Trigonometric Ratios And Identities – JEE Main MathematicsPractice Questions & PYQs

The infinite geometric series $\cos^2 x + \cos^4 x + \cos^6 x + \ldots$ features a first term $a = \cos^2 x$ and a common ratio $r = \cos^2 x$. The sum $S = \frac{a}{1-r}$ simplifies to $\frac{\cos^2 x}{1 - \cos^2 x} = \cot^2 x$. By the logarithm rule $e^{\ln 2 \cdot A} = 2^A$, the original expression becomes $2^{\cot^2 x}$. The quadratic equation
$t^2 - 9t + 8 = 0$
factors into $(t-1)(t-8) = 0$, leading to $t = 1$ or $t = 8$.

Matching $2^{\cot^2 x}$ with these roots implies $\cot^2 x = 0$ or $\cot^2 x = 3$. Given the restriction $0 < x < \frac{\pi}{2}$, the case where $\cot x = 0$ is rejected, leaving $\cot x = \sqrt{3}$ as the valid relationship. Dividing the expression $\frac{2\sin x}{\sin x + \sqrt{3}\cos x}$ by $\sin x$ simplifies it to $\frac{2}{1 + \sqrt{3}\cot x}$. Substituting $\cot x = \sqrt{3}$ yields
$\frac{2}{1 + \sqrt{3}(\sqrt{3})} = \frac{2}{1+3} = \frac{2}{4} = \frac{1}{2}$

The fundamental relationship between the sine and cosine of an angle, specifically the identity $(\sin \theta + \cos \theta)^2 = 1 + \sin 2\theta$, allows for the determination of $\sin 2\theta$ from the provided equation. By squaring both sides of $\sin \theta + \cos \theta = \frac{1}{2}$, we arrive at the following:
$1 + \sin 2\theta = \frac{1}{4}$
This yields $\sin 2\theta = -\frac{3}{4}$.

With $\sin 2\theta$ established, trigonometric double-angle and triple-angle identities provide the values for the remaining terms. Applying $\cos 4\theta = 1 - 2 \sin^2 2\theta$ results in $\cos 4\theta = 1 - 2(-\frac{3}{4})^2$, which simplifies to $-\frac{1}{8}$. Similarly, expressing $\sin 6\theta$ as $\sin 2\theta(3 - 4 \sin^2 2\theta)$ allows us to substitute $-\frac{3}{4}$ for $\sin 2\theta$, producing $(-\frac{3}{4})(3 - 4(\frac{9}{16}))$, which resolves to $-\frac{9}{16}$.

Substitution of these components into the expression $16(\sin 2\theta + \cos 4\theta + \sin 6\theta)$ results in $16(-\frac{3}{4} - \frac{1}{8} - \frac{9}{16})$. Distributing the factor of 16 across the sum leads to $-12 - 2 - 9$, which simplifies to $-23$.

We simplify the given equation by applying the sum-to-product identity $\cos x + \cos y = 2 \cos(\frac{x+y}{2}) \cos(\frac{x-y}{2})$ and the double-angle formula $\cos(x+y) = 2 \cos^2(\frac{x+y}{2}) - 1$. Substituting these identities into the expression results in $2 \cos(\frac{x+y}{2}) \cos(\frac{x-y}{2}) - 2 \cos^2(\frac{x+y}{2}) + 1 = \frac{3}{2}$. By isolating the trigonometric terms and multiplying the entire equation by 2, we obtain $4 \cos(\frac{x+y}{2}) \cos(\frac{x-y}{2}) - 4 \cos^2(\frac{x+y}{2}) = 1$.

