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CBSE Sample Paper 2023-24 · Section B · 2 marks

Q21: Value of sin⁡−1(cos⁡33π5)\sin^{-1}\left(\cos\frac{33\pi}{5}\right)⁠, or the domain of sin⁡−1(x2−4)\sin^{-1}(x^2 - 4)

Sample 2023-24Ch 2 · Inverse Trigonometric Functions
The question

(a) Find the value of sin⁡−1(cos⁡(33π5))\sin^{-1}\left(\cos\left(\dfrac{33\pi}{5}\right)\right)⁠.

OR

(b) Find the domain of sin⁡−1(x2−4)\sin^{-1}(x^2 - 4)⁠.

Exam answer

(a)

sin⁡−1(cos⁡33π5)=sin⁡−1cos⁡(6π+3π5)=sin⁡−1(cos⁡3π5)\begin{aligned} &\sin^{-1}\left(\cos\frac{33\pi}{5}\right) \\ &= \sin^{-1}\cos\left(6\pi + \frac{3\pi}{5}\right) \\ &= \sin^{-1}\left(\cos\frac{3\pi}{5}\right) \end{aligned}
Why?

Cosine repeats after every full turn of 2π2\pi⁠, so whole turns can be dropped:

cos⁡(2nπ+θ)=cos⁡θ\cos(2n\pi + \theta) = \cos\theta

Split the angle: 33π5=30π5+3π5\frac{33\pi}{5} = \frac{30\pi}{5} + \frac{3\pi}{5}⁠, and 30π5=6π\frac{30\pi}{5} = 6\pi is three full turns.

30π5=6π(30÷5=6)6π=3×2πcos⁡(6π+3π5)=cos⁡3π5\begin{aligned} \frac{30\pi}{5} &= 6\pi \quad (30 \div 5 = 6) \\ 6\pi &= 3 \times 2\pi \\ \cos\left(6\pi + \frac{3\pi}{5}\right) &= \cos\frac{3\pi}{5} \end{aligned}
=sin⁡−1sin⁡(π2−3π5)=sin⁡−1sin⁡(−π10)\begin{aligned} &= \sin^{-1}\sin\left(\frac{\pi}{2} - \frac{3\pi}{5}\right) \\ &= \sin^{-1}\sin\left(-\frac{\pi}{10}\right) \end{aligned}
Why?

sin⁡−1\sin^{-1} undoes a sine, not a cosine, so first turn the cosine into a sine:

cos⁡θ=sin⁡(π2−θ)\cos\theta = \sin\left(\frac{\pi}{2} - \theta\right)

With θ=3π5\theta = \frac{3\pi}{5}⁠. The common denominator is 10: π2=5π10\frac{\pi}{2} = \frac{5\pi}{10} and 3π5=6π10\frac{3\pi}{5} = \frac{6\pi}{10}⁠.

π2−3π5=5π10−6π10=5π−6π10=−π10\begin{aligned} \frac{\pi}{2} - \frac{3\pi}{5} &= \frac{5\pi}{10} - \frac{6\pi}{10} \\ &= \frac{5\pi - 6\pi}{10} \\ &= -\frac{\pi}{10} \end{aligned}

Since −π10∈[−π2,π2]-\frac{\pi}{10} \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]⁠,

Why?

sin⁡−1(sin⁡θ)=θ\sin^{-1}(\sin\theta) = \theta is true only when θ\theta lies in the principal value range [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]⁠. The angle −π10-\frac{\pi}{10} does, so sin⁡−1\sin^{-1} gives it back.

Watch out: 33π5\frac{33\pi}{5} and 3π5\frac{3\pi}{5} are both outside this range, so neither can be the answer.

sin⁡−1(cos⁡33π5)=−π10\sin^{-1}\left(\cos\frac{33\pi}{5}\right) = -\frac{\pi}{10}

(b)

sin⁡−1(x2−4)\sin^{-1}(x^2 - 4) is defined when

−1≤x2−4≤1-1 \le x^2 - 4 \le 1
Why?

sin⁡−1u\sin^{-1}u is defined only for −1≤u≤1-1 \le u \le 1⁠, because a sine never goes outside [−1,1][-1, 1]⁠. Here u=x2−4u = x^2 - 4⁠.

⇒3≤x2≤5\Rightarrow 3 \le x^2 \le 5
Why?

Add 4 to all three parts. Adding the same number to every part keeps the inequality signs: −1+4=3-1 + 4 = 3 and 1+4=51 + 4 = 5⁠.

⇒3≤∣x∣≤5\Rightarrow \sqrt3 \le |x| \le \sqrt5
Why?

All three parts are positive, so we may take the square root of each part. Remember that x2=∣x∣\sqrt{x^2} = |x|⁠, not xx⁠, because xx may be negative.

⇒x∈[−5,−3]∪[3,5]\Rightarrow x \in [-\sqrt5, -\sqrt3] \cup [\sqrt3, \sqrt5]
Why?

∣x∣≥3|x| \ge \sqrt3 means x≤−3x \le -\sqrt3 or x≥3x \ge \sqrt3⁠.

∣x∣≤5|x| \le \sqrt5 means −5≤x≤5-\sqrt5 \le x \le \sqrt5⁠.

Both must hold, which leaves one piece on each side of 0.

Watch out: [3,5][\sqrt3, \sqrt5] alone misses the negative values. For example x=−2x = -2 gives (−2)2−4=0(-2)^2 - 4 = 0⁠, which is allowed.

So, the required domain is [−5,−3]∪[3,5][-\sqrt5, -\sqrt3] \cup [\sqrt3, \sqrt5]⁠.

Where this comes from

Chapter 2, Inverse Trigonometric Functions: Inverse trig: principal value; domain.

Question type: (a) An inverse of its own trig function

How to spot it: click to see

(a) Type 5. An inverse of its own trig function: move the angle into the range first

Spot it: sin⁡−1(sin⁡…)\sin^{-1}(\sin\ldots)⁠, cos⁡−1(cos⁡…)\cos^{-1}(\cos\ldots) or tan⁡−1(tan⁡…)\tan^{-1}(\tan\ldots) with a given angle.

Here: 33π5\frac{33\pi}{5} is far outside the range. Drop 6π6\pi to get cos⁡3π5\cos\frac{3\pi}{5}⁠, then write it as sin⁡(π2−3π5)=sin⁡(−π10)\sin\left(\frac{\pi}{2} - \frac{3\pi}{5}\right) = \sin\left(-\frac{\pi}{10}\right)⁠; −π10-\frac{\pi}{10} is inside the range.

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