Let alpha$\alpha$ and beta$\beta$ respectively be the maximum and the minimum values of the function f(theta) = 4left(sin^4left(frac7pi2 -thetaright) + sin^4 (11pi +theta)right) -2left(sin^6left(frac3pi2 -thetaright) + sin^6 (9pi -theta)right), theta in mathbfR$f(\theta) = 4\left(\sin^4\left(\frac{7\pi}{2} -\theta\right) + \sin^4 (11\pi +\theta)\right) -2\left(\sin^6\left(\frac{3\pi}{2} -\theta\right) + \sin^6 (9\pi -\theta)\right), \theta \in \mathbf{R}$. Then alpha + 2beta$\alpha + 2\beta$ is equal to :

### Related Formula sin(A - B) = sin A cos B - cos A sin B$$\sin(A - B) = \sin A \cos B - \cos A \sin B$$ sin 2A = 2 sin A cos A$$\sin 2A = 2 \sin A \cos A$$ ### Core Logic csc 10^circ - sqrt3 sec 10^circ = frac1sin 10^circ - fracsqrt3cos 10^circ$$\csc 10^{\circ} - \sqrt{3} \sec 10^{\circ} = \frac{1}{\sin 10^{\circ}} - \frac{\sqrt{3}}{\cos 10^{\circ}}$$ = fraccos 10^circ - sqrt3sin 10^circsin 10^circ cos 10^circ$$= \frac{\cos 10^{\circ} - \sqrt{3}\sin 10^{\circ}}{\sin 10^{\circ} \cos 10^{\circ}}$$ ### Step 1: Sine Transformation Multiply and divide the numerator by 2 to inject standard trig values: = frac2 left(frac12cos 10^circ - fracsqrt32sin 10^circright)sin 10^circ cos 10^circ$$= \frac{2 \left(\frac{1}{2}\cos 10^{\circ} - \frac{\sqrt{3}}{2}\sin 10^{\circ}\right)}{\sin 10^{\circ} \cos 10^{\circ}}$$ Substitute sin 30^circ = 1/2$\sin 30^{\circ} = 1/2$ and cos 30^circ = sqrt3/2$\cos 30^{\circ} = \sqrt{3}/2$: = frac2 (sin 30^circ cos 10^circ - cos 30^circ sin 10^circ)sin 10^circ cos 10^circ$$= \frac{2 (\sin 30^{\circ} \cos 10^{\circ} - \cos 30^{\circ} \sin 10^{\circ})}{\sin 10^{\circ} \cos 10^{\circ}}$$ ### Step 2: Apply Multiple Angle identities The numerator becomes 2 sin(30^circ - 10^circ) = 2 sin 20^circ$2 \sin(30^{\circ} - 10^{\circ}) = 2 \sin 20^{\circ}$. Multiply and divide the denominator by 2 to construct sin 2theta$\sin 2\theta$: = frac4 sin 20^circ2 sin 10^circ cos 10^circ$$= \frac{4 \sin 20^{\circ}}{2 \sin 10^{\circ} \cos 10^{\circ}}$$ = frac4 sin 20^circsin 20^circ$$= \frac{4 \sin 20^{\circ}}{\sin 20^{\circ}}$$ = 4$= 4$ ### Pattern Recognition Any expression structured as A csc theta - B sec theta$A \csc \theta - B \sec \theta$ instantly signals a fraction collapse to 2 sin(alpha - theta) / sin(2theta)$2 \sin(\alpha - \theta) / \sin(2\theta)$ via multiplication by 2. When constants are 1$1$ and sqrt3$\sqrt{3}$, the anchor is always 30^circ$30^{\circ}$ or 60^circ$60^{\circ}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Trigonometric Functions

### Related Formula cos(A+B)cos(A-B) = cos^2 A - sin^2 B$$\cos(A+B)\cos(A-B) = \cos^2 A - \sin^2 B$$ sin(A+B)sin(A-B) = sin^2 A - sin^2 B$$\sin(A+B)\sin(A-B) = \sin^2 A - \sin^2 B$$ ### Core Logic Evaluate numerator: cos^2 48^circ - sin^2 12^circ$\cos^2 48^\circ - \sin^2 12^\circ$ Let A = 48^circ$A = 48^\circ$ and B = 12^circ$B = 12^\circ$. Using the identity: cos(48^circ + 12^circ)cos(48^circ - 12^circ) = cos 