Trigonometry for JEE Main and Advanced – May Series

Using the property ##\tan(90^\circ - \theta) = \cot \theta##, we can pair ##\tan 1^\circ## with ##\tan 89^\circ## to get ##\tan 1^\circ \cot 1^\circ = 1##. All terms pair up except ##\tan 45^\circ##, which is also 1, resulting in a total product of 1.

By the AM-GM inequality, ##\frac{9 \tan^2 x + 4 \cot^2 x}{2} \ge \sqrt{9 \tan^2 x \cdot 4 \cot^2 x}##. Thus, ##9 \tan^2 x + 4 \cot^2 x \ge 2 \sqrt{36} = 12##.

For an expression of the form ##a \sin x + b \cos x##, the range is ##[-\sqrt{a^2+b^2}, \sqrt{a^2+b^2}]##. Here, ##\sqrt{3^2+4^2} = 5##, so the range is ##[-5, 5]##.

Group the terms: ##(\cos 12^\circ + \cos 156^\circ) + (\cos 84^\circ + \cos 228^\circ)##. Applying ##\cos A + \cos B = 2 \cos\frac{A+B}{2} \cos\frac{A-B}{2}##, we get ##2 \cos 84^\circ \cos 72^\circ + 2 \cos 156^\circ \cos 72^\circ = 2 \cos 72^\circ (\cos 84^\circ + \cos 156^\circ) = 4 \cos 72^\circ \cos 120^\circ \cos 36^\circ##. Since ##\cos 120^\circ = -1/2## and ##2 \sin 18^\circ \cos 36^\circ = 1/2##, the result is ##-1/2##.

The equation ##\tan^2 \theta = 3## implies ##\tan^2 \theta = \tan^2(\pi/3)##. The general solution for ##\tan^2 \theta = \tan^2 \alpha## is ##\theta = n\pi \pm \alpha##.

Using the identity ##\sin \theta \sin(60^\circ - \theta) \sin(60^\circ + \theta) = \frac{1}{4} \sin 3\theta## with ##\theta = 10^\circ##, we get ##\sin 10^\circ \sin 50^\circ \sin 70^\circ = \frac{1}{4} \sin 30^\circ = 1/8##. Including the existing ##\sin 30^\circ = 1/2##, the product is ##1/16##.

##f(x) = (\sin^2 x + \cos^2 x)^2 - 2 \sin^2 x \cos^2 x = 1 - \frac{1}{2} \sin^2 2x = 1 - \frac{1}{4}(1 - \cos 4x) = \frac{3}{4} + \frac{1}{4} \cos 4x##. The period of ##\cos 4x## is ##2\pi/4 = \pi/2##.

The equation ##\sin x = \cos x## implies ##\tan x = 1##. In the interval ##[0, 2\pi]##, ##\tan x = 1## at ##x = \pi/4## and ##x = 5\pi/4##.

The domain of ##\sin^{-1} x## and ##\cos^{-1} x## is ##[-1, 1]##. In this domain, ##\sin^{-1} x + \cos^{-1} x = \pi/2##. Since ##x \in [-1, 1]##, ##\tan^{-1} x \in [-\pi/4, \pi/4]##. Thus, the range is ##\pi/2 + [-\pi/4, \pi/4] = [\pi/4, 3\pi/4]##.

Using the formula ##\prod_{k=0}^{n-1} \cos(2^k A) = \frac{\sin(2^n A)}{2^n \sin A}## with ##n=3## and ##A=\pi/7##, we get ##\frac{\sin(8\pi/7)}{8 \sin(\pi/7)} = \frac{-\sin(\pi/7)}{8 \sin(\pi/7)} = -1/8##.

For any triangle, ##A+B+C = \pi##. From the identity ##\tan(A+B+C) = \frac{\sum \tan A - \prod \tan A}{1 - \sum \tan A \tan B}##, since ##\tan \pi = 0##, it follows that ##\sum \tan A = \prod \tan A##.

Dividing by ##\sqrt{2}##, we get ##\sin(x - \pi/4) = 1/\sqrt{2}##. The smallest positive solutions for ##x - \pi/4## are ##\pi/4## and ##3\pi/4##, giving ##x = \pi/2## and ##x = \pi##. The smallest is ##\pi/2##.

