Trig Function Values 2Eb9F5

1. **State the problem:** We need to analyze the function $$y = 4\cos x + 3\sin x$$ for $x$ from $0^\circ$ to $270^\circ$ in $30^\circ$ intervals. 2. **Construct the table of values:** We calculate $y$ at each $x$ using the formula: $$y = 4\cos x + 3\sin x$$ | $x$ (degrees) | $\cos x$ | $\sin x$ | $y = 4\cos x + 3\sin x$ | |---------------|----------|----------|-------------------------| | $0^\circ$ | 1 | 0 | $4(1) + 3(0) = 4$ | | $30^\circ$ | $\frac{\sqrt{3}}{2} \approx 0.866$ | $\frac{1}{2} = 0.5$ | $4(0.866) + 3(0.5) = 3.464 + 1.5 = 4.964$ | | $60^\circ$ | $\frac{1}{2} = 0.5$ | $\frac{\sqrt{3}}{2} \approx 0.866$ | $4(0.5) + 3(0.866) = 2 + 2.598 = 4.598$ | | $90^\circ$ | 0 | 1 | $4(0) + 3(1) = 3$ | | $120^\circ$ | $-\frac{1}{2} = -0.5$ | $\frac{\sqrt{3}}{2} \approx 0.866$ | $4(-0.5) + 3(0.866) = -2 + 2.598 = 0.598$ | | $150^\circ$ | $-\frac{\sqrt{3}}{2} \approx -0.866$ | $\frac{1}{2} = 0.5$ | $4(-0.866) + 3(0.5) = -3.464 + 1.5 = -1.964$ | | $180^\circ$ | $-1$ | 0 | $4(-1) + 3(0) = -4$ | | $210^\circ$ | $-\frac{\sqrt{3}}{2} \approx -0.866$ | $-\frac{1}{2} = -0.5$ | $4(-0.866) + 3(-0.5) = -3.464 - 1.5 = -4.964$ | | $240^\circ$ | $-\frac{1}{2} = -0.5$ | $-\frac{\sqrt{3}}{2} \approx -0.866$ | $4(-0.5) + 3(-0.866) = -2 - 2.598 = -4.598$ | | $270^\circ$ | 0 | $-1$ | $4(0) + 3(-1) = -3$ | 3. **Graphing the function:** Plot the points $(x, y)$ from the table on a coordinate plane with $x$ in degrees and $y$ values as calculated. 4. **Solving the equations using the graph:** \u03b1. Solve $4\cos x + 3\sin x = 0$: From the table and graph, find $x$ where $y=0$. Between $120^\circ$ ($y=0.598$) and $150^\circ$ ($y=-1.964$), the function crosses zero. Use the formula: $$4\cos x + 3\sin x = 0 \implies 4\cos x = -3\sin x \implies \tan x = -\frac{4}{3}$$ Calculate: $$x = \arctan\left(-\frac{4}{3}\right)$$ This gives principal value $x \approx -53.13^\circ$, add $180^\circ$ for the second quadrant: $$x \approx 126.87^\circ$$ This matches the zero crossing between $120^\circ$ and $150^\circ$. \u03b2. Solve $8\cos x + 6\sin x = 100 - x$: This is a transcendental equation and cannot be solved exactly by graphing $y=8\cos x + 6\sin x$ and $y=100 - x$ on the same axes. Since $100 - x$ is very large compared to the range of $8\cos x + 6\sin x$ (which varies roughly between $-10$ and $10$), there is no solution in $0^\circ \leq x \leq 270^\circ$. \u03b3. Solve $8\cos x + 6\sin x - 9 = 0$: Rewrite as: $$8\cos x + 6\sin x = 9$$ The maximum value of $8\cos x + 6\sin x$ is: $$\sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10$$ Since $9 < 10$, solutions exist. Use the formula: $$8\cos x + 6\sin x = 9$$ Divide both sides by 10: $$\frac{8}{10}\cos x + \frac{6}{10}\sin x = \frac{9}{10}$$ Or: $$0.8\cos x + 0.6\sin x = 0.9$$ Rewrite as: $$R\cos(x - \alpha) = 0.9$$ where $$R = 1$$ and $$\cos \alpha = 0.8, \sin \alpha = 0.6$$ So $$x - \alpha = \pm \arccos(0.9)$$ Calculate: $$\alpha = \arccos(0.8) \approx 36.87^\circ$$ $$\arccos(0.9) \approx 25.84^\circ$$ Therefore, $$x = 36.87^\circ \pm 25.84^\circ$$ Solutions: $$x_1 = 36.87^\circ + 25.84^\circ = 62.71^\circ$$ $$x_2 = 36.87^\circ - 25.84^\circ = 11.03^\circ$$ Check if these are within $0^\circ$ to $270^\circ$ (they are). **Final answers:** - For $4\cos x + 3\sin x = 0$, $x \approx 126.87^\circ$ - For $8\cos x + 6\sin x = 100 - x$, no solution in $0^\circ \leq x \leq 270^\circ$ - For $8\cos x + 6\sin x - 9 = 0$, $x \approx 11.03^\circ$ and $62.71^\circ$