Q.Find the simplified form of $\cos^{-1}\left(\frac{3}{5}\cos x+\frac{4}{5}\sin x\right)$, where $x\in\left[\frac{-3\pi}{4},\frac{\pi}{4}\right]$.

Concept: write the combination as a single cosine, then apply the principal range of $\cos^{-1}$ carefully.

Let $\alpha=\cos^{-1}\tfrac35=\tan^{-1}\tfrac43\approx 0.927$, so $\cos\alpha=\tfrac35$ and $\sin\alpha=\tfrac45$. Then

so the expression equals $\cos^{-1}\big(\cos(x-\alpha)\big)$. …

Concept and Intuition

When you see something like $\frac{3}{5}\cos x + \frac{4}{5}\sin x$, your first instinct should be: this is a cosine of a shifted angle. Why? Because $\cos(A - B) = \cos A \cos B + \sin A \sin B$. If we let $\cos \phi = \frac{3}{5}$ and $\sin \phi = \frac{4}{5}$, then the expression becomes $\cos x \cos \phi + \sin x \sin \phi = \cos(x - \phi)$.

So the problem reduces to finding $\cos^{-1}(\cos(x - \phi))$, where $\phi = \tan^{-1}\frac{4}{3}$. But $\cos^{-1}(\cos \theta)$ is not simply $\theta$ — it gives the principal value, which lies in $[0, \pi]$. So we must check where $x - \phi$ falls, given the range of $x$, and adjust accordingly.

Step-by-step solution

1. Identify the angle $\phi$ such that $\cos\phi = \frac{3}{5}$ and $\sin\phi = \frac{4}{5}$.

Since $\left(\frac{3}{5}\right)^2 + \left(\frac{4}{5}\right)^2 = 1$, such an angle exists. We have $\tan\phi = \frac{4}{3}$, so $\phi = \tan^{-1}\frac{4}{3}$. Both sine and cosine are positive, so $\phi$ lies in the first quadrant: $0 < \phi < \frac{\pi}{2}$.

2. Rewrite the given expression.

3. Determine the range of $x - \phi$ given $x \in \left[-\frac{3\pi}{4}, \frac{\pi}{4}\right]$.

First, note $\phi = \tan^{-1}\frac{4}{3} \approx 0.9273$ rad, which is between $\frac{\pi}{4} \approx 0.785$ and $\frac{\pi}{2} \approx 1.571$. So $\phi \in \left(\frac{\pi}{4}, \frac{\pi}{2}\right)$.

Now compute the endpoints:

• When $x = -\frac{3\pi}{4}$:

  $x - \phi = -\frac{3\pi}{4} - \phi$. Since $\phi > \frac{\pi}{4}$, we have $-\frac{3\pi}{4} - \phi < -\frac{3\pi}{4} - \frac{\pi}{4} = -\pi$. So the lower bound is less than $-\pi$.

• When $x = \frac{\pi}{4}$:

  $x - \phi = \frac{\pi}{4} - \phi$. Since $\phi > \frac{\pi}{4}$, this is negative. Specifically, $\frac{\pi}{4} - \phi < 0$. And since $\phi < \frac{\pi}{2}$, we have $\frac{\pi}{4} - \phi > \frac{\pi}{4} - \frac{\pi}{2} = -\frac{\pi}{4}$.

So $x - \phi$ ranges from somewhere below $-\pi$ up to somewhere between $-\frac{\pi}{4}$ and $0$. In interval notation:

But more precisely, the entire interval lies within $(-\pi, 0)$? Let's check: the upper bound is negative, the lower bound is $-\frac{3\pi}{4} - \phi$. Since $\phi < \frac{\pi}{2}$, the lower bound $> -\frac{3\pi}{4} - \frac{\pi}{2} = -\frac{5\pi}{4}$. But is it always $> -\pi$? For that we need $-\frac{3\pi}{4} - \phi > -\pi \implies \phi < \pi - \frac{3\pi}{4} = \frac{\pi}{4}$, which is false because $\phi > \frac{\pi}{4}$. So the lower bound is actually less than $-\pi$. Therefore $x - \phi$ straddles $-\pi$: part of the interval is below $-\pi$, part above.

