Given that g(x)=5sin2x+12cos2x, express g(x) in the form Rain(2x+alpha), where R>0 and alpha is an acute angle. Find the minimum value of 26/g(x)+15

Step 1: Express $g(x)$ in the form $R\sin(2x+\alpha)$. We are given $g(x) = 5\sin(2x) + 12\cos(2x)$. We want to express this in the form $R\sin(2x+\alpha)$, where $R > 0$ and $\alpha$ is an acute angle. Using the compound angle formula, $R\sin(2x+\alpha) = R(\sin(2x)\cos\alpha + \cos(2x)\sin\alpha)$. $$ R\sin(2x+\alpha) = (R\cos\alpha)\sin(2x) + (R\sin\alpha)\cos(2x) $$ Comparing this to $5\sin(2x) + 12\cos(2x)$, we have: $$ R\cos\alpha = 5 \quad \text{(1)} $$ $$ R\sin\alpha = 12 \quad \text{(2)} $$ To find $R$, square and add (1) and (2): $$ (R\cos\alpha)^2 + (R\sin\alpha)^2 = 5^2 + 12^2 $$ $$ R^2(\cos^2\alpha + \sin^2\alpha) = 25 + 144 $$ Since $\cos^2\alpha + \sin^2\alpha = 1$: $$ R^2 = 169 $$ Since $R > 0$: $$ R = \sqrt{169} = 13 $$ To find $\alpha$, divide (2) by (1): $$ \frac{R\sin\alpha}{R\cos\alpha} = \frac{12}{5} $$ $$ \tan\alpha = \frac{12}{5} $$ Since $\alpha$ is an acute angle, we find its value: $$ \alpha = \arctan\left(\frac{12}{5}\right) \approx 67.38^\circ $$ Therefore, $g(x)$ can be expressed as: $$ g(x) = 13\sin(2x + \arctan\left(\frac{12}{5}\right)) $$ or approximately $$ g(x) = 13\sin(2x + 67.38^\circ) $$ Step 2: Find the minimum value of $\frac{26}{g(x)} + 15$. Substitute the expression for $g(x)$ into the given expression: $$ \frac{26}{g(x)} + 15 = \frac{26}{13\sin(2x+\alpha)} + 15 $$ $$ = \frac{2}{\sin(2x+\alpha)} + 15 $$ The range of the sine function is $-1 \le \sin(2x+\alpha) \le 1$. For the expression $\frac{2}{\sin(2x+\alpha)} + 15$ to be minimum, the term $\frac{2}{\sin(2x+\alpha)}$ must be minimum. This occurs when $\sin(2x+\alpha)$ is negative and has the largest possible magnitude, which is $\sin(2x+\alpha) = -1$. Substitute $\sin(2x+\alpha) = -1$ into the expression: $$ \text{Minimum value} = \frac{2}{-1} + 15 $$ $$ = -2 + 15 $$ $$ = 13 $$ The minimum value of $\frac{26}{g(x)} + 15$ is $\boxed{13}$. 3 done, 2 left today. You're making progress.