By substituting $1 = \cos^2(\frac{x-y}{2}) + \sin^2(\frac{x-y}{2})$, the equation transforms into $(\cos(\frac{x-y}{2}) - 2 \cos(\frac{x+y}{2}))^2 + \sin^2(\frac{x-y}{2}) = 0$. Because the sum of two squared terms is zero, each squared expression must equal zero independently. This requires $\sin(\frac{x-y}{2}) = 0$, meaning $x=y$, and $\cos(\frac{x-y}{2}) - 2 \cos(\frac{x+y}{2}) = 0$. With $x=y$, the condition simplifies to $\cos(0) - 2 \cos(x) = 0$, establishing that $\cos x = \frac{1}{2}$, which also implies $\cos y = \frac{1}{2}$. Given $0 < x < \pi$, we calculate $\sin x = \sqrt{1 - \cos^2 x} = \sqrt{1 - \frac{1}{4}} = \frac{\sqrt{3}}{2}$. Summing the calculated values yields $\sin x + \cos y = \frac{\sqrt{3}}{2} + \frac{1}{2} = \frac{1+\sqrt{3}}{2}$.

Three terms $a, b, c$ form an arithmetic progression if $2b = a + c$, which allows us to define the terms as $x = \frac{1}{2} \left( \tan \frac{\pi}{9} + \tan \frac{7\pi}{18} \right)$ and $2y = \tan \frac{\pi}{9} + \tan \frac{5\pi}{18}$. To evaluate the magnitude $|x - 2y|$, we simplify the trigonometric terms using complementary angle identities, where $\tan \frac{7\pi}{18} = \cot \frac{\pi}{9}$ and $\tan \frac{5\pi}{18} = \cot \frac{2\pi}{9}$. Substituting these into the expression gives $|x - 2y| = \left| \frac{1}{2} \tan \frac{\pi}{9} + \frac{1}{2} \cot \frac{\pi}{9} - \tan \frac{\pi}{9} - \cot \frac{2\pi}{9} \right|$, which simplifies to $\left| \frac{1}{2} \cot \frac{\pi}{9} - \frac{1}{2} \tan \frac{\pi}{9} - \cot \frac{2\pi}{9} \right|$. Recognizing the identity $\cot \theta - \tan \theta = 2 \cot 2\theta$, we can rewrite $\frac{1}{2} (\cot \frac{\pi}{9} - \tan \frac{\pi}{9})$ as $\cot \frac{2\pi}{9}$, leading to the final calculation $| \cot \frac{2\pi}{9} - \cot \frac{2\pi}{9} |$, which equals $0$.

The expression $\cot \frac{\pi}{24}$ relies on the trigonometric identity $\cot \theta = \frac{1+\cos 2\theta}{\sin 2\theta}$. Applying this with $\theta = \frac{\pi}{24}$ results in $\frac{1+\cos(\frac{\pi}{12})}{\sin(\frac{\pi}{12})}$, utilizing the values $\cos \frac{\pi}{12} = \frac{\sqrt{3}+1}{2\sqrt{2}}$ and $\sin \frac{\pi}{12} = \frac{\sqrt{3}-1}{2\sqrt{2}}$. This leads to the substitution $\frac{1+\frac{\sqrt{3}+1}{2\sqrt{2}}}{\frac{\sqrt{3}-1}{2\sqrt{2}}}$, which simplifies to $\frac{2\sqrt{2}+\sqrt{3}+1}{\sqrt{3}-1}$. Rationalizing this fraction by multiplying the numerator and denominator by $\sqrt{3}+1$ expands the expression to $\frac{2\sqrt{6}+2\sqrt{2}+3+\sqrt{3}+\sqrt{3}+1}{2}$. Final arithmetic across the terms yields $\sqrt{6}+\sqrt{2}+\sqrt{3}+2$.

To evaluate this product, we utilize the trigonometric symmetry property $\sin(\pi - \theta) = \sin(\theta)$ to simplify the argument angles. By applying this identity, we observe that $\sin(\frac{5\pi}{8})$ matches $\sin(\frac{3\pi}{8})$, $\sin(\frac{6\pi}{8})$ becomes $\sin(\frac{2\pi}{8})$, and $\sin(\frac{7\pi}{8})$ is equal to $\sin(\frac{\pi}{8})$, which transforms the expression into $2 \sin^2 (\frac{\pi}{8}) \sin^2 (\frac{2\pi}{8}) \sin^2 (\frac{3\pi}{8})$.

