TG EAMCET 2 May 2025 Shift 2 Paper

We have the expression: $5 \sin \theta + 3 \cos \left(\theta + \frac{\pi}{3}\right) + 3$Step 1: Expand the cosine term$3 \cos \left(\theta + \frac{\pi}{3}\right)$ can be written using the formula: $\cos(A + B) = \cos A \cos B - \sin A \sin B$. $\cos \frac{\pi}{3} = \frac{1}{2}$ and $\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}$.So, $3 \cos \left(\theta + \frac{\pi}{3}\right) = 3 \left( \cos \theta \cdot \frac{1}{2} - \sin \theta \cdot \frac{\sqrt{3}}{2} \right) = \frac{3}{2} \cos \theta - \frac{3\sqrt{3}}{2} \sin \theta$Step 2: Combine all termsNow put this back into the main expression:$5 \sin \theta + \frac{3}{2} \cos \theta - \frac{3\sqrt{3}}{2} \sin \theta + 3$Combine the $\sin \theta$ terms:$5 \sin \theta - \frac{3\sqrt{3}}{2} \sin \theta = \left(5 - \frac{3\sqrt{3}}{2}\right) \sin \theta$So the expression becomes: $\left(5 - \frac{3\sqrt{3}}{2}\right) \sin \theta + \frac{3}{2} \cos \theta + 3$Step 3: Find the maximum and minimum valuesThis is now in the form $a \sin \theta + b \cos \theta + c$, where $a = 5 - \frac{3\sqrt{3}}{2}, b = \frac{3}{2}, c = 3$.The largest and smallest values of $a \sin \theta + b \cos \theta$ are $R$ and $-R$, where $R = \sqrt{a^2 + b^2}$.Calculate $R: R = \sqrt{ \left(5 - \frac{3\sqrt{3}}{2}\right)^2 + \left(\frac{3}{2}\right)^2 }$Expanding: $= \sqrt{25 + \frac{27}{4} - 15\sqrt{3} + \frac{9}{4}} = \sqrt{25 - 15\sqrt{3} + 9} = \sqrt{34 - 15\sqrt{3}}$Step 4: Substitute the extreme valuesThe minimum value is $-R + 3 = \alpha$. The maximum value is $R + 3 = \beta$.So, $\alpha = -R + 3$ and $\beta = R + 3$.Step 5: Calculate the final resultFind $\alpha + \beta$ : $\alpha + \beta = (-R + 3) + (R + 3) = 6$Find $\alpha - \beta$ : $\alpha - \beta = (-R + 3) - (R + 3) = -2R$Now, $(\alpha - \beta)(\alpha + \beta - 6) = (-2R) \times (6 - 6) = (-2R) \times 0 = 0$