##\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}##.

The equation can be written as ##\sqrt{2} \sin(x + \pi/4) = \sqrt{2}##, which means ##\sin(x + \pi/4) = 1##. This implies ##x + \pi/4 = 2n\pi + \pi/2##, so ##x = 2n\pi + \pi/4##.

Expanding ##3 \cos(\theta + \pi/3)## and combining terms, the expression becomes ##6.5 \cos \theta - 1.5\sqrt{3} \sin \theta + 3##. The amplitude is ##\sqrt{6.5^2 + (1.5\sqrt{3})^2} = \sqrt{42.25 + 6.75} = 7##. Maximum value is ##7 + 3 = 10##.

Using the tan-half-angle substitution ##t = \tan(\theta/2)##, the equation becomes ##(a+c)t^2 - 2bt + (c-a) = 0##. Then ##\tan(\frac{\alpha+\beta}{2}) = \frac{2b/(a+c)}{1 - (c-a)/(c+a)} = b/a##. Finally, ##\cos(\alpha+\beta) = \frac{1-(b/a)^2}{1+(b/a)^2} = \frac{a^2-b^2}{a^2+b^2}##.

Using the identity ##\tan^{-1} x + \tan^{-1} y = \pi + \tan^{-1}(\frac{x+y}{1-xy})## when ##xy > 1##, we have ##\tan^{-1} 2 + \tan^{-1} 3 = \pi + \tan^{-1}(-1) = 3\pi/4##. Adding ##\tan^{-1} 1 = \pi/4## gives ##\pi##.

The equation simplifies to ##1 + \sin x = 2 \cos^2 x = 2(1 - \sin^2 x)##, leading to ##2 \sin^2 x + \sin x - 1 = 0##. Roots are ##\sin x = 1/2## and ##\sin x = -1##. ##\sin x = 1/2## gives 2 solutions. ##\sin x = -1## is rejected because ##\tan x## is undefined there.

Note that ##\sin^2(7\pi/8) = \sin^2(\pi/8)## and ##\sin^2(5\pi/8) = \sin^2(3\pi/8) = \cos^2(\pi/8)##. The expression becomes ##2(\sin^2 \pi/8 + \cos^2 \pi/8) = 2(1) = 2##.

In the second quadrant, ##\sin \theta## is positive. Given ##\tan \theta = -4/3##, we can imagine a triangle with opposite 4 and adjacent 3. The hypotenuse is 5, so ##\sin \theta = 4/5##.

Using ##2 \sin A \cos B = \sin(A+B) + \sin(A-B)##, we get ##\sin 8x + \sin 2x = \sin 8x + \sin 4x##. This implies ##\sin 4x = \sin 2x##, which means ##\sin 2x (2 \cos 2x - 1) = 0##. Thus ##2x = n\pi## or ##2x = 2n\pi \pm \pi/3##.

##\cot^{-1}(1+n+n^2) = \tan^{-1}(n+1) - \tan^{-1} n##. The telescoping sum is ##\tan^{-1} 20 - \tan^{-1} 1 = \tan^{-1}(19/21)##. Then ##\cot(\tan^{-1}(19/21)) = 21/19##.

##1 \text{ rad} \approx 57^\circ##, ##2 \text{ rad} \approx 114^\circ## (or ##\sin 66^\circ##), and ##3 \text{ rad} \approx 171^\circ## (or ##\sin 9^\circ##). Comparing the sine values, ##\sin 9^\circ < \sin 57^\circ < \sin 66^\circ##, so ##z < x < y##.

The range of ##\sin x## is ##[-1, 1]##. In the interval ##[-1, 1]## (radians), the cosine function is always positive because ##1 \text{ rad} < \pi/2 \text{ rad}##. Therefore, the expression inside the square root is always positive for all real ##x##.

The maximum value of ##\sin^{-1} \theta## is ##\pi/2##. For the sum to be ##3\pi/2##, each term must be exactly ##\pi/2##. This implies ##x=1, y=1, z=1##. Thus, ##1^{100} + 1^{100} + 1^{100} = 3##.