The number of solutions of the equation 2 θ (θ)/(2) + (5 θ)/(2) = 2 ^ 3 (5 θ)/(2)$\cos 2 \theta \cos \frac {\theta}{2} + \cos \frac {5 \theta}{2} = 2 \cos^ {3} \frac {5 \theta}{2}$ in [ - (π)/(2), (π)/(2) ]$\left[ - \frac {\pi}{2}, \frac {\pi}{2} \right]$ is:

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Core Logic

Step 1: Sine Transformation

Multiply and divide the numerator by 2 to inject standard trig values:

Substitute 30° = 1/2$\sin 30^{\circ} = 1/2$ and 30° = √(3)/2$\cos 30^{\circ} = \sqrt{3}/2$:

Step 2: Apply Multiple Angle identities

The numerator becomes 2 (30° - 10°) = 2 20°$2 \sin(30^{\circ} - 10^{\circ}) = 2 \sin 20^{\circ}$. Multiply and divide the denominator by 2 to construct 2θ$\sin 2\theta$:

= 4$= 4$

Pattern Recognition

Any expression structured as A θ - B θ$A \csc \theta - B \sec \theta$ instantly signals a fraction collapse to 2 (α - θ) / (2θ)$2 \sin(\alpha - \theta) / \sin(2\theta)$ via multiplication by 2. When constants are 1$1$ and √(3)$\sqrt{3}$, the anchor is always 30°$30^{\circ}$ or 60°$60^{\circ}$.

Chapter Mix

Class 11 Maths: Trigonometric Functions

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Core Logic

Let y = ( ⁻¹x)² + ( ⁻¹x)²$y = (\sin^{-1}x)^2 + (\cos^{-1}x)^2$.

Complete the square:

Step 1: Bound the Function

Given domain x in [ - √(3)2, 1√(2) ]$x \in \left[ -\frac{\sqrt{3}}{2}, \frac{1}{\sqrt{2}} \right]$. The range of t = ⁻¹x$t = \sin^{-1}x$ for this domain is [ -(π)/(3), (π)/(4) ]$\left[ -\frac{\pi}{3}, \frac{\pi}{4} \right]$. So the expression is f(t) = 2( t - (π)/(4) )² + (π²)/(8)$f(t) = 2\left( t - \frac{\pi}{4} \right)^2 + \frac{\pi^2}{8}$.

Step 2: Find the Maximum

The function f(t)$f(t)$ is a downward-opening distance squared logic? No, leading coefficient is positive, it's an upward parabola. Max value occurs at the boundary furthest from the vertex t = (π)/(4)$t = \frac{\pi}{4}$. The boundaries are -(π)/(3)$-\frac{\pi}{3}$ and (π)/(4)$\frac{\pi}{4}$. The distance from -(π)/(3)$-\frac{\pi}{3}$ to (π)/(4)$\frac{\pi}{4}$ is greatest. At t = -(π)/(3)$t = -\frac{\pi}{3}$:

Step 3: Calculate Required Value

Here m = 29, n = 36$m = 29, n = 36$. Check (29, 36) = 1$\gcd(29, 36) = 1$. m + n = 29 + 36 = 65$m + n = 29 + 36 = 65$.

Pattern Recognition

Expressions shaped like f(x)² + g(x)²$f(x)^2 + g(x)^2$ where f(x)+g(x) = C$f(x)+g(x) = C$ will always map to a simple parabola. Analyze strictly based on vertex distance in the restricted f(x)$f(x)$ domain bounds.

Chapter Mix

Class 12 Maths: Inverse Trigonometric Functions Class 11 Maths: Quadratic Equations

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Core Logic

Given x in (- 12√(6), 12√(6))$x \in \left(-\frac{1}{2\sqrt{6}}, \frac{1}{2\sqrt{6}}\right)$, the product (4x)(6x) = 24x² < 24((1)/(24)) = 1$(4x)(6x) = 24x^2 < 24\left(\frac{1}{24}\right) = 1$. Thus, the standard identity holds without additive phase shifts.

Step 1: Forming the Equation

Taking $\tan$ on both sides:

Cross multiply:

Step 2: Solving the Quadratic

Use the quadratic formula x = -b ± √(b² - 4ac)2a$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:

Step 3: Validating Roots against Domain

We need to check which roots fall within (- 12√(6), 12√(6))$\left(-\frac{1}{2\sqrt{6}}, \frac{1}{2\sqrt{6}}\right)$.

