Q.Find the principal value of $\sin^{-1}\left(\dfrac{1}{\sqrt{2}}\right)$.

Concept & Intuition

We have a composition of three functions: $f(x) = \sin^{-1}(\log_2(x^2/2))$.

• The outermost function, $\sin^{-1}(u)$, is defined only when $u \in [-1,1]$.
• Inside that, $\log_2(v)$ is defined only when $v > 0$.
• The innermost expression is $v = x^2/2$, which is always non‑negative, but we must enforce $v > 0$ for the log.

So the domain is the set of $x$ such that both conditions hold simultaneously. We solve step by step.

1. Inner condition: argument of $\log_2$ must be positive

So $x$ can be any real number except $0$.

2. Outer condition: argument of $\sin^{-1}$ must be in $[-1,1]$ Let $u = \log_2\!\left(\frac{x^2}{2}\right)$. We require

Since $\log_2$ is increasing, we can exponentiate with base $2$ (preserving inequalities):

That is,

3. Solve the two inequalities

   • Left inequality: $\frac{x^2}{2} \ge \frac12 \implies x^2 \ge 1 \implies |x| \ge 1$.
   • Right inequality: $\frac{x^2}{2} \le 2 \implies x^2 \le 4 \implies |x| \le 2$.

   Combining: $1 \le |x| \le 2$.

4. Combine with $x \neq 0$

   The condition $1 \le |x| \le 2$ already excludes $0$, so the only restriction is $|x| \in [1,2]$.

   In interval notation, this means $x \in [-2,-1] \cup [1,2]$.

We start with

Concept & Intuition:

Inverse trig equations often become algebraic after applying a trigonometric function to both sides. But we must be careful: the ranges of $\sin^{-1}$ and $\cos^{-1}$ restrict possible $x$ values. A classic trick is to isolate one inverse function and use the identity $\sin^{-1}u + \cos^{-1}u = \frac{\pi}{2}$ to rewrite the other term, making it easier to apply sine.

Step-by-step solution:

1. Rewrite $\cos^{-1}(3x)$ using the identity For any $u$ in $[-1,1]$, we have $\cos^{-1}u = \frac{\pi}{2} - \sin^{-1}u$. So

2. Simplify the equation

Bring $\frac{\pi}{2}$ to the right:

3. Take sine of both sides Let $\alpha = \sin^{-1}(4x)$ and $\beta = \sin^{-1}(3x)$. Then $\alpha + \beta = \frac{2\pi}{3}$. Taking sine:

Using the sine addition formula:

4. Express $\sin$ and $\cos$ in terms of $x$ $\sin\alpha = 4x$, $\sin\beta = 3x$. Since $\alpha, \beta \in [-\frac{\pi}{2}, \frac{\pi}{2}]$, we have $\cos\alpha = \sqrt{1 - (4x)^2}$ and $\cos\beta = \sqrt{1 - (3x)^2}$ (non-negative because cosine is non-negative on that interval). So:

5. Isolate one square root and square Move $3x\sqrt{1 - 16x^2}$ to the right:

Square both sides:

6. Simplify Left: $16x^2 - 144x^4$. Right: $\frac{3}{4} + 9x^2 - 144x^4 - 3\sqrt{3}\,x\sqrt{1 - 16x^2}$. Cancel $-144x^4$ from both sides:

So

7. Isolate the square root again

Square again:

8. Expand and solve Left: $27x^2 - 432x^4$. Right: $\frac{9}{16} - \frac{21}{2}x^2 + 49x^4$. Multiply through by 16 to clear denominators:

Bring all terms to one side:

i.e.

9. Solve the quadratic in $x^2$ Let $t = x^2$. Then

Discriminant: $\Delta = 600^2 - 4\cdot 7696 \cdot 9 = 360000 - 277056 = 82944$.

$\sqrt{82944} = 288$ (since $288^2 = 82944$).

So

Two possibilities:

10. Check domain and original equation

    Since $4x$ and $3x$ must be in $[-1,1]$, we need $|x| \le \frac{1}{4}$.

    $t_1 = \frac{3}{52} \approx 0.0577$ gives $x \approx \pm 0.240$, which is less than $0.25$ — okay.

    $t_2 = \frac{3}{148} \approx 0.02027$ gives $x \approx \pm 0.142$ — also okay.

    Now test in the original equation. The sum $\sin^{-1}(4x) + \sin^{-1}(3x)$ must equal $\frac{2\pi}{3} \approx 2.094$.

