Trigonometric Ratios And Identities – JEE Main MathematicsPractice Questions & PYQs

To evaluate $f_4(x) - f_6(x)$, we substitute the definition of the function into the expression, which gives $\frac{1}{4}(\sin^4 x+\cos^4 x) - \frac{1}{6}(\sin^6 x+\cos^6 x)$. To simplify these trigonometric powers, we utilize the fundamental identity $\sin^2 x + \cos^2 x = 1$. By algebraic expansion, $\sin^4 x + \cos^4 x$ can be written as $(\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x$, which simplifies to $1 - 2\sin^2 x \cos^2 x$. Similarly, $\sin^6 x + \cos^6 x$ can be expressed as $(\sin^2 x + \cos^2 x)(\sin^4 x - \sin^2 x \cos^2 x + \cos^4 x)$, reducing to $1 - 3\sin^2 x \cos^2 x$.

Substituting these simplified expressions into the original calculation gives $\frac{1}{4}(1 - 2\sin^2 x \cos^2 x) - \frac{1}{6}(1 - 3\sin^2 x \cos^2 x)$. By finding a common denominator of 12, the expression transforms into:
$\frac{3(1 - 2\sin^2 x \cos^2 x) - 2(1 - 3\sin^2 x \cos^2 x)}{12}$
Distributing the constants across the numerator yields $3 - 6\sin^2 x \cos^2 x - 2 + 6\sin^2 x \cos^2 x$. Because the trigonometric terms cancel each other out, the final result is $\frac{1}{12}$.

Simplifying trigonometric expressions effectively relies on unifying the terms into a single trigonometric ratio, such as tangent, which allows for algebraic factoring and the application of fundamental identities. Starting with the expression $\frac{\tan A}{1-\cot A} + \frac{\cot A}{1-\tan A}$, substituting $\cot A = \frac{1}{\tan A}$ transforms the expression into $\frac{\tan^2 A}{\tan A - 1} + \frac{1}{\tan A(1 - \tan A)}$, which is equivalent to $\frac{\tan^2 A}{\tan A - 1} - \frac{1}{\tan A(\tan A - 1)}$. Combining these over the common denominator $\tan A(\tan A - 1)$ results in the numerator $\tan^3 A - 1$. Factoring the numerator using the difference of cubes identity $a^3 - b^3 = (a-b)(a^2+ab+b^2)$ yields $\frac{(\tan A - 1)(\tan^2 A + \tan A + 1)}{\tan A(\tan A - 1)}$, which simplifies to $\tan A + 1 + \cot A$. Since $\tan A + \cot A$ can be written as $\frac{\sin A}{\cos A} + \frac{\cos A}{\sin A} = \frac{\sin^2 A + \cos^2 A}{\sin A \cos A} = \frac{1}{\sin A \cos A}$, or $\sec A \csc A$, the entire expression evaluates to $\sec A \csc A + 1$.

The range of trigonometric expressions involving powers of sine and cosine is effectively determined by reducing them to a single variable using double-angle identities. Simplifying $A = \sin^2 x + \cos^4 x$ begins by transforming the expression into $1 - \sin^2 x \cos^2 x$, and then applying the double-angle identity for sine, which yields
$A = 1 - \frac{1}{4} \sin^2(2x)$
Because $\sin^2(2x)$ is bounded between 0 and 1, the value of $A$ is limited by the extremes of this term. When $\sin^2(2x)$ is 0, $A$ reaches 1, and when $\sin^2(2x)$ is 1, $A$ reaches $\frac{3}{4}$, establishing the range as $\frac{3}{4} \leq A \leq 1$.

Expressing $2\alpha$ as the sum of $(\alpha+\beta)$ and $(\alpha-\beta)$ allows for a straightforward application of the angle addition identity for tangent. From the provided cosine value $\cos(\alpha+\beta) = \frac{4}{5}$, we recognize a right triangle relationship where the adjacent side is 4 and the hypotenuse is 5, implying an opposite side of 3 and thus $\tan(\alpha+\beta) = \frac{3}{4}$. Applying similar logic to the sine value $\sin(\alpha-\beta) = \frac{5}{13}$, we identify an opposite side of 5 and hypotenuse of 13, resulting in an adjacent side of 12 and $\tan(\alpha-\beta) = \frac{5}{12}$.

