📝 Inverse trigonometric functions arcsin arccos arctan (20 MCQs)

📖 From Calculus • 1. Basics before calculus • 20 questions available

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What is Inverse trigonometric functions arcsin arccos arctan?

Definition:
Inverse trigonometric functions, such as $\arcsin(x)$ , $\arccos(x)$ , and $\arctan(x)$ , return the angle whose sine, cosine, or tangent is $x$ , with principal value ranges: $[-\pi/2, \pi/2]$ , $[0, \pi]$ , and $(-\pi/2, \pi/2)$ , respectively.

Example:
$\arcsin(1/2) = \pi/6$ because $\sin(\pi/6) = 1/2$ ; $\arccos(0) = \pi/2$ ; $\arctan(1) = \pi/4$ .

Reason:
These functions are essential for solving trigonometric equations and in geometric applications like finding angles from side ratios.

Easy

Medium

Hard

📝 All Inverse trigonometric functions arcsin arccos arctan MCQs

Q1. If $\\sin \\theta = \\frac{1}{2}\$ and \ $\\theta\$ is in the first quadrant, what is \ $\\arcsin\\left(\\frac{1}{2}\\right)\$ ?

A.30° ✅

B.45°

C.60°

D.90°

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Since the sine of 30° equals 1/2 and the principal value of arcsine lies in \ $[-90°,90°]\$ , the inverse returns 30°. No other angle in the first quadrant gives the same sine value, making this the unique correct answer.

Q2. Given that \ $\\arctan x = \\frac{\\pi}{4}\$ , which of the following must be true about \ $x\$ ?

A.x = 1 ✅

B.x = 0

C.x = -1

D.x is undefined

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The tangent of \ $\\frac{\\pi}{4}\$ is 1, so the equation \ $\\tan(\\arctan x)=x\$ forces \ $x=1\$ . The other choices do not satisfy the given arctangent value, confirming the necessity of \ $x=1\$ .

Q3. Suppose \ $\\theta\$ satisfies \ $\\sin \\theta = \\frac{3}{5}\$ with \ $\\theta\$ in the second quadrant. What is the value of \ $\\arccos\\left(-\\frac{4}{5}\\right)\$ ?

A.\

\\frac{\\pi}{6}\

B.\

\\frac{2\\pi}{3}\

C.\

\\frac{3\\pi}{5}\

D.\

\\pi - \\arcsin\\frac{3}{5}\

✅

💡 Difficulty: hard | ✅ Correct: D

📖 Explanation: In the second quadrant, cosine is negative while sine is positive. The angle whose cosine is \ $-\\frac{4}{5}\$ therefore has sine \ $\\frac{3}{5}\$ . Its principal arccos value equals \ $\\pi - \\arcsin(\\frac{3}{5})\$ , matching option D.

Q4. If \ $\\arccos y = 0\$ , what is the value of \ $y\$ ?

A.0

B.1 ✅

C.-1

D.undefined

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The arccosine function returns the angle whose cosine equals the argument. Cosine of 0 radians is 1, so the only y satisfying \ $\\arccos y = 0\$ is \ $y=1\$ . No other value yields a zero arccosine.

Q5. For a real number \ $z\$ , the identity \ $\\arcsin z + \\arccos z = \\frac{\\pi}{2}\$ holds. Which of the following statements is a logical consequence?

A.Both arcsin and arccos are defined for all real \

z\

B.The sum is constant only when \

|z|\\le1\

. ✅

C.The identity fails for \

z=0\

D.None of the above.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The identity follows directly from the definitions of arcsine and arccosine, but it requires the argument to lie within their common domain \ $[-1,1]\$ . Hence the sum being constant is guaranteed only for \ $|z|\\le1\$ , making statement B correct.

Q6. If \ $\\arctan a + \\arctan b = \\frac{\\pi}{4}\$ and \ $ab=1\$ , which of the following must be true about \ $a\$ and \ $b\$ ?

A.a = b = 1

B.a = b = \\sqrt{2}-1

C.a = \\frac{1}{b} ✅

D.a = b = \\frac{\\sqrt{2}}{2}

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: The condition \ $ab=1\$ directly implies \ $a = \\frac{1}{b}\$ . While the sum condition imposes additional constraints, the only statement that must always hold regardless of those constraints is the reciprocal relationship, option C.

Q7. Compare the domains of \ $\\arcsin x\$ and \ $\\arccos x\$ . Which statement is correct?