    For $t_1$: $x = \frac{\sqrt{3}}{2\sqrt{13}}$ (positive root). Then $4x \approx 0.96$, $\sin^{-1}(0.96) \approx 1.287$, $3x \approx 0.72$, $\sin^{-1}(0.72) \approx 0.804$, sum $\approx 2.091$ — close but check exactly:

    Actually $\frac{\sqrt{3}}{2\sqrt{13}} = \sqrt{\frac{3}{52}}$, so $4x = 4\sqrt{\frac{3}{52}} = \sqrt{\frac{48}{52}} = \sqrt{\frac{12}{13}}$, $\sin^{-1}\sqrt{\frac{12}{13}}$ and $3x = 3\sqrt{\frac{3}{52}} = \sqrt{\frac{27}{52}} = \sqrt{\frac{27}{52}}$. Their sum is not $\frac{2\pi}{3}$ exactly — it fails the original equation because squaring introduced extraneous solutions.

    For $t_2$: $x = \frac{\sqrt{3}}{4\sqrt{7}}$ (positive root). Then $4x = \frac{\sqrt{3}}{\sqrt{7}} = \sqrt{\frac{3}{7}}$, $\sin^{-1}\sqrt{\frac{3}{7}} \approx 0.7137$, $3x = \frac{3\sqrt{3}}{4\sqrt{7}} = \sqrt{\frac{27}{112}}$, $\sin^{-1}\sqrt{\frac{27}{112}} \approx 0.5236$ (which is $\pi/6$). Sum = $0.7137 + 0.5236 = 1.2373$ rad? Wait, that’s not $2\pi/3$. Let’s check carefully:

    Actually $\sin^{-1}\sqrt{\frac{3}{7}} \approx \sin^{-1}(0.6547) \approx 0.715$, $\sin^{-1}\sqrt{\frac{27}{112}} \approx \sin^{-1}(0.491) \approx 0.513$, sum $\approx 1.228$ rad, not $2.094$. So this also seems off? Let’s re-evaluate.

    Correction: We must check the original equation $\sin^{-1}(4x) - \cos^{-1}(3x) = \pi/6$, not the transformed sum equation. For $x = \frac{\sqrt{3}}{4\sqrt{7}}$:

    $4x = \frac{\sqrt{3}}{\sqrt{7}} \approx 0.6547$, $\sin^{-1}(0.6547) \approx 0.715$ rad.

    $3x = \frac{3\sqrt{3}}{4\sqrt{7}} \approx 0.491$, $\cos^{-1}(0.491) \approx 1.060$ rad.

    Difference: $0.715 - 1.060 = -0.345$, not $\pi/6 \approx 0.5236$. So this fails too.

    Let’s instead test $x = \frac{\sqrt{3}}{2\sqrt{7}}$ (option A):

    $4x = \frac{2\sqrt{3}}{\sqrt{7}} \approx 1.309 > 1$ — invalid because $\sin^{-1}$ domain is $[-1,1]$. So A is out.

    Option C: $x = \frac{\sqrt{3}}{2\sqrt{13}} \approx 0.240$, $4x \approx 0.96$, $\sin^{-1}(0.96) \approx 1.287$, $3x \approx 0.72$, $\cos^{-1}(0.72) \approx 0.766$, difference $= 0.521 \approx \pi/6$ — this works!

    Option D: $x = \frac{\sqrt{3}}{4\sqrt{13}} \approx 0.120$, $4x \approx 0.48$, $\sin^{-1}(0.48) \approx 0.5$, $3x \approx 0.36$, $\cos^{-1}(0.36) \approx 1.203$, difference $\approx -0.703$ — fails.

    So the correct $x$ is $\frac{\sqrt{3}}{2\sqrt{13}}$, which is option (C).

Rearrange the given equation:

Let $\theta=\cos^{-1}\sqrt{1-x^2}$, so $\cos\theta=\sqrt{1-x^2}$. Taking the sine of both sides:

This gives the quadratic

The argument $2x$ must satisfy $|2x|\le 1$, so the admissible root is $x=\dfrac{\sqrt3-1}{2}$ (the other root $\dfrac{-1-\sqrt3}{2}\approx-1.37$ is rejected).

Then

The key to this problem is understanding that $f(x)$ is a composition: an inverse trigonometric function applied to a rational expression. To find $A$ (domain) and $B$ (range), you must work from the inside out, respecting the constraints of $\cos^{-1}$.

The inverse cosine function $\cos^{-1}(t)$ is defined only for $t \in [-1, 1]$, and its output (range) is $[0, \pi]$. So the domain $A$ is all $x$ such that $\frac{2 - x^2}{2 + x^2}$ lies in $[-1, 1]$, and the range $B$ is the set of all possible $\cos^{-1}$ values that actually occur as $x$ varies over $A$.

Let’s work through it step by step.

1. Find the domain $A$. The inner expression is $g(x) = \frac{2 - x^2}{2 + x^2}$. Since $2 + x^2 > 0$ for all real $x$, the denominator never vanishes. We need $-1 \le g(x) \le 1$. First, note that $g(x)$ is an even function (replace $x$ with $-x$ gives the same value), so we can consider $x \ge 0$ and then include negatives symmetrically. Check the lower bound: $g(x) \ge -1$?

which is always true. So the lower bound is automatically satisfied.

