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:

triangle

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:

river

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:

triangle2

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:

 

circle

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.

 

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