We know that $\sin(\frac{2\pi}{8}) = \sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$, so its square is $\frac{1}{2}$, and through the complementary angle relationship, $\sin(\frac{3\pi}{8}) = \cos(\frac{\pi}{8})$. Substituting these into our expression results in $2 \sin^2 (\frac{\pi}{8}) \cdot \frac{1}{2} \cdot \cos^2 (\frac{\pi}{8})$, which simplifies to $\sin^2 (\frac{\pi}{8}) \cos^2 (\frac{\pi}{8})$. By rearranging this as $\frac{1}{4} (2 \sin(\frac{\pi}{8}) \cos(\frac{\pi}{8}))^2$, we apply the double-angle identity for sine to yield $\frac{1}{4} \sin^2 (\frac{\pi}{4})$. Since $\sin^2 (\frac{\pi}{4}) = (\frac{1}{\sqrt{2}})^2 = \frac{1}{2}$, the final calculation becomes $\frac{1}{4} \cdot \frac{1}{2}$, resulting in $\frac{1}{8}$.

The equation $\cos^{4}\theta + \sin^{4}\theta + \lambda = 0$ rearranges to $\lambda = -(\sin^{4}\theta + \cos^{4}\theta)$. This trigonometric expression simplifies by completing the square, noting that $\sin^{4}\theta + \cos^{4}\theta$ is equivalent to $(\sin^{2}\theta + \cos^{2}\theta)^{2} - 2\sin^{2}\theta\cos^{2}\theta$. Substituting the identity $\sin^{2}\theta + \cos^{2}\theta = 1$ and converting $2\sin^{2}\theta\cos^{2}\theta$ into $\frac{\sin^{2}2\theta}{2}$ results in $\lambda = -(1 - \frac{\sin^{2}2\theta}{2})$, which simplifies further to $\lambda = \frac{\sin^{2}2\theta}{2} - 1$. Given that $\sin^{2}2\theta$ varies between $0$ and $1$, the term $\frac{\sin^{2}2\theta}{2}$ must fall within the range $[0, \frac{1}{2}]$, placing $\lambda$ in the interval $[-1, -\frac{1}{2}]$.

Simplifying trigonometric expressions using double-angle identities allows for the direct determination of the values for $\alpha$ and $\beta$, which leads to the evaluation of the final tangent expression. Since $\alpha$ and $\beta$ are in the first quadrant, all trigonometric ratios remain positive, allowing us to resolve square roots without ambiguity. For the first term, applying the identity $1+\cos2\alpha = 2\cos^2\alpha$ to the denominator results in $\frac{\sqrt{2}\sin\alpha}{\sqrt{2}\cos\alpha}$, which simplifies to $\tan\alpha = 1/7$. Similarly, the second expression simplifies using $1-\cos2\beta = 2\sin^2\beta$ to obtain $\sqrt{\sin^2\beta} = \sin\beta = 1/\sqrt{10}$.

With $\sin\beta = 1/\sqrt{10}$, a right triangle with an opposite side of 1 and a hypotenuse of $\sqrt{10}$ has an adjacent side of 3, implying $\tan\beta = 1/3$. Using the double-angle formula $\tan2\beta = \frac{2\tan\beta}{1-\tan^2\beta}$, we calculate $\tan2\beta = \frac{2(1/3)}{1-(1/3)^2} = 3/4$. Finally, substituting these values into the tangent addition formula $\tan(\alpha+2\beta) = \frac{\tan\alpha+\tan2\beta}{1-\tan\alpha\tan2\beta}$ gives $\frac{1/7 + 3/4}{1 - (1/7)(3/4)} = \frac{25/28}{25/28} = 1$.

