡>0\cos>0 on (−π2,π2)\left(-\frac\pi2,\frac\pi2\right)Step 2Class XI, Trigonometric Functions
sin⁡θ=tan⁡θcos⁡θ\sin\theta = \tan\theta\cos\theta⁠; Pythagoras in the triangleStep 3Class XI, Trigonometric Functions
Option values at x=±12x = \pm\frac12Step 4, CheckRecomputed in the site's verification script
Answer (D)ResultNCERT answer key, Miscellaneous Exercise on Chapter 2
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