The positive root is x₁ = √(396) - 10√(3)48$x_1 = \frac{\sqrt{396} - 10\sqrt{3}}{48}$. Since √(396) > √(300) = 10√(3)$\sqrt{396} > \sqrt{300} = 10\sqrt{3}$, x₁ > 0$x_1 > 0$, making it a valid positive fraction.

The negative root is x₂ = -√(396) - 10√(3)48$x_2 = \frac{-\sqrt{396} - 10\sqrt{3}}{48}$. This is extremely negative and falls well outside the tight bound of - 12√(6)$-\frac{1}{2\sqrt{6}}$.

Only 1 solution exists in the specified domain.

Pattern Recognition

Checking the valid domain for ab < 1$ab < 1$ prevents phantom solutions. A quadratic always yields two roots, but the strict interval provided in the question stems directly from the convergence limits of the inverse tangent addition identity, aggressively discarding the far-flung negative root.

Chapter Mix

Class 12 Maths: Inverse Trigonometric Functions Class 11 Maths: Quadratic Equations

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Core Logic

Evaluate numerator: ² 48^° - ² 12^°$\cos^2 48^\circ - \sin^2 12^\circ$ Let A = 48^°$A = 48^\circ$ and B = 12^°$B = 12^\circ$. Using the identity: (48^° + 12^°) (48^° - 12^°) = 60^° 36^°$\cos(48^\circ + 12^\circ)\cos(48^\circ - 12^\circ) = \cos 60^\circ \cos 36^\circ$

Evaluate denominator: ² 24^° - ² 6^°$\sin^2 24^\circ - \sin^2 6^\circ$ Let A = 24^°$A = 24^\circ$ and B = 6^°$B = 6^\circ$. Using the identity: (24^° + 6^°) (24^° - 6^°) = 30^° 18^°$\sin(24^\circ + 6^\circ)\sin(24^\circ - 6^\circ) = \sin 30^\circ \sin 18^\circ$

Step 1: Substituting Standard Trigonometric Values

We need standard values: 60^° = (1)/(2)$\cos 60^\circ = \frac{1}{2}$ 36^° = √(5) + 14$\cos 36^\circ = \frac{\sqrt{5} + 1}{4}$ 30^° = (1)/(2)$\sin 30^\circ = \frac{1}{2}$ 18^° = √(5) - 14$\sin 18^\circ = \frac{\sqrt{5} - 1}{4}$

Substitute into the expression:

Step 2: Rationalizing the Denominator

Rationalize by multiplying numerator and denominator by (√(5) + 1)$(\sqrt{5} + 1)$:

Step 3: Final Mapping

Comparing 3 + √(5)2$\frac{3 + \sqrt{5}}{2}$ to the given form α + β√(5)2$\frac{\alpha + \beta\sqrt{5}}{2}$:

α = 3$\alpha = 3$ β = 1$\beta = 1$

Both are natural numbers. Their sum is α + β = 3 + 1 = 4$\alpha + \beta = 3 + 1 = 4$.

Pattern Recognition

Difference of squares of sines/cosines is a massive neon sign to use product-to-sum composite identities. Always commit 18^°$\sin 18^\circ$ and 36^°$\cos 36^\circ$ to memory as they form complementary golden ratio components in geometry and algebra.

Chapter Mix

Class 11 Maths: Trigonometric Functions

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Tangent sum formula:

Core Logic

Given (α + β) = -(1)/(10) (α + β) = -√(99) = -3√(11)$\cos(\alpha + \beta) = -\frac{1}{10} \implies \tan(\alpha + \beta) = -\sqrt{99} = -3\sqrt{11}$. Given (α - β) = (3)/(8) (α - β) = 3√(55) = 3√(5)√(11)$\sin(\alpha - \beta) = \frac{3}{8} \implies \tan(\alpha - \beta) = \frac{3}{\sqrt{55}} = \frac{3}{\sqrt{5}\sqrt{11}}$.

Step 1: Simplify Tangent of 2 Alpha

Step 2: Compare Constants

Comparing with 3(1 - r√(5))√(11)(s + √(5))$\frac{3(1 - r\sqrt{5})}{\sqrt{11}(s + \sqrt{5})}$:

Pattern Recognition

Express 2α$2\alpha$ as (α+β) + (α-β)$(\alpha+\beta) + (\alpha-\beta)$ to apply standard compound tangent formula directly.

Chapter Mix

Class 11 Maths: Trigonometry