60^circ cos 36^circ$\cos(48^\circ + 12^\circ)\cos(48^\circ - 12^\circ) = \cos 60^\circ \cos 36^\circ$ Evaluate denominator: sin^2 24^circ - sin^2 6^circ$\sin^2 24^\circ - \sin^2 6^\circ$ Let A = 24^circ$A = 24^\circ$ and B = 6^circ$B = 6^\circ$. Using the identity: sin(24^circ + 6^circ)sin(24^circ - 6^circ) = sin 30^circ sin 18^circ$\sin(24^\circ + 6^\circ)\sin(24^\circ - 6^\circ) = \sin 30^\circ \sin 18^\circ$ ### Step 1: Substituting Standard Trigonometric Values We need standard values: cos 60^circ = frac12$\cos 60^\circ = \frac{1}{2}$ cos 36^circ = fracsqrt5 + 14$\cos 36^\circ = \frac{\sqrt{5} + 1}{4}$ sin 30^circ = frac12$\sin 30^\circ = \frac{1}{2}$ sin 18^circ = fracsqrt5 - 14$\sin 18^\circ = \frac{\sqrt{5} - 1}{4}$ Substitute into the expression: fraccos 60^circ cos 36^circsin 30^circ sin 18^circ = frac (1/2) cdot (fracsqrt5 + 14) (1/2) cdot (fracsqrt5 - 14) $$\frac{\cos 60^\circ \cos 36^\circ}{\sin 30^\circ \sin 18^\circ} = \frac{ (1/2) \cdot (\frac{\sqrt{5} + 1}{4}) }{ (1/2) \cdot (\frac{\sqrt{5} - 1}{4}) }$$ ### Step 2: Rationalizing the Denominator = fracsqrt5 + 1sqrt5 - 1$$= \frac{\sqrt{5} + 1}{\sqrt{5} - 1}$$ Rationalize by multiplying numerator and denominator by (sqrt5 + 1)$(\sqrt{5} + 1)$: = frac(sqrt5 + 1)^2(sqrt5)^2 - 1^2 = frac5 + 1 + 2sqrt55 - 1$$= \frac{(\sqrt{5} + 1)^2}{(\sqrt{5})^2 - 1^2} = \frac{5 + 1 + 2\sqrt{5}}{5 - 1}$$ = frac6 + 2sqrt54 = frac3 + sqrt52$$= \frac{6 + 2\sqrt{5}}{4} = \frac{3 + \sqrt{5}}{2}$$ ### Step 3: Final Mapping Comparing frac3 + sqrt52$\frac{3 + \sqrt{5}}{2}$ to the given form fracalpha + betasqrt52$\frac{\alpha + \beta\sqrt{5}}{2}$: alpha = 3$\alpha = 3$ beta = 1$\beta = 1$ Both are natural numbers. Their sum is alpha + beta = 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 sin 18^circ$\sin 18^\circ$ and cos 36^circ$\cos 36^\circ$ to memory as they form complementary golden ratio components in geometry and algebra. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Trigonometric Functions

### Related Formula Tangent sum formula: tan 2alpha = tan[(alpha + beta) + (alpha - beta)] = fractan(alpha + beta) + tan(alpha - beta)1 - tan(alpha + beta)tan(alpha - beta)$$\tan 2\alpha = \tan[(\alpha + \beta) + (\alpha - \beta)] = \frac{\tan(\alpha + \beta) + \tan(\alpha - \beta)}{1 - \tan(\alpha + \beta)\tan(\alpha - \beta)}$$ ### Core Logic Given cos(alpha + beta) = -frac110 implies tan(alpha + beta) = -sqrt99 = -3sqrt11$\cos(\alpha + \beta) = -\frac{1}{10} \implies \tan(\alpha + \beta) = -\sqrt{99} = -3\sqrt{11}$. Given sin(alpha - beta) = frac38 implies tan(alpha - beta) = frac3sqrt55 = frac3sqrt5sqrt11$\sin(\alpha - \beta) = \frac{3}{8} \implies \tan(\alpha - \beta) = \frac{3}{\sqrt{55}} = \frac{3}{\sqrt{5}\sqrt{11}}$. ### Step 1: Simplify Tangent of 2 Alpha tan 2alpha = frac-3sqrt11 + frac3sqrt5sqrt111 + 3sqrt11 cdot frac3sqrt5sqrt11 = fracfrac-3sqrt55 + 3sqrt55fracsqrt5 + 9sqrt5 = frac3(1 - 11sqrt5)sqrt11(9 + sqrt5)$$\tan 