If x + ² x = 1$\sin x + \sin^2 x = 1$, x in (0, (π)/(2))$x \in \left(0, \frac{\pi}{2}\right)$, then ( ¹² x + ¹² x) + 3 ( ¹⁰ x + ¹⁰ x + ⁸ x + ⁸ x) + ( ⁶ x + ⁶ x)$(\cos^{12} x + \tan^{12} x) + 3 (\cos^{10} x + \tan^{10} x + \cos^8 x + \tan^8 x) + (\cos^6 x + \tan^6 x)$ is equal to

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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

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.

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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

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Core Logic

Substitute the double-angle formula into the given equation to form a quadratic in θ$\cos\theta$:

Step 1: Solve the Quadratic

Factorize the quadratic equation:

Wait, 2 × 2√(3) = 12$2 \times 2\sqrt{3} = 12$. The factors of 12$12$ that sum to 8$8$ are 6$6$ and 2$2$.

This gives:

Since -1 ≤ θ ≤ 1$-1 \leq \cos\theta \leq 1$, we reject θ = -√(3)$\cos\theta = -\sqrt{3}$.

Step 2: Count Solutions in Interval

We need solutions for θ = - 1√(3)$\cos\theta = -\frac{1}{\sqrt{3}}$ in the interval [-3π, 2π]$[-3\pi, 2\pi]$. The period of cosine is 2π$2\pi$. The equation θ = k$\cos\theta = k$ (where -1 < k < 0$-1 < k < 0$) has 2$2$ solutions per 2π$2\pi$ interval. Intervals: [0, 2π]$[0, 2\pi]$: 2$2$ solutions (in Quadrants II and III). [-2π, 0]$[-2\pi, 0]$: 2$2$ solutions. [-3π, -2π]$[-3\pi, -2\pi]$: 1$1$ solution (in Quadrant II equivalent, which is Quadrant III when going backwards. Specifically, from -3π$-3\pi$ to -2π$-2\pi$ covers the top half of the circle. Wait, [-3π, -2π]$[-3\pi, -2\pi]$ goes from 180^°$180^\circ$ to 360^°$360^\circ$ logically, i.e., quadrants III and IV. $\cos$ is negative in Quadrant III. So exactly 1$1$ solution). Total solutions = 2 + 2 + 1 = 5$2 + 2 + 1 = 5$.

Pattern Recognition

Mapping phase intervals chunk by chunk (2π$2\pi$ cycles yield 2$2$ roots for | x|<1$|\cos x|<1$) prevents overcounting when domain bounds don't cleanly align with full periods.

Chapter Mix

Class 11 Maths: Trigonometric Functions

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Core Logic

First, simplify the trigonometric arguments by reducing angles: ((7π)/(2) - θ) = - θ$\sin\left(\frac{7\pi}{2} - \theta\right) = -\cos\theta$ (11π + θ) = - θ$\sin(11\pi + \theta) = -\sin\theta$ ((3π)/(2) - θ) = - θ$\sin\left(\frac{3\pi}{2} - \theta\right) = -\cos\theta$ (9π - θ) = θ$\sin(9\pi - \theta) = \sin\theta$

Step 1: Simplify the Function

Substitute these back into f(θ)$f(\theta)$:

Applying the algebraic identities:

Step 2: Convert to Double Angle

Multiply and divide the second term by 2 to use (2θ)$\sin(2\theta)$:

Step 3: Find Maximum and Minimum

Since 0 ≤ ²(2θ) ≤ 1$0 \leq \sin^2(2\theta) \leq 1$: Max value (α$\alpha$): Occurs when ²(2θ) = 0$\sin^2(2\theta) = 0$. α = 2 - 0 = 2$\alpha = 2 - 0 = 2$. Min value (β$\beta$): Occurs when ²(2θ) = 1$\sin^2(2\theta) = 1$. β = 2 - (1)/(2) = (3)/(2)$\beta = 2 - \frac{1}{2} = \frac{3}{2}$.

Step 4: Evaluate the Target Expression

We need α + 2β$\alpha + 2\beta$:

Pattern Recognition

The expression a( ⁴ x + ⁴ x) - b( ⁶ x + ⁶ x)$a(\cos^4 x + \sin^4 x) - b(\cos^6 x + \sin^6 x)$ is a ubiquitous JEE template. Immediately swap to (1-2 ² x ² x)$(1-2\sin^2 x\cos^2 x)$ and (1-3 ² x ² x)$(1-3\sin^2 x\cos^2 x)$ for mass cancellation.

Chapter Mix

Class 11 Maths: Trigonometric Functions