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The principal value of ^-1( (7 /6)) is:

5 /6. Tests understanding of principal value branch for ^-1x.

What is the value of ^-1( (- /3))?

2 /3. Requires evaluating (- /3) and then applying ^-1(-x) formula.

Value of ^-1( (7 /6)) is:

2 /3 (7 /6) = ( + /6) = - ( /6) = -1/2 So, we need to find ^-1(-1/2). Let = -1/2. Since ^-1(-x) = - ^-1(x), = - ^-1(1/2) = - /3 = 2 /3.

Mathematics · Class 11-12 · CBSE

Inverse Trigonometric Functions: Solving $\cos^{-1}(\sin x)$

Inverse trigonometric functions, jaise ki $\\cos^{-1}x$ , humeinangle findkarne mein help karte hain jab humein trigonometric ratio pata ho. Inko solve karte waqt, humeinprincipal value branchka dhyaan rakhna hota hai.

1Understanding Inverse Cosine Function

The inverse cosine function, denoted as $\\cos^{-1}x$ or arccos $x$ , gives the angle whose cosine is $x$ . Iskiprincipal value branch $[0, \pi]$ hoti hai.

Definition

Principal Value Branch — The specific range of angles for which an inverse trigonometric function is defined and gives a unique output.

Principal Value Branches of Inverse Trig Functions

Function	Domain	Range (Principal Value Branch)
$\\sin^{-1}x$	$[-1, 1]$	$[-\pi/2, \pi/2]$
$\\cos^{-1}x$	$[-1, 1]$	$[0, \pi]$
$\\tan^{-1}x$	$(-\infty, \infty)$	$(-\pi/2, \pi/2)$

$\\sin^{-1}x$
Domain
$[-1, 1]$
Range (Principal Value Branch)
$[-\pi/2, \pi/2]$
$\\cos^{-1}x$
Domain
$[-1, 1]$
Range (Principal Value Branch)
$[0, \pi]$
$\\tan^{-1}x$
Domain
$(-\infty, \infty)$
Range (Principal Value Branch)
$(-\pi/2, \pi/2)$

2Converting $\sin$ to $\cos$

Jab humein $\\cos^{-1}(\sin x)$ jaisa expression solve karna ho, toh humein $\sin x$ ko $\cos$ ke terms mein convert karna padta hai. Iske liye humcomplementary angle identitiesuse karte hain.

Complementary Angle Identity

\sin \theta = \cos (\frac{\pi}{2} - \theta)

$\theta$ = angle in radians

Real-life example · Converting $\sin(4\pi/3)$

Humein $\sin(4\pi/3)$ ko $\cos$ mein convert karna hai. Using the identity, $\sin(4\pi/3) = \cos(\pi/2 - 4\pi/3)$ . Isse hum value ko $\\cos^{-1}$ ke domain mein la sakte hain.

3Solving $\cos^{-1}(\sin(4\pi/3))$ Step-by-Step

Is problem ko solve karne ke liye, humein pehle inner function $\sin(4\pi/3)$ ki value nikalni hogi, phir use $\cos$ mein convert karke $\\cos^{-1}$ apply karna hoga. Remember to keep theprincipal value branchin mind.

Steps to Solve $\cos^{-1}(\sin(4\pi/3))$

1
Evaluate $\sin(4\pi/3)$
Find the value of the inner sine function.
2
Convert to $\cos$
Use $\sin \theta = \cos(\pi/2 - \theta)$ to convert.
3
Simplify angle
Simplify the angle inside the cosine function.
4
Check range
Ensure the angle is within $[0, \pi]$ for $\\cos^{-1}$ .
5
Apply $\\cos^{-1}$
Use $\\cos^{-1}(\cos x) = x$ if $x \in [0, \pi]$ .

Unit Circle for Angles

Unit circle se hum angles aur unke trigonometric values ko visualize kar sakte hain.

The picture didn’t pass the label check, so here are its parts.

1$4\pi/3$ — Angle in the third quadrant
2$\sin(4\pi/3)$ — Negative y-coordinate
3$\pi/2 - 4\pi/3$ — Angle in the second quadrant

4Detailed Solution

Let's solve the given problem: $\\cos^{-1}(\sin(4\pi/3))$ . Pehle $\sin(4\pi/3)$ ki value nikalte hain.

Derivation — Solving $\cos^{-1}(\sin(4\pi/3))$

1
$\sin(4\pi/3) = \sin(\pi + \pi/3)$
Angle ko simplify kiya
2
$= -\sin(\pi/3)$
$\sin(\pi + \theta) = -\sin \theta$
3
$= -\frac{\sqrt{3}}{2}$
Value of $\sin(\pi/3)$
4
$\\cos^{-1}(\sin(4\pi/3)) = \\cos^{-1}(-\frac{\sqrt{3}}{2})$
Substitute the value
5
$\text{Let } \cos \theta = -\frac{\sqrt{3}}{2}$
Find the angle $\theta$
6
$\theta = \pi - \\cos^{-1}(\frac{\sqrt{3}}{2})$
For negative values, $\\cos^{-1}(-x) = \pi - \\cos^{-1}(x)$
7
$\theta = \pi - \frac{\pi}{6}$
$\cos(\pi/6) = \sqrt{3}/2$
8
$\theta = \frac{5\pi}{6}$
Simplify the angle

Common mistake: Common mistake hai ki students

\\cos^{-1}(\cos x) = x

directly apply kar dete hain bina principal value branch check kiye. Always ensure

x \in [0, \pi]

for

\\cos^{-1}

5Exam Corner

Important for exams

Inverse trigonometric functions kiprincipal value branchesyaad rakho.
$\sin x$ ko $\cos x$ mein convert karne ke liyecomplementary identitiesuse karo.
$\\cos^{-1}(-x) = \pi - \\cos^{-1}(x)$ formula important hai.
$\\cos^{-1}(\cos \theta) = \theta$ only if $\theta \in [0, \pi]$ .
Unit circle se values aur quadrants ko visualize karna helpful hota hai.

Exam question · NEET 2019 style

Value of

\\cos^{-1}(\sin(7\pi/6))

is:

A. $\pi/3$
B. $\pi/6$
C. $2\pi/3$
D. $5\pi/6$

Remember it: CAST Rule: All Students Take Calculus (A-All positive in 1st, S-Sin positive in 2nd, T-Tan positive in 3rd, C-Cos positive in 4th quadrant).

Did you know? Inverse trigonometric functions ka concept 17th century mein mathematicians like Euler aur Bernoulli ne develop kiya tha, jo calculus mein bahut useful prove hua.

Remember

$\\cos^{-1}x$ kiprincipal value branch $[0, \pi]$ hoti hai.
$\sin \theta$ ko $\cos(\pi/2 - \theta)$ meinconvertkar sakte hain.
$\\cos^{-1}(-x) = \pi - \\cos^{-1}(x)$ ekkey identityhai.
Always check if the angle is in theprincipal value branchbefore applying $\\cos^{-1}(\cos x) = x$ .

Check yourself

Question 1
The principal value of $\\cos^{-1}(\cos(7\pi/6))$ is:
Question 2
What is the value of $\\cos^{-1}(\sin(-\pi/3))$ ?