A.Both have domain \

(-\\infty,\\infty)\

B.Both have domain \

[-1,1]\

. ✅

C.\

\\arcsin\

domain is larger.

D.\

\\arccos\

domain is larger.

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Both inverse functions are defined only for inputs whose absolute value does not exceed one, because they invert the sine and cosine functions restricted to \ $[-1,1]\$ . Consequently, their domains coincide as \ $[-1,1]\$ , validating option B.

Q8. Evaluate which of the following statements correctly differentiates \ $\\arcsin x\$ and \ $\\arctan x\$ .

A.\

(\\arcsin x)&#039; = \\frac{1}{1+x^2}\

and \

(\\arctan x)&#039; = \\frac{1}{\\sqrt{1-x^2}}\

B.Both derivatives are \

\\frac{1}{1+x^2}\

C.\

(\\arcsin x)&#039; = \\frac{1}{\\sqrt{1-x^2}}\

and \

(\\arctan x)&#039; = \\frac{1}{1+x^2}\

. ✅

D.Derivatives are swapped.

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The derivative of arcsine follows from implicit differentiation of \ $\\sin(y)=x\$ , giving \ $1/\\sqrt{1-x^2}\$ . The derivative of arctangent comes from \ $\\tan(y)=x\$ , yielding \ $1/(1+x^2)\$ . Option C states these formulas correctly.

Q9. Consider the composite function \ $f(x)=\\arcsin(\\sin x)\$ . Which of the following best describes its behavior on the interval \ $[-\\pi, \\pi]\$ ?

A.f(x)=x for all x in interval.

B.f(x)=\\pi - x for \

x\\in[\\frac{\\pi}{2},\\pi]\

C.f(x)= -\\pi - x for \

x\\in[-\\pi,-\\frac{\\pi}{2}]\

D.Piecewise defined as \

f(x)=x\

for \

|x|\\le\\frac{\\pi}{2}\

and \

f(x)=\\pi - x\

for \

x\\in[\\frac{\\pi}{2},\\pi]\

and \

-\\pi - x\

for \

x\\in[-\\pi,-\\frac{\\pi}{2}]\

. ✅

💡 Difficulty: hard | ✅ Correct: D

📖 Explanation: The arcsine function returns the principal value in \ $[-\\frac{\\pi}{2},\\frac{\\pi}{2}]\$ . When \ $x\$ lies outside this interval, the sine repeats, and the inverse maps back to the principal range, producing the piecewise expression given in option D.

Q10. Which inverse trigonometric function has a range of \ $(0,\\pi)\$ ?

A.arcsin

B.arctan

C.arccot

D.arccos ✅

💡 Difficulty: easy | ✅ Correct: D

📖 Explanation: By definition, the arccosine function outputs angles between 0 and \ $\\pi\$ (exclusive of the endpoints for the principal value). The other listed inverses have ranges that are symmetric about the origin or lie in different intervals, confirming arccos as the correct choice.

Q11. If \ $g(x)=\\arctan\\left(\\frac{2x}{1-x^2}\\right)\$ , for \ $|x|<1\$ , what is \ $g(x)\$ equal to?

A.2\\arctan x ✅

B.\\arcsin x

C.\\arccos x

D.\\arctan x

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Using the tangent double‑angle identity \ $\\tan(2\\theta)=\\frac{2\\tan\\theta}{1-\\tan^2\\theta}\$ , we see that \ $\\frac{2x}{1-x^2}=\\tan(2\\arctan x)\$ . Hence the arctangent of that expression returns \ $2\\arctan x\$ .

Q12. Determine the limit \ $\\displaystyle\\lim_{x\\to 0}\\frac{\\arcsin x - \\arctan x}{x^3}\$ .

A.\\frac{1}{6}

B.\\frac{1}{3}

C.\\frac{1}{2} ✅

D.0

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Expanding both functions in Maclaurin series: \ $\\arcsin x = x + \\frac{x^3}{6}+\\dots\$ and \ $\\arctan x = x - \\frac{x^3}{3}+\\dots\$ . Their difference is \ $\\frac{x^3}{2}+\\dots\$ . Dividing by \ $x^3\$ and taking the limit yields \ $\\frac{1}{2}\$ .

Q13. Explain why \ $\\arcsin(\\sin \\theta)\$ does not always equal \ $\\theta\$ .

A.Because sine is not injective.

B.Because arcsin is undefined for some values.

C.Because domains differ.

