Welcome to the IIT JEE Trigonometry Practice Quiz – May Series. This quiz is specifically curated for engineering aspirants looking to strengthen their grasp on trigonometric concepts through application-based problems.
This session covers a variety of topics including:
- Compound Angles and Multiple Angle Identities
- Trigonometric Equations and General Solutions
- Inverse Trigonometric Functions (ITF)
- Properties and Solutions of Triangles
The difficulty level is set to Moderate, focusing on multi-step reasoning and common JEE patterns. Ensure you have a pen and paper ready for calculations. Good luck!
Q1. The value of the product ##\cos 20^\circ \cos 40^\circ \cos 80^\circ## is equal to:
Using the product formula ##\prod_{k=0}^{n-1} \cos(2^k \theta) = \frac{\sin(2^n \theta)}{2^n \sin \theta}##, with ##\theta = 20^\circ## and ##n=3##, we get ##\frac{\sin 160^\circ}{8 \sin 20^\circ} = \frac{\sin 20^\circ}{8 \sin 20^\circ} = 1/8##.
Q2. The general solution of the equation ##\sqrt{3} \sin x + \cos x = 2## is:
Divide by 2: ##\frac{\sqrt{3}}{2} \sin x + \frac{1}{2} \cos x = 1##, which is ##\cos(x - \pi/3) = 1##. Thus ##x - \pi/3 = 2n\pi##, leading to ##x = 2n\pi + \pi/3##.
Q3. If ##\tan \alpha = 1/2## and ##\tan \beta = 1/3##, then the value of ##\alpha + \beta## (where ##\alpha, \beta## are acute) is:
Using the compound angle formula ##\tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} = \frac{1/2 + 1/3}{1 - (1/2)(1/3)} = \frac{5/6}{5/6} = 1##. Since they are acute, ##\alpha + \beta = \pi/4##.
Q4. The range of the function ##f(x) = \sin^6 x + \cos^6 x## is:
##\sin^6 x + \cos^6 x = (\sin^2 x + \cos^2 x)^3 - 3 \sin^2 x \cos^2 x (\sin^2 x + \cos^2 x) = 1 - \frac{3}{4} \sin^2 2x##. Since ##0 \le \sin^2 2x \le 1##, the range is ##[1 - 3/4, 1] = [1/4, 1]##.
Q5. The number of real solutions of the equation ##\sin x = x/10## is:
By plotting ##y = \sin x## and ##y = x/10##, we see they intersect at ##x=0##, 3 times in the positive region (since ##3\pi < 10 < 4\pi##), and 3 times in the negative region. Total = 1 + 3 + 3 = 7.
Q6. The value of ##\tan 20^\circ + \tan 40^\circ + \sqrt{3} \tan 20^\circ \tan 40^\circ## is:
Since ##\tan(20^\circ + 40^\circ) = \tan 60^\circ = \sqrt{3}##, we have ##\frac{\tan 20^\circ + \tan 40^\circ}{1 - \tan 20^\circ \tan 40^\circ} = \sqrt{3}##, which simplifies to ##\tan 20^\circ + \tan 40^\circ + \sqrt{3} \tan 20^\circ \tan 40^\circ = \sqrt{3}##.
Q7. In a triangle ABC, if ##a \cos A = b \cos B##, then the triangle is:
##a \cos A = b \cos B \implies \sin A \cos A = \sin B \cos B \implies \sin 2A = \sin 2B##. This implies ##2A = 2B## or ##2A = \pi - 2B##, so ##A=B## (isosceles) or ##A+B = \pi/2## (right-angled).
Q8. The maximum value of ##3 \sin x - 4 \cos x + 7## is:
The expression ##a \sin x + b \cos x## has a maximum value of ##\sqrt{a^2 + b^2}##. Here, ##\sqrt{3^2 + (-4)^2} = 5##. Thus, the maximum value of the function is ##5 + 7 = 12##.
