Occasionally, it might be useful to do as the title here suggests.

Two examples that spring to mind include:

• solving $a\cdot\cos\theta\pm b\cdot\sin\theta=c$ for $\theta$ (relative velocity example with $-$ below)
• maximising $a\cdot\cos\theta\pm b\cdot\sin\theta$ without the use of calculus

### $a\cdot \cos\theta- b\cdot\sin\theta$

Note first of all the similarity between: $\displaystyle a\cdot \cos\theta-b\cdot \sin \theta\sim \sin\phi\cos\theta-\cos\phi\sin\theta$.

This identity is in the Department of Education formula booklet.

The only problem is that $a$ and $b$ are not necessarily sines and cosines respectively. Consider them, however, as opposites and adjacents to an angle in a right-angled-triangle as shown: Using Pythagoras Theorem, the hypotenuse is $\sqrt{a^2+b^2}$ and so if we multiply our expression by $\displaystyle \frac{\sqrt{a^2+b^2}}{\sqrt{a^2+b^2}}$ then we have something: $\displaystyle \frac{\sqrt{a^2+b^2}}{\sqrt{a^2+b^2}}\cdot \left(a\cdot \cos\theta- b\cdot\sin\theta\right)$ $\displaystyle=\sqrt{a^2+b^2}\cdot \left(\frac{a}{\sqrt{a^2+b^2}}\cos\theta-\frac{b}{\sqrt{a^2+b^2}}\sin\theta\right)$ $=\sqrt{a^2+b^2}\cdot \left(\sin\phi\cos\theta-\cos\phi\sin\theta\right)=\sqrt{a^2+b^2}\sin(\phi-\theta)$.

Similarly, we have $a\cdot\cos\theta+b\cdot \sin\theta=\sqrt{a^2+b^2}\sin(\phi+\theta)$,

where $\displaystyle\sin\phi=\frac{a}{\sqrt{a^2+b^2}}$.

### Example

A girl wishes to cross a river, which is 20 m wide, in a boat. The river flows with speed 5 m/s. The boat can travel at 4 m/s in still water. The girl wishes to land on the opposite side at a point which is 21 m downstream. Find the time taken.

[Example 4.8, Fundamental Applied Maths, 2nd Edition, Oliver Murphy]

Solution: The scenario looks as follows: The boat must head out at an $\theta$ such that $V_B=V_{BR}+V_R$ is in the direction as shown.

If $|V_{BR}|=4$, then $V_{BR}=4\cos\theta\, \hat{\textbf{\i}}+4\sin \theta\,\hat{\textbf{\j}}$.

Therefore $V_B=5\hat{\textbf{\i}}+4\cos\theta\, \hat{\textbf{\i}}+4\sin \theta\,\hat{\textbf{\j}}=(5+4\cos\theta)\,\hat{\textbf{\i}}+4\sin \theta\,\hat{\textbf{\j}}$.

For this to be in the appropriate direction, the direction, the $\hat{\textbf{\j}}$-component divided by the $\hat{\textbf{\i}}$-component, must be in the ratio $\displaystyle \frac{20}{21}$ (see the image above): $\displaystyle \frac{4\sin\theta}{5+4\cos\theta}=\frac{20}{21}$ $\Rightarrow 84\sin \theta=100+80\cos\theta$ $\Rightarrow 84\sin \theta-80\cos\theta=100$ $\Rightarrow 21 \sin\theta-20\cos\theta=25$.

Note $21\sin \theta-20\cos\theta\sim \cos\phi\sin\theta-\sin\phi\cos\theta=\sin(\theta-\phi)$. Take 21 as an adjacent and 20 as an opposite to $\phi$. Pythagoras Theorem shows that the hypotenuse is 29: Now we can divide both sides of the equation by $29$ (no need to multiply by $\frac{29}{29}$ with the right-hand side): $\displaystyle\frac{21}{29}\sin\theta-\frac{20}{29}\cos\theta=\frac{25}{29}$ $\displaystyle\Rightarrow \cos\phi\sin\theta-\sin\phi\cos\theta=\frac{25}{29}$ $\displaystyle\Rightarrow \sin(\theta-\phi)=\frac{25}{29}$ $\Rightarrow \theta-\phi=\sin^{-1}\left(\frac{25}{29}\right)$ $\Rightarrow \theta=\sin^{-1}(25/29)+\phi$.

Now $\sin^{-1}(25/29)\approx 59.55^\circ$ but there is another solution in the second quadrant: That is another solution is $180^\circ-59.55^\circ=120.45^\circ$. With $\varphi\approx 43.60^\circ$ we have two solutions: $\theta=59.55^\circ+43.60^\circ=103.2^\circ$, and $\theta=120.45^\circ+43.60^\circ=164.1^\circ$.

Therefore we have two possible starting angles for $V_{BR}$ and so two possible times for $V_B$. From above, $V_B=(5+4\cos\theta)\,\hat{\textbf{\i}}+4\sin \theta\,\hat{\textbf{\j}}$.

The first solution yields: $V_B=4.087\,\hat{\textbf{\i}}+3.894\,\hat{\textbf{\i}}$.

Using Pythagoras theorem, this velocity has speed: $|V_B|=\sqrt{4.087^2+3.894^2}=5.645$,

and using Pythagoras Theorem again, we know the distance is 29, so the first time is: $\displaystyle t_1=\frac{s}{|V_B|}=\frac{29}{5.645}\approx 5.137$ s.

Similarly, the other answer is $t_2\approx 18.25$ s.