Now the upper bound: $g(x) \le 1$?

which is also always true (equality at $x=0$). So $g(x)$ is always between $-1$ and $1$ for every real $x$.

Therefore, the domain $A$ is all real numbers: $A = \mathbb{R}$.

2. Find the range $B$. Since $A = \mathbb{R}$, the range $B$ is the set of all values $f(x) = \cos^{-1}(g(x))$ as $x$ runs over $\mathbb{R}$. Let’s examine what values $g(x)$ can take. As $x \to \pm\infty$,

At $x = 0$, $g(0) = \frac{2}{2} = 1$.

Since $g(x)$ is continuous, it takes all values between $-1$ and $1$ as $x$ varies from $0$ to $\infty$ (and symmetrically for negative $x$). In fact, for $x \ge 0$, $g(x)$ decreases from $1$ at $x=0$ to $-1$ as $x \to \infty$, covering every value in $(-1, 1]$ exactly once.

So the output of $\cos^{-1}$ is:

• When $g = 1$, $\cos^{-1}(1) = 0$.
• When $g = -1$, $\cos^{-1}(-1) = \pi$.
• As $g$ runs through all values in $(-1, 1)$, $\cos^{-1}(g)$ runs through all values in $(0, \pi)$. Therefore, the range $B$ is $[0, \pi]$.

3. Find $A \cap B$. $A = \mathbb{R}$ and $B = [0, \pi]$. Their intersection is simply $[0, \pi]$, since every number in $[0, \pi]$ is a real number. But note the options: (A) $[0, \sqrt{2})$, (B) $[0, \pi/2)$, (C) $[0, \pi)$, (D) $(-\pi/2, \pi/2)$. The interval $[0, \pi]$ is not exactly listed — option (C) is $[0, \pi)$, which is $[0, \pi)$ (open at $\pi$). Wait — is $\pi$ included in the range? Yes, because $g(x) = -1$ is achieved only in the limit as $x \to \pm\infty$, but not at any finite $x$. So $\cos^{-1}(-1) = \pi$ is not actually attained; the range is $[0, \pi)$, not $[0, \pi]$. Let’s verify: $g(x) = -1$ would require $2 - x^2 = -2 - x^2 \implies 2 = -2$, impossible. So $\pi$ is a supremum, not a maximum. Hence $B = [0, \pi)$. Therefore, $A \cap B = [0, \pi)$.

We are asked for the principal argument (the angle in $(-\pi, \pi]$) of

The key idea: convert $1+i$ and $1-i$ to polar form, then use exponent rules to combine them into one complex number whose argument is easy to read.

1. Polar forms of $1+i$ and $1-i$

• $1+i$ has modulus $\sqrt{1^2+1^2} = \sqrt{2}$ and argument $\frac{\pi}{4}$ (since it lies on the line $y=x$ in the first quadrant). So $1+i = \sqrt{2}\, e^{i\pi/4}$.
• $1-i$ has modulus $\sqrt{2}$ and argument $-\frac{\pi}{4}$ (fourth quadrant). So $1-i = \sqrt{2}\, e^{-i\pi/4}$.

2. Rewrite the fraction using these forms

3. Simplify the powers

• Numerator: $(\sqrt{2})^{2025} e^{i\cdot 2025\pi/4}$
• Denominator: $(\sqrt{2})^{2022} e^{-i\cdot 2022\pi/4}$

So the fraction becomes

4. Combine the exponent

$2025+2022 = 4047$, so the angle is $\frac{4047\pi}{4}$.

5. Reduce the angle modulo $2\pi$

We want the principal argument, so subtract multiples of $2\pi = \frac{8\pi}{4}$ until the angle lies in $(-\pi, \pi]$.

Divide $4047$ by $8$:

$8 \times 506 = 4048$, so $4047 = 8 \times 506 - 1$.

Thus

Since $506\cdot 2\pi$ is a full multiple of $2\pi$, the effective angle is $-\frac{\pi}{4}$.

6. But wait — we must also consider the modulus factor

The modulus $(\sqrt{2})^3 = 2\sqrt{2}$ is positive, so it does not affect the argument. The complex number is $2\sqrt{2}\, e^{-i\pi/4}$, whose argument is $-\pi/4$.

7. Check the options

$-\pi/4$ is option (A). However, let’s double-check: Did we handle the sign in the exponent correctly?

Original denominator exponent: $(e^{-i\pi/4})^{2022} = e^{-i\cdot 2022\pi/4}$. When we bring it to the numerator, the exponent becomes $+i\cdot 2022\pi/4$. So the total exponent is

which we reduced to $-i\pi/4$. That seems correct.

Thus the answer is $-\pi/4$.