Q9. The value of ##\tan^{-1}(1) + \tan^{-1}(2) + \tan^{-1}(3)## is:
Using ##\tan^{-1} x + \tan^{-1} y = \pi + \tan^{-1} \frac{x+y}{1-xy}## for ##xy > 1##: ##\tan^{-1}(2) + \tan^{-1}(3) = \pi + \tan^{-1} \frac{5}{1-6} = \pi + \tan^{-1}(-1) = \pi - \pi/4 = 3\pi/4##. Adding ##\tan^{-1}(1) = \pi/4## gives ##\pi##.
Q10. The principal value of ##\sin^{-1}(\sin 10)## is:
The value 10 radians lies in the range ##[3\pi - \pi/2, 3\pi + \pi/2]##. For ##\sin^{-1}(\sin x)## where ##x## is near ##3\pi##, the value is ##3\pi - x##. Since ##3\pi \approx 9.42##, ##3\pi - 10 \approx -0.58##, which is within ##[-\pi/2, \pi/2]##.
Q11. If ##\sin x + \sin^2 x = 1##, then the value of ##\cos^2 x + \cos^4 x## is:
From ##\sin x + \sin^2 x = 1##, we get ##\sin x = 1 - \sin^2 x = \cos^2 x##. Squaring both sides, ##\sin^2 x = \cos^4 x##. Therefore, ##\cos^2 x + \cos^4 x = \sin x + \sin^2 x = 1##.
Q12. The period of the function ##f(x) = | \sin x | + | \cos x |## is:
The function is symmetric and repeats every ##\pi/2## because ##f(x + \pi/2) = | \sin(x+\pi/2) | + | \cos(x+\pi/2) | = | \cos x | + | -\sin x | = f(x)##. No smaller period exists.
Q13. The value of ##\cos \frac{\pi}{7} \cos \frac{2\pi}{7} \cos \frac{4\pi}{7}## is:
Using the formula ##\frac{\sin(2^n \theta)}{2^n \sin \theta}## with ##\theta = \pi/7## and ##n=3##, we get ##\frac{\sin(8\pi/7)}{8 \sin(\pi/7)} = \frac{-\sin(\pi/7)}{8 \sin(\pi/7)} = -1/8##.
Q14. The value of ##\sum_{n=1}^{\infty} \tan^{-1} \left( \frac{1}{1 + n + n^2} \right)## is:
The term is ##\tan^{-1}(n+1) - \tan^{-1}(n)##. This is a telescoping series: ##(\tan^{-1} 2 - \tan^{-1} 1) + (\tan^{-1} 3 - \tan^{-1} 2) + \dots##. The sum up to ##k## is ##\tan^{-1}(k+1) - \tan^{-1} 1##. As ##k \to \infty##, it becomes ##\pi/2 - \pi/4 = \pi/4##.
Q15. If ##\alpha, \beta## are the roots of ##a \cos \theta + b \sin \theta = c##, then ##\tan(\alpha + \beta)## is:
Using ##\tan(\theta/2)## substitution, we find ##\tan(\frac{\alpha+\beta}{2}) = b/a##. Then ##\tan(\alpha+\beta) = \frac{2 \tan((\alpha+\beta)/2)}{1 - \tan^2((\alpha+\beta)/2)} = \frac{2(b/a)}{1 - (b/a)^2} = \frac{2ab}{a^2 - b^2}##.
Q16. The domain of the function ##f(x) = \sqrt{\log_{1/2} (\sin x)}## for ##x \in [0, 2\pi]## is:
For the square root, ##\log_{1/2} (\sin x) \ge 0##, which means ##0 < \sin x \le (1/2)^0 = 1##. This is satisfied whenever ##\sin x## is positive. In ##[0, 2\pi]##, this occurs for ##x \in (0, \pi)##.
Q17. The value of ##\tan 75^\circ - \cot 75^\circ## is:
##\tan 75^\circ = 2 + \sqrt{3}## and ##\cot 75^\circ = 2 - \sqrt{3}##. Subtracting them gives ##(2 + \sqrt{3}) - (2 - \sqrt{3}) = 2\sqrt{3}##.
