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

Two examples that spring to mind include:

- solving for (relative velocity example with below)
- maximising
*without*the use of calculus

Note first of all the similarity between:

.

This identity is in the Department of Education formula booklet.

The only problem is that and 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 and so if we multiply our expression by then we have something:

.

Similarly, we have

,

where .

### 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 such that is in the direction as shown.

If

, then .

Therefore

.

For this to be in the appropriate direction, the direction, the -component divided by the -component, must be in the ratio (see the image above):

.

Note . Take 21 as an adjacent and 20 as an opposite to . Pythagoras Theorem shows that the hypotenuse is 29:

Now we can divide both sides of the equation by (no need to multiply by with the right-hand side):

.

Now but there is another solution in the second quadrant:

That is another solution is . With we have two solutions:

, and

.

Therefore we have two possible starting angles for and so two possible times for . From above,

.

The first solution yields:

.

Using Pythagoras theorem, this velocity has speed:

,

and using Pythagoras Theorem again, we know the distance is 29, so the first time is:

s.

Similarly, the other answer is s.

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