Recognizing that $\frac{3\pi}{8}$ and $\frac{\pi}{8}$ are complementary angles allows us to replace $\cos\left(\frac{3\pi}{8}\right)$ with $\sin\left(\frac{\pi}{8}\right)$ and $\sin\left(\frac{3\pi}{8}\right)$ with $\cos\left(\frac{\pi}{8}\right)$. Substituting these into the original expression gives $\cos^3\left(\frac{\pi}{8}\right)\sin\left(\frac{\pi}{8}\right) + \sin^3\left(\frac{\pi}{8}\right)\cos\left(\frac{\pi}{8}\right)$, which simplifies by factoring out $\sin\left(\frac{\pi}{8}\right)\cos\left(\frac{\pi}{8}\right)$ to produce $\sin\left(\frac{\pi}{8}\right)\cos\left(\frac{\pi}{8}\right) \left( \cos^2\left(\frac{\pi}{8}\right) + \sin^2\left(\frac{\pi}{8}\right) \right)$. Since the trigonometric identity for the sum of squares is equal to 1, the expression reduces to $\sin\left(\frac{\pi}{8}\right)\cos\left(\frac{\pi}{8}\right)$, which is equivalent to $\frac{1}{2}\sin\left(\frac{\pi}{4}\right)$ and yields a final value of $\frac{1}{2\sqrt{2}}$.

The double-angle formulas for trigonometric squares provide a direct path to simplifying these expressions by converting $\sin^2 \theta$ to $\frac{1-\cos(2\theta)}{2}$ and $\cos^2 \theta$ to $\frac{1+\cos(2\theta)}{2}$. Applying these identities to $M = \cos^2\left(\frac{\pi}{16}\right) - \sin^2\left(\frac{\pi}{8}\right)$ yields the expression $\left(\frac{1+\cos(\pi/8)}{2}\right) - \left(\frac{1-\cos(\pi/4)}{2}\right)$. When these terms are combined, the constant values cancel out, leaving $\frac{1}{2}\cos\left(\frac{\pi}{8}\right) + \frac{1}{2}\cos\left(\frac{\pi}{4}\right)$. Given that $\cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$, the final form becomes $\frac{1}{2\sqrt{2}} + \frac{1}{2}\cos\left(\frac{\pi}{8}\right)$.

Infinite series of the form $\sum_{n=0}^{\infty } r^n$ converge to $\frac{1}{1-r}$ provided the magnitude of the common ratio $r$ is less than 1. For the expression $x = \sum_{n=0}^{\infty } (-1)^n \tan^{2n} \theta$, the first term is 1 and the common ratio is $-\tan^2 \theta$, yielding
$x = \frac{1}{1 - (-\tan^2 \theta)} = \frac{1}{1 + \tan^2 \theta} = \cos^2 \theta$
Applying the same principle to the expression for $y = \sum_{n=0}^{\infty } \cos^{2n} \theta$ with a common ratio of $\cos^2 \theta$ leads to
$y = \frac{1}{1 - \cos^2 \theta} = \frac{1}{\sin^2 \theta}$
Substituting the derived value $\cos^2 \theta = x$ into the denominator shows that $y = \frac{1}{1 - x}$, which simplifies to $y(1 - x) = 1$.

The general principle that an expression of the form $a\sin\theta + b\cos\theta$ achieves a maximum value of $\sqrt{a^2 + b^2}$ simplifies this calculation significantly. Expanding the term $5\sin(\theta - \frac{\pi}{6})$ produces $5(\sin\theta \frac{\sqrt{3}}{2} - \cos\theta \frac{1}{2})$, transforming the full expression into $\frac{5\sqrt{3}}{2}\sin\theta + \frac{1}{2}\cos\theta$ after combining the cosine terms. Substituting these coefficients into the maximum value formula results in $\sqrt{(\frac{5\sqrt{3}}{2})^2 + (\frac{1}{2})^2}$, which calculates to $\sqrt{\frac{75}{4} + \frac{1}{4}}$ and simplifies to $\sqrt{19}$.

