Questions: Inverse Trigonometric Functions
Inverse Trigonometric Functions26 min readLesson 4 of 5
Q1. Find the principal value of sin inverse of -1/2.
Answer (English): The principal branch of sine inverse is from minus pi/2 to pi/2. Since sine of minus pi/6 is minus 1/2, and minus pi/6 lies inside the branch, the principal value is minus pi/6.
Hinglish: sin inverse ka branch [-pi/2, pi/2] hai. sin(-pi/6) = -1/2, aur -pi/6 branch ke andar hai, isliye principal value -pi/6 hai.
Q2. Evaluate cos inverse of -1/2 using the principal branch.
Answer (English): The branch for cosine inverse is [0, pi]. Since cosine of 2pi/3 is minus 1/2, the principal value is 2pi/3. Writing pi/3 would ignore the minus sign, because cos pi/3 is plus 1/2.
Hinglish: cos inverse ka branch [0, pi] hai. cos(2pi/3) = -1/2, isliye answer 2pi/3. Sign ignore karke pi/3 likhna galti hogi.
Q3. Evaluate tan inverse of 1 plus cot inverse of 1.
Answer (English): The complementary identity for tangent and cotangent gives pi/2 for every real input. So tan inverse 1 plus cot inverse 1 is pi/2. Checking directly, tan inverse 1 is pi/4 and cot inverse 1 is also pi/4, and the sum is pi/2.
Hinglish: tan inverse x + cot inverse x = pi/2 har real x ke liye. Yahan dono 1 hain, to answer pi/2. Direct check: pi/4 + pi/4 = pi/2.
Q4. Find the domain of sin inverse of (2x - 1).
Answer (English): The argument must lie between -1 and 1. So -1 <= 2x - 1 <= 1. Adding 1 to all parts gives 0 <= 2x <= 2, and dividing by 2 gives 0 <= x <= 1. The domain is the closed interval [0, 1].
Hinglish: Argument ko -1 aur 1 ke beech rakho: -1 <= 2x - 1 <= 1. Isse 0 <= 2x <= 2 aur phir 0 <= x <= 1 milta hai. Domain [0, 1].
Q5. Simplify sin inverse of sin of 5pi/6.
Answer (English): The inner angle 5pi/6 is outside the branch [-pi/2, pi/2]. But sin(5pi/6) equals sin(pi/6), and pi/6 is inside the branch. Therefore the value is pi/6, not 5pi/6.
Hinglish: 5pi/6 branch ke andar nahi hai. Par sin(5pi/6) = sin(pi/6), aur pi/6 andar hai, isliye answer pi/6. Bina check kiye 5pi/6 likhna galat hai.
Q6. Show that tan inverse 1/2 plus tan inverse 1/3 equals pi/4.
Answer (English): Apply the tangent sum formula. The sum is tan inverse of ((1/2 + 1/3)/(1 - (1/2)(1/3))), that is tan inverse of ((5/6)/(5/6)), which is tan inverse of 1. The principal value of tan inverse 1 is pi/4. The product of the two inputs is 1/6, which is below 1, so the formula applies directly.
Hinglish: tan inverse sum formula lagao. (1/2 + 1/3) / (1 - 1/6) = (5/6)/(5/6) = 1. tan inverse 1 = pi/4. Product 1/6 < 1 hai, isliye formula valid hai.
Q7. Prove cos inverse of x plus cos inverse of -x equals pi for x in [-1, 1].
Answer (English): Use the negative angle result cos inverse(-x) = pi - cos inverse x. Adding cos inverse x to both sides gives cos inverse x plus cos inverse(-x) = pi. The identity holds for every x in the closed interval [-1, 1], including the endpoints where both values are defined.
Hinglish: Negative angle result: cos inverse(-x) = pi - cos inverse x. Isko jodo to sum pi aata hai. Ye identity [-1, 1] ke har x ke liye chalti hai.
Q8. Solve tan inverse 2x plus tan inverse 3x equals pi/4.
Answer (English): Take tangent on both sides. The left side gives (5x)/(1 - 6x^2) and the right side gives 1, so 6x^2 + 5x - 1 = 0. Factoring gives (6x - 1)(x + 1) = 0, so x = 1/6 or x = -1. For x = -1 both inverse tangent values are negative, so the sum cannot equal pi/4. The only solution is x = 1/6.
Hinglish: Tangent lene par (5x)/(1 - 6x^2) = 1, isliye 6x^2 + 5x - 1 = 0. Roots 1/6 aur -1. x = -1 par dono tan inverse negative hote hain, sum pi/4 nahi ho sakta. Answer x = 1/6.
Q9. Find the value of tan inverse of tan of 3pi/4.
Answer (English): The branch of tangent inverse is the open interval from minus pi/2 to pi/2, and 3pi/4 is outside it. Since tan(3pi/4) equals tan(-pi/4), and -pi/4 lies inside the branch, the value is -pi/4.
Hinglish: 3pi/4 branch ke bahar hai. tan(3pi/4) = tan(-pi/4), aur -pi/4 andar hai, isliye answer -pi/4.
Q10. Write sin inverse x in terms of cos inverse x, and give the valid interval.
Answer (English): The complementary identity gives sin inverse x = pi/2 - cos inverse x for x in the closed interval [-1, 1]. Both sides are defined on exactly that interval, and at the endpoints pi/2 - pi = -pi/2 and pi/2 - 0 = pi/2 match the sine inverse values at minus 1 and 1.
Hinglish: sin inverse x + cos inverse x = pi/2, isliye sin inverse x = pi/2 - cos inverse x. Interval [-1, 1] hai. Endpoints par bhi values match karti hain.
Mistakes to avoid in this chapter
Three errors appear again and again. Writing sin inverse as 1 over sin is a notation mistake, not a value mistake. Forgetting the branch when evaluating sin inverse of a sine or tan inverse of a tangent gives answers outside the allowed interval. Dropping a candidate solution without checking its sign gives an answer that cannot satisfy the original equation. Before every answer, ask which branch the question uses, then check that the final angle sits inside it. A useful last step is to substitute the answer back into the original expression; if the value does not match the branch, the answer is wrong even when the algebra is clean. Keep a one-line check at the end of every answer: state the branch and confirm the final angle lies inside it.
Quiz below: five multiple choice questions on principal values, complementary identities, domains and simplification. Attempt them after the formula sheet and the two topic pages.