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

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

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 $-\pi/4$.

We start with the given complex number:

The core idea is to first simplify the right-hand side into the standard form $a + ib$, so we can read off $x$ and $y$. Then we compute $\tan^{-1}(y/x)$, which gives an angle whose tangent is $y/x$. Finally, we plug that angle into the cosecant expression.

1. Simplify the denominator

   $(2 - i)^2 = 4 - 4i + i^2 = 4 - 4i - 1 = 3 - 4i$.

2. Divide the numerator by this result

Multiply numerator and denominator by the conjugate of the denominator, $3 + 4i$:

3. Compute the denominator

   $(3 - 4i)(3 + 4i) = 9 + 16 = 25$.

4. Compute the numerator

   $(1 + 7i)(3 + 4i) = 1\cdot3 + 1\cdot4i + 7i\cdot3 + 7i\cdot4i$

   $= 3 + 4i + 21i + 28i^2$

   $= 3 + 25i - 28$ (since $i^2 = -1$)

   $= -25 + 25i$.

5. So the complex number becomes

Hence $x = -1$, $y = 1$.

6. Find $\tan^{-1}(y/x)$

The principal value of $\tan^{-1}(-1)$ is $-\pi/4$. But note: the point $(-1, 1)$ lies in the second quadrant (since $x$ negative, $y$ positive). The principal value of $\tan^{-1}$ gives an angle in $(-\pi/2, \pi/2)$, which is $-\pi/4$ — but that corresponds to the point $(-1, -1)$, not $(-1, 1)$.

The problem writes $\tan^{-1}(y/x)$ without specifying a branch. In standard Indian exam usage, $\tan^{-1}$ denotes the principal value, which lies in $(-\pi/2, \pi/2)$. So $\tan^{-1}(-1) = -\pi/4$.

But then the expression becomes $\csc(-\pi/4 - \pi/4) = \csc(-\pi/2) = -1$, which is option (C). However, let's check if the intended interpretation uses the actual angle of the complex number.

If we instead take the argument of $(-1, i)$ as $3\pi/4$, then $\tan^{-1}(y/x)$ would be $3\pi/4$ (though this is not the principal value). Then the expression becomes $\csc(3\pi/4 - \pi/4) = \csc(\pi/2) = 1$, option (A).

Which one is correct? The problem likely expects the principal value, but let's verify by a direct substitution approach.

7. Alternate approach: compute $\tan^{-1}(y/x)$ using the complex number's argument

   Since $x + iy = -1 + i$, its argument $\theta$ satisfies $\tan\theta = y/x = -1$, and $\theta$ is in the second quadrant, so $\theta = 3\pi/4$. Many textbooks define $\tan^{-1}(y/x)$ as the argument of the complex number when $x$ and $y$ are given, not the principal value. In fact, the notation $\tan^{-1}(y/x)$ in the context of a complex number $x+iy$ often means the angle whose tangent is $y/x$ and which lies in the correct quadrant.

   Tip

   When $x$ and $y$ are given from a complex number, $\tan^{-1}(y/x)$ is usually taken as the argument of that complex number, i.e., $\arg(x+iy)$. For $(-1,1)$, that's $3\pi/4$.

   So we take $\tan^{-1}(y/x) = 3\pi/4$.

8. Now evaluate the cosecant

Thus the expression simplifies to $1$.