2\alpha = \frac{-3\sqrt{11} + \frac{3}{\sqrt{5}\sqrt{11}}}{1 + 3\sqrt{11} \cdot \frac{3}{\sqrt{5}\sqrt{11}}} = \frac{\frac{-3\sqrt{55} + 3}{\sqrt{55}}}{\frac{\sqrt{5} + 9}{\sqrt{5}}} = \frac{3(1 - 11\sqrt{5})}{\sqrt{11}(9 + \sqrt{5})}$$ ### Step 2: Compare Constants Comparing with frac3(1 - rsqrt5)sqrt11(s + sqrt5)$\frac{3(1 - r\sqrt{5})}{\sqrt{11}(s + \sqrt{5})}$: r = 11, quad s = 9 implies r + s = 20$$r = 11, \quad s = 9 \implies r + s = 20$$ ### Pattern Recognition Express 2alpha$2\alpha$ as (alpha+beta) + (alpha-beta)$(\alpha+\beta) + (\alpha-\beta)$ to apply standard compound tangent formula directly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Trigonometry

### Related Formula cos 2theta = 2cos^2theta - 1$$\cos 2\theta = 2\cos^2\theta - 1$$ ### Core Logic Substitute the double-angle formula into the given equation to form a quadratic in costheta$\cos\theta$: sqrt3(2cos^2theta - 1) + 8costheta + 3sqrt3 = 0$$\sqrt{3}(2\cos^2\theta - 1) + 8\cos\theta + 3\sqrt{3} = 0$$ 2sqrt3cos^2theta + 8costheta + 2sqrt3 = 0$$2\sqrt{3}\cos^2\theta + 8\cos\theta + 2\sqrt{3} = 0$$ ### Step 1: Solve the Quadratic Factorize the quadratic equation: 2sqrt3cos^2theta + 2costheta + 6costheta + 2sqrt3 = 0$$2\sqrt{3}\cos^2\theta + 2\cos\theta + 6\cos\theta + 2\sqrt{3} = 0$$ Wait, 2 times 2sqrt3 = 12$2 \times 2\sqrt{3} = 12$. The factors of 12$12$ that sum to 8$8$ are 6$6$ and 2$2$. 2costheta(sqrt3costheta + 1) + 2sqrt3(sqrt3costheta + 1) = 0$$2\cos\theta(\sqrt{3}\cos\theta + 1) + 2\sqrt{3}(\sqrt{3}\cos\theta + 1) = 0$$ (sqrt3costheta + 1)(2costheta + 2sqrt3) = 0$$(\sqrt{3}\cos\theta + 1)(2\cos\theta + 2\sqrt{3}) = 0$$ This gives: costheta = -frac1sqrt3 quad textor quad costheta = -sqrt3$$\cos\theta = -\frac{1}{\sqrt{3}} \quad \text{or} \quad \cos\theta = -\sqrt{3}$$ Since -1 leq costheta leq 1$-1 \leq \cos\theta \leq 1$, we reject costheta = -sqrt3$\cos\theta = -\sqrt{3}$. ### Step 2: Count Solutions in Interval We need solutions for costheta = -frac1sqrt3$\cos\theta = -\frac{1}{\sqrt{3}}$ in the interval [-3pi, 2pi]$[-3\pi, 2\pi]$. The period of cosine is 2pi$2\pi$. The equation costheta = k$\cos\theta = k$ (where -1 < k < 0$-1 < k < 0$) has 2$2$ solutions per 2pi$2\pi$ interval. Intervals: [0, 2pi]$[0, 2\pi]$: 2$2$ solutions (in Quadrants II and III). [-2pi, 0]$[-2\pi, 0]$: 2$2$ solutions. [-3pi, -2pi]$[-3\pi, -2\pi]$: 1$1$ solution (in Quadrant II equivalent, which is Quadrant III when going backwards. Specifically, from -3pi$-3\pi$ to -2pi$-2\pi$ covers the top half of the circle. Wait, [-3pi, -2pi]$[-3\pi, -2\pi]$ goes from 180^circ$180^\circ$ to 360^circ$360^\circ$ logically, i.e., quadrants III and IV. cos$\cos$ 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 (2pi$2\pi$ cycles yield 2$2$ roots for |cos x|<1$|\cos x|<1$) prevents overcounting when domain bounds don't cleanly align with full periods. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Trigonometric Functions