D.Both A and C. ✅

💡 Difficulty: easy | ✅ Correct: D

📖 Explanation: The sine function repeats its values every \ $2\\pi\$ and is symmetric about \ $\\pi/2\$ , so it fails the one‑to‑one requirement needed for an inverse. Consequently, \ $\\arcsin\$ can only recover the original angle when \ $\\theta\$ lies within its principal range, making statements A and C jointly correct.

Q14. Which of the following best describes the relationship between \ $\\arccot x\$ and \ $\\arctan x\$ ?

A.\\arccot x = \\frac{\\pi}{2} - \\arctan x ✅

B.\\arccot x = \\arctan x

C.\\arccot x = \\pi - \\arctan x

D.No simple relationship

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: By definition, cotangent is the reciprocal of tangent, and the principal values are chosen so that \ $\\arccot x\$ plus \ $\\arctan x\$ sum to \ $\\frac{\\pi}{2}\$ . This identity holds for all real \ $x\$ , confirming option A.

Q15. A particle moves so that its angle \ $\\theta(t)\$ satisfies \ $\\theta(t)=\\arctan\\left(\\frac{t}{\\sqrt{1-t^2}}\\right)\$ for \ $|t|<1\$ . Which of the following statements about \ $\\theta(t)\$ is true?

A.\\theta(t)=\\arcsin t ✅

B.\\theta(t)=\\arccos t

C.\\theta(t)=\\arctan t

D.\\theta(t)=\\arcsin(\\sqrt{1-t^2})

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Using the identity \ $\\tan(\\arcsin t)=\\frac{t}{\\sqrt{1-t^2}}\$ , the given expression is precisely \ $\\arctan\$ of that ratio, which equals \ $\\arcsin t\$ . Hence \ $\\theta(t)\$ simplifies to \ $\\arcsin t\$ .

Q16. Which inverse trigonometric function is odd?

A.arcsin

B.arccos

C.arctan ✅

D.arccot

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: An odd function satisfies \ $f(-x)=-f(x)\$ . Both \ $\\arcsin\$ and \ $\\arctan\$ have this property, but among the listed choices, \ $\\arctan\$ is explicitly odd, making option C the correct answer.

Q17. If \ $\\alpha = \\arcsin\\left(\\frac{3}{5}\\right)\$ and \ $\\beta = \\arccos\\left(\\frac{4}{5}\\right)\$ , what is \ $\\alpha + \\beta\$ ?

A.\\frac{\\pi}{2} ✅

B.\\frac{\\pi}{3}

C.\\frac{\\pi}{4}

D.\\pi

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Since \ $\\sin\\alpha = \\frac{3}{5}\$ and \ $\\cos\\beta = \\frac{4}{5}\$ , we have \ $\\sin\\alpha = \\cos\\beta\$ . For complementary angles, the sum equals \ $\\frac{\\pi}{2}\$ . Therefore \ $\\alpha+\\beta = \\frac{\\pi}{2}\$ .

Q18. Let \ $h(x)=\\arcsin x + \\arccos x\$ . Which of the following statements about the derivative \ $h'(x)\$ is correct for \ $-1<x<1\$ ?

A.h'(x)=0 ✅

B.h'(x)=\\frac{2}{\\sqrt{1-x^2}}

C.h'(x)=-\\frac{2}{\\sqrt{1-x^2}}

D.Derivative does not exist

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Differentiating each term gives \ $(\\arcsin x)' = 1/\\sqrt{1-x^2}\$ and \ $(\\arccos x)' = -1/\\sqrt{1-x^2}\$ . Adding them cancels the terms, leaving a derivative of zero throughout the open interval, confirming option A.

Q19. What is the principal value range of \ $\\arctan x\$ ?

A.(-\\pi,\\pi)

B.(-\\frac{\\pi}{2},\\frac{\\pi}{2}) ✅

C.[0,\\pi]

D.(0,2\\pi)

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The arctangent function is defined to return the unique angle whose tangent equals \ $x\$ and lies between \ $-\\frac{\\pi}{2}\$ and \ $\\frac{\\pi}{2}\$ (excluding the endpoints). This interval is the standard principal range for \ $\\arctan\$ .

Q20. The inverse function of \ $\\sec x\$ (restricted to its principal domain) is denoted by:

A.arcsec x ✅

B.arccot x

C.arcsin x

D.arccos x

💡 Difficulty: medium | ✅ Correct: A

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