4. Use the identity $\cos^{-1}(\cos \theta) = |\theta|$ when $\theta \in [-\pi, 0]$?

Actually, the standard formula: for $\theta \in [-\pi, 0]$, $\cos^{-1}(\cos \theta) = -\theta$ (since $-\theta \in [0, \pi]$). For $\theta < -\pi$, we first add $2\pi$ to bring it into $[-\pi, \pi]$? Let's be systematic.

The principal value of $\cos^{-1}$ always lies in $[0, \pi]$. So $\cos^{-1}(\cos \theta)$ equals:

• $\theta$ if $\theta \in [0, \pi]$
• $-\theta$ if $\theta \in [-\pi, 0]$
• For $\theta$ outside $[-\pi, \pi]$, reduce modulo $2\pi$ into $[-\pi, \pi]$ first, then apply the above.

5. Find where $x - \phi$ lies relative to $-\pi$.

We need to find the $x$ in $\left[-\frac{3\pi}{4}, \frac{\pi}{4}\right]$ for which $x - \phi = -\pi$. Solve:

Since $\phi \approx 0.927$, $\phi - \pi \approx -2.214$ rad, which is about $-126.8^\circ$. Compare with $-\frac{3\pi}{4} \approx -2.356$ rad. So $\phi - \pi \approx -2.214 > -2.356$, meaning the crossover point lies inside the interval.

Thus:

• For $x \in \left[-\frac{3\pi}{4},\; \phi - \pi\right]$, we have $x - \phi \leq -\pi$.
• For $x \in \left[\phi - \pi,\; \frac{\pi}{4}\right]$, we have $x - \phi \geq -\pi$ (and still negative, since upper bound is negative).

6. Simplify piecewise.

• Case 1: $x \in \left[-\frac{3\pi}{4},\; \phi - \pi\right]$. Here $x - \phi \leq -\pi$. Add $2\pi$ to bring into $[-\pi, \pi]$: $\theta = x - \phi + 2\pi$. Since $x - \phi \in [-\frac{3\pi}{4} - \phi,\; -\pi]$, adding $2\pi$ gives $\theta \in [-\frac{3\pi}{4} - \phi + 2\pi,\; \pi]$. The lower bound: $-\frac{3\pi}{4} - \phi + 2\pi = \frac{5\pi}{4} - \phi$. Since $\phi < \frac{\pi}{2}$, $\frac{5\pi}{4} - \phi > \frac{5\pi}{4} - \frac{\pi}{2} = \frac{3\pi}{4}$, so $\theta \in [\text{something} > \frac{3\pi}{4},\; \pi] \subset [0, \pi]$. Hence $\cos^{-1}(\cos(x - \phi)) = \theta = x - \phi + 2\pi$. …

Method: Simplifying $\cos^{-1}(a\cos x+b\sin x)$ via a phase shift

The technique for collapsing a linear combination of $\cos x$ and $\sin x$ inside an inverse function is to rewrite it as a single cosine, then reduce carefully using the given interval for $x$.

Steps

Step 1: Write the combination as one cosine.

With $a^2+b^2=1$, pick $\phi$ so that $\cos\phi=a$, $\sin\phi=b$; then

(If $a^2+b^2\ne1$, factor out $R=\sqrt{a^2+b^2}$ first.)

Step 2: Reduce to $\cos^{-1}(\cos(\text{angle}))$.

The expression becomes $\cos^{-1}\!\big(\cos(x-\phi)\big)$.

Step 3 (the decisive step): bring the inside angle into $[0,\pi]$. …

Common Mistakes

Mistake 1: Asserting $\cos^{-1}(\cos(x-\phi))=\phi-x$ across the whole interval without checking the reduced angle.

Why it's wrong: $\cos^{-1}(\cos u)=u$ only for $u\in[0,\pi]$. Here $\phi=\cos^{-1}\tfrac35\approx0.927$, and at the endpoint $x=-\tfrac{3\pi}{4}$ the angle $\phi-x\approx3.28$ exceeds $\pi$, so the simple formula fails on part of the interval. Correct approach: substitute the interval endpoints, detect where the angle crosses $\pi$, and give a piecewise answer.

Mistake 2: Reading $\tfrac35$ and $\tfrac45$ as $\sin\phi$ and $\cos\phi$ in the wrong order. …