Using the identity $\tan 2\alpha = \tan((\alpha+\beta) + (\alpha-\beta)) = \frac{\tan(\alpha+\beta) + \tan(\alpha-\beta)}{1 - \tan(\alpha+\beta)\tan(\alpha-\beta)}$, we substitute the derived values to obtain $\frac{\frac{3}{4} + \frac{5}{12}}{1 - (\frac{3}{4})(\frac{5}{12})}$. This expression simplifies the numerator to $\frac{9+5}{12} = \frac{14}{12} = \frac{7}{6}$ and the denominator to $1 - \frac{15}{48} = \frac{33}{48}$. Dividing these two fractions, $\frac{7}{6} \cdot \frac{48}{33}$ results in the final value of $\frac{56}{33}$.

Expanding the squares $(\cos \alpha + \cos \beta + \cos \gamma)^2$ and $(\sin \alpha + \sin \beta + \sin \gamma)^2$ allows us to combine terms using the fundamental Pythagorean identity $\sin^2 \theta + \cos^2 \theta = 1$. Adding these two squared expressions together results in:
$(\cos \alpha + \cos \beta + \cos \gamma)^2 + (\sin \alpha + \sin \beta + \sin \gamma)^2 = 3 + 2[\cos (\beta - \gamma) + \cos (\gamma - \alpha) + \cos (\alpha - \beta)]$
Substituting the provided value of $-\frac{3}{2}$ into this expression yields $3 + 2(-\frac{3}{2})$, which equals zero. Because the sum of two squared real quantities can only be zero if each individual quantity is zero, both $\cos \alpha + \cos \beta + \cos \gamma = 0$ and $\sin \alpha + \sin \beta + \sin \gamma = 0$ must be true statements.

Squaring both sides of the equation $\cos x + \sin x = \frac{1}{2}$ reveals the relationship between the trigonometric terms through the identity $\cos^2 x + \sin^2 x + 2 \sin x \cos x = \frac{1}{4}$. Substituting the identity $\sin^2 x + \cos^2 x = 1$ and the double-angle formula $\sin 2x = 2 \sin x \cos x$ simplifies the expression to $1 + \sin 2x = \frac{1}{4}$, which establishes that $\sin 2x = -\frac{3}{4}$. Utilizing the tangent identity $\sin 2x = \frac{2 \tan x}{1 + \tan^2 x}$ allows for the construction of the equation $\frac{2 \tan x}{1 + \tan^2 x} = -\frac{3}{4}$. Clearing the denominator and rearranging terms results in the quadratic equation $3 \tan^2 x + 8 \tan x + 3 = 0$. Applying the quadratic formula leads to the solution $\tan x = \frac{-8 \pm \sqrt{64 - 36}}{6}$, which reduces to $\frac{-4 \pm \sqrt{7}}{3}$. Among the possible roots, the expression $\frac{-(4 - \sqrt{7})}{3}$ corresponds to this result.

Finding the value of $\cos \frac{\alpha - \beta}{2}$ starts by squaring and adding the given equations for $\sin \alpha + \sin \beta$ and $\cos \alpha + \cos \beta$. This approach utilizes the trigonometric identity $\cos(\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta$ along with the Pythagorean identity $\sin^2 x + \cos^2 x = 1$ to consolidate the terms. Performing this operation yields the following equation:

$(\sin \alpha + \sin \beta)^2 + (\cos \alpha + \cos \beta)^2 = \left(-\frac{21}{65}\right)^2 + \left(-\frac{27}{65}\right)^2$

Expanding the left side results in $2 + 2\cos(\alpha - \beta) = \frac{441 + 729}{4225}$, which simplifies to $2(1 + \cos(\alpha - \beta)) = \frac{1170}{4225}$. Applying the half-angle formula $1 + \cos \theta = 2\cos^2 \frac{\theta}{2}$, the equation becomes $4\cos^2 \frac{\alpha - \beta}{2} = \frac{1170}{4225}$. Dividing both sides by 4 leads to $\cos^2 \frac{\alpha - \beta}{2} = \frac{1170}{16900} = \frac{9}{130}$.