Q18. The number of solutions of ##\tan x + \sec x = 2 \cos x## in the interval ##[0, 2\pi]## is:
Multiply by ##\cos x##: ##\sin x + 1 = 2 \cos^2 x = 2(1 - \sin^2 x)##. Thus ##2 \sin^2 x + \sin x - 1 = 0##, which factors to ##(2 \sin x - 1)(\sin x + 1) = 0##. ##\sin x = 1/2## gives ##x = \pi/6, 5\pi/6##. ##\sin x = -1## gives ##x = 3\pi/2##, but ##\sec x## is undefined there. So, 2 solutions.
Q19. The maximum value of ##\sin^2 \theta + \cos^4 \theta## is:
Let ##y = \sin^2 \theta + \cos^4 \theta = 1 - \cos^2 \theta + \cos^4 \theta##. Let ##u = \cos^2 \theta \in [0, 1]##. Then ##y = u^2 - u + 1##. This parabola opens upwards with vertex at ##u=1/2## (value 3/4). At endpoints ##u=0, 1##, ##y=1##. Max value is 1.
Q20. The value of ##\cos^2 10^\circ - \cos 10^\circ \cos 50^\circ + \cos^2 50^\circ## is:
Using ##\cos^2 \theta = (1+\cos 2\theta)/2## and ##2 \cos A \cos B = \cos(A+B) + \cos(A-B)##, the expression simplifies to ##\frac{1}{2} [1 + \cos 20^\circ + 1 + \cos 100^\circ - (\cos 60^\circ + \cos 40^\circ)] = 1 + \frac{1}{2} [\cos 20^\circ + \cos 100^\circ - 1/2 - \cos 40^\circ] = 1 - 1/4 = 3/4##.
Q21. In triangle ABC, if ##\sin A + \sin B + \sin C = \frac{3\sqrt{3}}{2}##, then the triangle is:
In any triangle, ##\sin A + \sin B + \sin C \le \frac{3\sqrt{3}}{2}##. Equality holds only when ##A = B = C = 60^\circ##, meaning the triangle is equilateral.
Q22. The value of ##\sin 10^\circ \sin 30^\circ \sin 50^\circ \sin 70^\circ## is:
##\sin 30^\circ = 1/2##. The rest is ##\sin 10^\circ \sin(60-10)^\circ \sin(60+10)^\circ = \frac{1}{4} \sin 30^\circ = 1/8##. Multiplying by the initial ##\sin 30^\circ## gives ##(1/2) \times (1/8) = 1/16##.
Q23. The smallest positive root of the equation ##\tan x - x = 0## lies in the interval:##(0, \pi/2)##
For ##x \in (0, \pi/2)##, ##\tan x > x##. There is no root in ##(\pi/2, \pi)## as ##\tan x## is negative. In ##(\pi, 3\pi/2)##, ##\tan x## goes from 0 to ##\infty## while ##x## is positive, allowing for an intersection.
Q24. The number of solutions of ##\sin^2 \theta + 3 \cos \theta = 3## in ##[0, 2\pi]## is:
##(1 - \cos^2 \theta) + 3 \cos \theta = 3 \implies \cos^2 \theta - 3 \cos \theta + 2 = 0 \implies (\cos \theta - 1)(\cos \theta - 2) = 0##. Since ##\cos \theta \neq 2##, we have ##\cos \theta = 1##. In ##[0, 2\pi]##, solutions are ##0## and ##2\pi##. Total = 2.
Q25. The value of ##\tan 1^\circ \tan 2^\circ \tan 3^\circ \dots \tan 89^\circ## is:
Terms can be paired as ##\tan \theta \cdot \tan(90-\theta) = \tan \theta \cdot \cot \theta = 1##. Pairing ##\tan 1^\circ## with ##\tan 89^\circ##, ##\tan 2^\circ## with ##\tan 88^\circ##, etc., leaves ##\tan 45^\circ = 1## in the middle. All pairs multiply to 1.
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