Review question

# For which $0^\circ < \theta < 500^\circ$ is $3\cos \theta + 4\sin \theta > 2$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R8709

## Solution

$3\cos \theta + 4\sin \theta$ may be written in the form $r\cos(\theta - \alpha)$, where $r > 0$.

1. Calculate the value of $r$, and show that one value of $\alpha$ is approximately $53.1^\circ$.

We have the compound angle identity, $\cos (\theta -\alpha) = \cos\theta\cos\alpha+\sin\theta\sin\alpha.$

Thus $r\cos (\theta -\alpha) = (r\cos\alpha)\cos\theta+(r\sin\alpha)\sin\theta.$

Comparing this with $3\cos \theta + 4\sin \theta$ gives us that $r\cos\alpha = 3, r\sin\alpha = 4.$

Squaring and adding these two equations gives us that $r^2 = 25$, so $r = 5$.

Dividing the two equations gives us that $\tan \alpha = \dfrac{4}{3}$, and one solution of this is $\alpha \approx 53.1^\circ$.

1. Hence show that one solution of $3\cos \theta + 4\sin \theta = 2$ is approximately $120^\circ$.

Using the results above, we can write the equation as $5\cos(\theta - 53.1^\circ) = 2.$ Two of the solutions to this are $\theta - 53.1^\circ = \cos^{-1} 0.4 \quad\implies\quad \theta \approx 119.5^\circ$ and $\theta - 53.1^\circ = -\cos^{-1} 0.4 \quad\implies\quad \theta \approx -13.3^\circ.$

Thus $120^\circ$ is an approximate solution to the equation.

1. State all the other solutions for $0^\circ \leq \theta \leq 500^\circ.$

As well as the above solutions, we can find others by adding integer multiples of $360^\circ$. We find that $119.5^\circ$, $346.7^\circ$ and $479.5^\circ$ are all the solutions within the given range.

1. Hence give the positive values of $\theta$ less than $500^\circ$ for which $3\cos \theta + 4\sin \theta > 2$.

A graph of the function $y=3\cos \theta + 4\sin \theta$ looks like this.

Note that the function switches between being above and below $y=2$ at each of the intersections found in part (c). When $\theta = 0$, the function is greater than $2$.

This must mean that $3\cos \theta + 4\sin \theta > 2$ on the intervals $0 < \theta < 119.5^\circ$ and $346.7^\circ < \theta < 479.5^\circ$.