Taking the square root suggests a value of $\pm \frac{3}{\sqrt{130}}$. To determine the correct sign, we analyze the provided interval $\pi < \alpha - \beta < 3\pi$. Dividing this inequality by 2 gives $\frac{\pi}{2} < \frac{\alpha - \beta}{2} < \frac{3\pi}{2}$. Because the cosine function is negative in both the second and third quadrants, the result must be the negative root.

Squaring the expression $u$ using the identity $(x+y)^2 = x^2 + y^2 + 2xy$ results in the expanded form $u^2 = (a^2 \cos^2 \theta + b^2 \sin^2 \theta) + (a^2 \sin^2 \theta + b^2 \cos^2 \theta) + 2\sqrt{(a^2 \cos^2 \theta + b^2 \sin^2 \theta)(a^2 \sin^2 \theta + b^2 \cos^2 \theta)}$. The sum of the first two bracketed terms simplifies to $a^2 + b^2$, while the product within the square root simplifies to $(a^2-b^2)^2 \frac{\sin^2 2\theta}{4} + a^2 b^2$, leading to the expression $u^2 = a^2 + b^2 + 2\sqrt{(a^2-b^2)^2 \frac{\sin^2 2\theta}{4} + a^2 b^2}$.

Since the range of $\sin^2 2\theta$ is $[0, 1]$, we can determine the extreme values of $u^2$ by evaluating the expression at these boundaries. Setting $\sin^2 2\theta = 0$ gives a minimum value of $u^2 = a^2 + b^2 + 2ab = (a+b)^2$, whereas setting $\sin^2 2\theta = 1$ results in a maximum value of $u^2 = a^2 + b^2 + 2\sqrt{\frac{(a^2+b^2)^2}{4}} = 2(a^2 + b^2)$. Subtracting the minimum value from the maximum value yields $2a^2 + 2b^2 - (a^2 + 2ab + b^2)$, which simplifies to $a^2 - 2ab + b^2$, or $(a-b)^2$.

The squared magnitudes of the vectors $\vec{a}_k = (\tan \theta_k) \hat{i} + \hat{j}$ and $\vec{b}_k$ simplify to $|\vec{a}_k|^2 = \tan^2 \theta_k + 1 = \sec^2 \theta_k$ and $|\vec{b}_k|^2 = \csc^2 \theta_k$. Determining the required ratio involves evaluating the sums $\sum_{k=1}^n \sec^2 \theta_k$ and $\sum_{k=1}^n \csc^2 \theta_k$. By applying the trigonometric identity $\sec^2 \theta + \csc^2 \theta = 4 \csc^2 2\theta$, we can relate the terms in both summations. If $A = \sum_{k=1}^n \sec^2 \theta_k$ and $B = \sum_{k=1}^n \csc^2 \theta_k$, then the sum $A + B$ is given by $\sum_{k=1}^n (\sec^2 \theta_k + \csc^2 \theta_k) = \sum_{k=1}^n 4 \csc^2 2\theta_k$.

Since $\theta_k = \frac{2^{k-1} \pi}{2^n + 1}$, the double angle is $\theta_{k+1} = 2\theta_k$, which transforms the summation into $4 \sum_{k=1}^n \csc^2 \theta_{k+1}$. This expanded expression is $4(\csc^2 \theta_2 + \csc^2 \theta_3 + \dots + \csc^2 \theta_{n+1})$. Observing that $\theta_{n+1} = \frac{2^n \pi}{2^n + 1} = \pi - \theta_1$, we find that $\csc^2 \theta_{n+1} = \csc^2 \theta_1$. Therefore, the sum $4(\csc^2 \theta_2 + \csc^2 \theta_3 + \dots + \csc^2 \theta_n + \csc^2 \theta_1)$ is simply $4B$. Given that $A + B = 4B$, it follows that $A = 3B$, which confirms that the value of the ratio is 3.