If 10 ⁴θ + 15 ⁴θ = 6$10\sin^4\theta + 15\cos^4\theta = 6$, then the value of (27 ⁶θ + 8 ⁶θ)/(16 ⁸θ)$\frac{27\csc^6\theta + 8\sec^6\theta}{16\sec^8\theta}$ is:

Related Formula

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

Related Formula

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

Related Formula

Compound angle formulas:

Core Logic

Rearranging terms to form standard identities:

Step 1: Finding Trigonometric Values

Given θ = - 12√(2)$\cot\theta = -\frac{1}{2\sqrt{2}}$ and θ in ((π)/(2), π)$\theta \in (\frac{\pi}{2}, \pi)$. This implies θ = 2√(2)3$\sin\theta = \frac{2\sqrt{2}}{3}$.

We need to evaluate (θ)/(2) - (θ)/(2)$\cos\frac{\theta}{2} - \sin\frac{\theta}{2}$. We know:

Since (π)/(2) < θ < π$\frac{\pi}{2} < \theta < \pi$, we have (π)/(4) < (θ)/(2) < (π)/(2)$\frac{\pi}{4} < \frac{\theta}{2} < \frac{\pi}{2}$. In this quadrant, (θ)/(2) > (θ)/(2)$\sin\frac{\theta}{2} > \cos\frac{\theta}{2}$. Therefore, (θ)/(2) - (θ)/(2) = -√(1 - θ)$\cos\frac{\theta}{2} - \sin\frac{\theta}{2} = -\sqrt{1 - \sin\theta}$.

Step 2: Final Calculation

Notice that 3 - 2√(2) = (√(2) - 1)²$3 - 2\sqrt{2} = (\sqrt{2} - 1)^2$.

Pattern Recognition

Recognize the expanded forms of (A-B)$\sin(A-B)$ and (A-B)$\cos(A-B)$ hiding within the products. Also, carefully determine the sign of (θ/2) - (θ/2)$\cos(\theta/2) - \sin(\theta/2)$ based on the half-angle quadrant.

Chapter Mix

Class 11 Maths: Trigonometric Functions

Related Formula

The expression A x + B x + C$A\sin x + B\cos x + C$ lies in the range:

Core Logic

Let f(θ) = ²θ - 6 θ θ + 3 ²θ + 2$f(\theta) = \cos^2\theta - 6\sin\theta\cos\theta + 3\sin^2\theta + 2$. Convert all quadratic terms to multiple angles:

Step 1: Simplifying Expression

Step 2: Finding Extremes

The expression is of the form C + A 2θ + B 2θ$C + A\sin 2\theta + B\cos 2\theta$, where C=4, A=-3, B=-1$C=4, A=-3, B=-1$. The minimum value is C - √(A² + B²)$C - \sqrt{A^2 + B^2}$:

The maximum value is C + √(A² + B²) = 4 + √(10)$C + \sqrt{A^2 + B^2} = 4 + \sqrt{10}$. Therefore, the least value is 4 - √(10)$4 - \sqrt{10}$.

Pattern Recognition

Every quadratic trigonometric expression in θ$\sin\theta$ and θ$\cos\theta$ must be linearly transformed into (2θ)$\sin(2\theta)$ and (2θ)$\cos(2\theta)$ terms to utilize the standard bound formula.

Chapter Mix

Class 11 Maths: Trigonometric Functions

Related Formula

Core Logic

Evaluate the denominator first: D = 20^° 40^° 60^° 80^°$D = \cos 20^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ$. Using θ (60^°-θ) (60^°+θ)$\cos \theta \cos(60^\circ-\theta) \cos(60^\circ+\theta)$ with θ = 20^°$\theta = 20^\circ$:

Step 1: Simplify Numerator

Multiply and divide by 2:

Step 2: Combine Terms

Pattern Recognition

The expression √(3)cosecθ - θ$\sqrt{3}\operatorname{cosec}\theta - \sec\theta$ simplifies instantly to 4$4$ by creating a sine difference angle identity. Denominators of 20^° 40^° 80^°$\cos20^\circ\cos40^\circ\cos80^\circ$ universally yield (1)/(8)$\frac{1}{8}$.

Chapter Mix

Class 11 Maths: Trigonometric Functions