Recognizing the trigonometric product $\sin 10^{\circ} \sin 50^{\circ} \sin 70^{\circ}$ reveals a pattern consistent with the identity $\sin \theta \sin(60^{\circ} - \theta) \sin(60^{\circ} + \theta) = \frac{1}{4} \sin 3\theta$. Applying this formula with $\theta = 10^{\circ}$, the terms combine to form $\frac{1}{4} \sin(3 \times 10^{\circ})$, or $\frac{1}{4} \sin 30^{\circ}$. Since $\sin 30^{\circ} = \frac{1}{2}$, this specific product simplifies to $\frac{1}{8}$. Multiplying this result by the remaining $\sin 30^{\circ}$ factor in the original expression gives $\frac{1}{2} \times \frac{1}{8}$, which results in $\frac{1}{16}$.

The trigonometric expression $y=\sin x \sin (x+2)-\sin ^{2}(x+1)$ can be simplified by applying the product-to-sum identity $\sin A \sin B = \frac{1}{2}[\cos(A-B) - \cos(A+B)]$ to the first term and the half-angle identity $\sin^2 \theta = \frac{1-\cos 2\theta}{2}$ to the second term. This substitution results in $y = \frac{1}{2}[\cos 2 - \cos(2x+2)] - \frac{1-\cos(2x+2)}{2}$, where the variable terms containing $x$ cancel each other out, leaving the constant value $y = \frac{\cos 2 - 1}{2}$. By utilizing the double-angle identity $\cos 2 = 1 - 2\sin^2 1$, the equation reduces to $y = -\sin^2 1$. Since $1$ radian is a first-quadrant angle, $\sin 1$ is positive, which makes $y$ a negative constant. A horizontal line defined by a negative $y$-coordinate lies entirely below the $x$-axis, passing through the third quadrant, where both coordinates are negative, and the fourth quadrant, where $x$ is positive and $y$ is negative.

Recognizing the core trigonometric identity $(\sin\theta \pm \cos\theta)^2 = 1 \pm \sin 2\theta$ allows for an efficient reduction of the given expression, as it transforms the powers of sine and cosine into manageable terms. Applying this, the expression $3(\sin\theta - \cos\theta)^4 + 6(\sin\theta + \cos\theta)^2 + 4\sin^6\theta$ becomes $3(1 - \sin 2\theta)^2 + 6(1 + \sin 2\theta) + 4\sin^6\theta$. Expanding these squares leads to $3(1 - 2\sin 2\theta + \sin^2 2\theta) + 6 + 6\sin 2\theta + 4\sin^6\theta$, where the linear terms involving $\sin 2\theta$ cancel entirely, simplifying the equation to $9 + 3\sin^2 2\theta + 4\sin^6\theta$.

Substituting $\sin^2 2\theta = 4\sin^2\theta\cos^2\theta$ gives $9 + 12\sin^2\theta\cos^2\theta + 4\sin^6\theta$. Expressing $\sin^2\theta$ as $1 - \cos^2\theta$ leads to $9 + 12(1 - \cos^2\theta)\cos^2\theta + 4(1 - \cos^2\theta)^3$. Expanding the cubic term $4(1 - \cos^2\theta)^3$ into $4(1 - 3\cos^2\theta + 3\cos^4\theta - \cos^6\theta)$ and distributing the $12\cos^2\theta$ terms results in a total cancellation of all intermediate $\cos^2\theta$ and $\cos^4\theta$ terms, leaving the final expression as $13 - 4\cos^6\theta$.

The expression simplifies through the repeated application of the double angle identity $\sin(2\theta) = 2 \sin \theta \cos \theta$, which generates a telescopic chain. Multiplying and dividing the product by $2^9$ allows us to combine the terms $2^9 \sin \frac{\pi}{2^{10}} \cos \frac{\pi}{2^{10}}$ into $2^8 \sin \frac{\pi}{2^9}$. As this process continues, the entire chain of products collapses into $\sin \frac{\pi}{2}$ in the numerator while leaving $2^9$ in the denominator. With $\sin \frac{\pi}{2}$ equaling 1, the result is $\frac{1}{2^9}$, which evaluates to $\frac{1}{512}$.

