If π/2 ≤ x ≤ 3π/4 then cos⁻¹(12/13 cos x + 5/13 sin x) is equal to
(1) x - tan⁻¹(4/3)
(2) x + tan⁻¹(

Q: Wait, why can't I just cancel $\cos^{-1}$ and $\cos$ and write $x - \alpha$ immediately?

That is the most common trap in inverse trigonometry! The function $\cos^{-1}(\cos \theta)$ is only equal to $\theta$ if $\theta$ is within the principal branch $[0, \pi]$. If $\theta$ is outside this range, the identity does not hold directly. You must always verify the domain of the argument before simplifying.

Q: How do I know whether to use $\cos(x - \alpha)$ or $\sin(x - \alpha)$?

Look at the structure of the expression. We have $\frac{12}{13}\cos x + \frac{5}{13}\sin x$. Since the identity $\cos(A - B) = \cos A \cos B + \sin A \sin B$ matches this structure (where $A=x$ and $B=\alpha$), it is natural to use the cosine identity. If the expression were $\frac{12}{13}\sin x + \frac{5}{13}\cos x$, you would use the sine identity.

Q: What if the range of $x$ was different, say $x \in [\pi, \frac{3\pi}{2}]$? Would the answer change?

Absolutely! If the range of $x$ changed, the range of $x - \alpha$ would also change. If $x - \alpha$ fell outside $[0, \pi]$, you would need to use the properties of the cosine function (like $\cos(2\pi - \theta) = \cos \theta$ or $\cos(-\theta) = \cos \theta$) to shift the angle back into the principal branch before simplifying. Always check your boundaries!

Question

If $\frac{\pi}{2}\le x\le\frac{3\pi}{4}$ then $\cos^{-1}(\frac{12}{13}\cos~x+\frac{5}{13}\sin~x)$ is equal to

Options:

(A) $x-	an^{-1}\frac{4}{3}$

(B) $x+	an^{-1}\frac{4}{5}$

(C) $x-	an^{-1}\frac{5}{12}$

(D) $x+	an^{-1}\frac{5}{12}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup The expression provided is $\cos^{-1}(\frac{12}{13}\cos x + \frac{5}{13}\sin x)$. This is a classic JEE Advanced problem that tests algebraic manipulation, geometric intuition, and domain constraints. The first observation is the coefficients $\frac{12}{13}$ and $\frac{5}{13}$. Note that the sum of their squares is: $$(\frac{12}{13})^2 + (\frac{5}{13})^2 = \frac{144}{169} + \frac{25}{169} = \frac{169}{169} = 1$$ This confirms that these coefficients represent the coordinates of a point on the unit circle, signaling the use of an auxiliary angle. ### Building the Geometric Bridge We interpret these coefficients as sides of a right-angled triangle. With a base of $12$ and a perpendicular of $5$, the hypotenuse is $\sqrt{12^2 + 5^2} = 13$. Let us define an auxiliary angle $\alpha$ such that: $$\cos \alpha = \frac{12}{13}, \quad \sin \alpha = \frac{5}{13}$$ This implies $ an \alpha = \frac{5}{12}$. By introducing $\alpha$, the expression transforms into: $$\cos^{-1}(\cos x \cos \alpha + \sin x \sin \alpha)$$ ### The Power of Compound Angles The argument inside the inverse cosine function follows the identity $\cos(x - \alpha) = \cos x \cos \alpha + \sin x \sin \alpha$. Thus, the expression simplifies to: $$\cos^{-1}(\cos(x - \alpha))$$ Many students incorrectly assume this equals $x - \alpha$ immediately. However, we must verify if $x - \alpha$ lies within the principal value branch of the inverse cosine function, which is $[0, \pi]$. ### The Domain Check: The Final Hurdle We are given the constraint $\frac{\pi}{2} \le x \le \frac{3\pi}{4}$. We must determine the range of $x - \alpha$. Since $ an \alpha = \frac{5}{12}$ and $ an(\frac{\pi}{4}) = 1$, we know that $0 < \alpha < \frac{\pi}{4}$. Subtracting $\alpha$ from the given interval for $x$: The lower bound becomes $\frac{\pi}{2} - \alpha$. Since $\alpha < \frac{\pi}{4}$, this is greater than $\frac{\pi}{4}$, which is $> 0$. The upper bound becomes $\frac{3\pi}{4} - \alpha$. Since $\alpha > 0$, this is less than $\frac{3\pi}{4}$, which is $< \pi$. ### Final Calculation Because the interval for $x - \alpha$ lies entirely within $(0, \pi)$, the identity $\cos^{-1}(\cos(x - \alpha)) = x - \alpha$ holds true. Substituting back the value of $\alpha$, we arrive at the final result: **$$x - an^{-1}\left(\frac{5}{12} ight)$$**

Animated Solution for Mathematics - Inverse Trigonometric Functions: If $\frac{π}{2} \leq x \leq \frac{3 π}{4}$ then $cos^{- 1} (\frac{12}{13} cos x + \frac{5}{13} sin x)$ is equal to

Select Answer:

Visualized Solution

Analyzing the Expression

The Inner Expression

The Auxiliary Angle Triangle

Defining $α$

Substitution

Compound Angle Formula

The Principal Branch Trap

Analyzing the Given Domain

Estimating $α$

Shifting the Interval

Verifying the Bounds

Final Simplification

Summary & Key Takeaway

The Sigma Insight: Properties of Inverse Trigonometric Functions

Analyzing the Setup

Building the Geometric Bridge

The Power of Compound Angles

The Domain Check

Final Calculation

Similar Questions

Considering the principal values of the inverse trigonometric functions, $sin^{- 1} (\frac{3}{2} x + \frac{1}{2} 1 - x^{2}), - \frac{1}{2} < x < \frac{1}{2}$ , is equal to

$1 + x^{2} [{x cos (cot^{- 1} x) + sin (cot^{- 1} x)}^{2} - 1]^{1/2} =$

$2 π - (sin^{- 1} \frac{4}{5} + sin^{- 1} \frac{5}{13} + sin^{- 1} \frac{16}{65})$ is equal to:

If $sin^{- 1} (x - \frac{x ^{2}}{2} + \frac{x ^{3}}{4} - \dots) + cos^{- 1} (x^{2} - \frac{x ^{4}}{2} + \frac{x ^{6}}{4} - \dots) = \frac{π}{2}$ for $0 < ∣ x ∣ < 2$ , then $x$ equals

If $cos^{- 1} (\frac{2}{3 x}) + cos^{- 1} (\frac{3}{4 x}) = \frac{π}{2} (x > \frac{3}{4})$ , then x is equal to :

$tan^{- 1} (\frac{1 + 3}{3 + 3}) + sec^{- 1} (\frac{8 + 4 3}{6 + 3 3})$ is equal to

The value of $sin^{- 1} (\frac{12}{13}) - sin^{- 1} (\frac{3}{5})$ is equal to :

Prove that $cos tan^{- 1} sin cot^{- 1} x = \frac{x ^{2} + 1}{x ^{2} + 2}$ .

The value of $tan (2 tan^{- 1} (\frac{3}{5}) + sin^{- 1} (\frac{5}{13}))$ is equal to:

If $cos^{- 1} x - cos^{- 1} \frac{y}{2} = α$ , then $4 x^{2} - 4 x y cos α + y^{2}$ is equal to

Animated Solution for Mathematics - Inverse Trigonometric Functions: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If 2π​≤x≤43π​ then cos−1(1312​cos x+135​sin x) is equal to