Questions: Trigonometric Functions
Trigonometric Functions26 min readLesson 6 of 7
Q1. Convert 150 degrees into radian measure.
Answer: Multiply by pi over 180. So 150 degrees becomes 150 pi over 180, which simplifies to 5 pi by 6 radians. The simplification is part of the answer; writing 150 pi by 180 without reducing loses a mark in many schemes.
Hinglish: Degree ko radian me badalne ke liye pi/180 se multiply karo. 150 pi/180 ko simplify karke 5 pi/6 milta hai. Answer simplify karna zaroori hai.
Q2. A circle has radius 7 cm and an arc subtends an angle of 2 radians at the centre. Find the arc length.
Answer: The angle is already in radians, so apply l = r theta directly. The length is 7 times 2, which is 14 cm. If the angle had been given in degrees, the first step would have been a conversion.
Hinglish: Angle radian me hai, isliye l = r theta seedha lagao. l = 7 x 2 = 14 cm. Agar angle degree me hota to pehle convert karna padta.
Q3. Find sin x, given cos x = 3/5 and x in the fourth quadrant.
Answer: Use the Pythagorean identity: sin squared x = 1 minus 9 by 25, which is 16 by 25. So sine is plus or minus 4 by 5. In the fourth quadrant sine is negative, so sin x equals minus 4 by 5. The quadrant decides the sign; the identity alone cannot.
Hinglish: Identity se sin squared x = 16/25 milta hai, to sin x = plus/minus 4/5. Fourth quadrant me sine negative hota hai, isliye answer -4/5. Sign quadrant se aata hai.
Q4. Evaluate sin(31 pi by 3).
Answer: Strip whole turns first. The angle 31 pi by 3 equals 10 pi plus pi by 3, and 10 pi is five full turns, which change nothing. So the value is sin of pi by 3, which is square root of 3 by 2.
Hinglish: Pehle poore turns hatao. 31 pi/3 = 10 pi + pi/3, aur 10 pi ka koi effect nahi. Isliye sin(pi/3) = square root 3 / 2.
Q5. Find the value of cos 15 degrees.
Answer: Write 15 as 45 minus 30 and use the cosine subtraction formula. Then cos 15 equals cos 45 cos 30 plus sin 45 sin 30. Substituting the standard values gives (square root of 3 plus 1) over (2 root 2). The plus sign appears because the formula for a difference has a plus in the middle.
Hinglish: 15 = 45 - 30 likho aur cosine difference formula lagao: cos 45 cos 30 + sin 45 sin 30. Values daalne par (root 3 + 1)/(2 root 2) milta hai. Difference formula me plus sign aata hai.
Q6. Prove that sin 2x equals 2 sin x cos x.
Answer: Start from the addition result for sine: the sine of a sum is sin x cos y plus cos x sin y. Set y equal to x. Both products then collapse to sin x cos x, and adding them gives twice that product. This is the cleanest proof and it also shows why the double angle formula has no extra term.
Hinglish: Sine addition formula me y ki jagah x rakho. Dono products sin x cos x ban jaate hain, aur unko jodne par 2 sin x cos x milta hai. Yahi double angle formula ka proof hai, koi extra term nahi aata.
Q7. If sin x = 5/13 and x is in the second quadrant, find cos x and tan x.
Answer: From the identity, cos squared x equals 1 minus 25 by 169, that is 144 by 169, so cosine is plus or minus 12 by 13. The second quadrant makes cosine negative, so cos x equals minus 12 by 13. Then tan x equals sine over cosine, which is 5 by 13 divided by minus 12 by 13, that is minus 5 by 12.
Hinglish: Identity se cos squared = 144/169, to cos = plus/minus 12/13. Second quadrant me cos negative, isliye cos = -12/13. tan = sin/cos = -5/12.
Q8. Prove that cos 2x = 1 - 2 sin squared x.
Answer: Start from cos 2x equal to cos squared x minus sin squared x. Replace cos squared x by 1 minus sin squared x. The expression becomes 1 minus sin squared x minus sin squared x, which is 1 minus 2 sin squared x. This form is useful when the rest of the question contains sines only.
Hinglish: cos 2x = cos squared x - sin squared x se shuru karo. cos squared ko 1 - sin squared se replace karo, to 1 - 2 sin squared x milta hai. Ye form tab kaam aata hai jab sawal me sirf sine ho.
Q9. Find the general solution of sin x = 0.
Answer: On the unit circle, sine is the y coordinate, which is zero exactly on the x axis. That happens at 0, pi, 2 pi and also at minus pi, minus 2 pi and so on. Combined, the solutions are all integer multiples of pi, so x equals n pi where n is any integer.
Hinglish: Unit circle par sine y coordinate hai, jo x axis par zero hota hai. Ye 0, pi, 2 pi aur negative multiples par hota hai, isliye x = n pi, jahan n koi bhi integer ho.
Q10. Prove that (1 - cos 2x) / sin 2x = tan x.
Answer: Use the double angle forms. The numerator 1 minus cos 2x equals 2 sin squared x, and the denominator sin 2x equals 2 sin x cos x. The fraction becomes 2 sin squared x over 2 sin x cos x, and cancelling 2 sin x gives sin x over cos x, which is tan x. The cancellation is valid wherever sin x is not zero.
Hinglish: Double angle forms lagao: 1 - cos 2x = 2 sin squared x, aur sin 2x = 2 sin x cos x. Fraction me cancel karne par sin x / cos x = tan x. Jahan sin x zero ho wahan cancel nahi hota, isliye condition likhna achha hai.
Common mistakes on this chapter
Forgetting to convert degrees before using a radian formula is the first. Ignoring the quadrant when taking a square root is the second. Writing cos 2x equal to 2 cos x when the correct form is 2 cos squared x minus 1 is the third. Using the tangent addition formula without checking that the angles avoid odd multiples of pi by 2 is the fourth. Read the final answer against all four before closing the page.
Quiz
Quiz below: five multiple choice questions on radian measure, quadrant signs, identities and graphs. Attempt them after studying the notes above, then move to the previous year page.