Evaluating expressions involving squared cosine terms and their products often points toward trigonometric identities specifically designed for angular sums. A useful relationship states that for any two angles $A$ and $B$, the expression $\cos^2 A + \cos^2 B - \cos A \cos B$ simplifies to the constant $\frac{3}{4}$ whenever the sum of those angles equals 60°. Since the angles provided in the problem are 10° and 50°, their sum is exactly 60°, which satisfies the necessary condition for this identity. Because the expression matches the structure and the angle sum requirement of this rule, the entire calculation simplifies immediately to $\frac{3}{4}$.

Representing $2\alpha$ as the sum $(\alpha+\beta) + (\alpha-\beta)$ allows us to use the tangent addition identity to solve the problem directly. We first derive the tangent values from the provided ratios: $\cos(\alpha+\beta) = \frac{3}{5}$ implies $\tan(\alpha+\beta) = \frac{4}{3}$, while $\sin(\alpha-\beta) = \frac{5}{13}$ implies $\tan(\alpha-\beta) = \frac{5}{12}$.

Applying the formula $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$, we substitute these values to find $\tan(2\alpha) = \frac{\frac{4}{3} + \frac{5}{12}}{1 - (\frac{4}{3})(\frac{5}{12})}$. Using a common denominator of $36$ for the entire expression simplifies this to $\frac{48+15}{36-20}$, which results in $\frac{63}{16}$.

Expressing all trigonometric terms through the substitution $t = \cos^2 x$ simplifies the original expression into a standard algebraic equation, using the identities $\tan^2 x = \frac{1-t}{t}$ and $\cos 2x = 2t - 1$. Substituting these expressions results in the following equation:
$5\left(\frac{1-t}{t} - t\right) = 2(2t - 1) + 9$
Multiplying both sides by $t$ transforms the relationship into $5(1 - t - t^2) = t(4t + 7)$, which rearranges into the quadratic form $9t^2 + 12t - 5 = 0$. Factoring the quadratic yields $(3t - 1)(3t + 5) = 0$, and since $t = \cos^2 x$ must be non-negative, the only valid solution is $t = \frac{1}{3}$. Determining the final value requires calculating $\cos 2x = 2\left(\frac{1}{3}\right) - 1 = -\frac{1}{3}$ and then utilizing the double-angle formula $\cos 4x = 2\cos^2 2x - 1$, which equals $2\left(-\frac{1}{3}\right)^2 - 1 = \frac{2}{9} - 1 = -\frac{7}{9}$.

Expressing the function in terms of a single trigonometric ratio allows for a straightforward analysis of its range. By applying the double-angle identity $2\cos^2 x = 1 + \cos 2x$, the term $2\cos^4 x$ transforms into $\frac{1}{2}(1 + \cos 2x)^2$, while $\sin^2 2x$ is replaced with $1 - \cos^2 2x$. Substituting these into the original formula creates the expression $4 + \frac{1}{2}(1 - \cos^2 2x) - \frac{1}{2}(1 + \cos 2x)^2$. After algebraic simplification, this reduces to the quadratic form $-\cos^2 2x - \cos 2x + 4$.

Completing the square on this quadratic yields $\frac{17}{4} - (\cos 2x + \frac{1}{2})^2$. Because the value of $\cos 2x$ is bounded between $-1$ and $1$, the maximum value $M$ occurs at the vertex where $\cos 2x = -\frac{1}{2}$, resulting in $M = \frac{17}{4}$. The minimum value $m$ occurs at the boundary point $\cos 2x = 1$, which gives $m = -1 - 1 + 4 = 2$. The difference $M - m$ is therefore $\frac{17}{4} - 2$, which equals $\frac{9}{4}$.