Further examples on the addition formula
Here's another example to work through. This time you need to draw two right-angled triangles to help you with your working.
Question
If \(\sin p = \frac{3}{5}\) and \(\tan q = \frac{5}{{12}}\) where \(0 \le p \le \frac{\pi }{2}\) and \(0 \le q \le \frac{\pi }{2}\), find the exact value of \(\sin (p - q)\).
From \(\sin p = \frac{3}{5}\)
From \(\tan q = \frac{5}{{12}}\)
Use Pythagoras' theorem to work out the third side of each of these triangles. Then insert these values into your working as follows.
\(\sin (p - q) = \sin p\cos q - \cos p\sin q\)
\(= \frac{3}{5} \times \frac{{12}}{{13}} - \frac{4}{5} \times \frac{5}{{13}}\)
\(= \frac{{36}}{{65}} - \frac{{20}}{{65}}\)
\(= \frac{{16}}{{65}}\)
You can use the formulae to expand and simplify expressions such as \(\cos \left( {\frac{\pi }{2} - 3x} \right)\).
\(\cos \left( {\frac{\pi }{2} - 3x} \right) = \cos \frac{\pi }{2}\cos 3x + \sin \frac{\pi }{2}\sin 3x\)
\(= 0 \times \cos 3x + 1 \times \sin 3x\)
\(= \sin 3x\)
To expand your knowledge, use the formulae to prove some of the standard formulae that you already know.
You can practise with the following:
\(\sin x^\circ = \cos (90 - x)^\circ\)
\(\sin x = \sin (\pi - x)\)
\(\cos x^\circ = - \cos (180 + x)^\circ\)
\(\cos x = \cos (2\pi - x)\)
\(\cos x = \cos ( - x)\)(Hint): \(\cos x = \cos (0 - ( - x))\)
\(\sin ( - x) = - \sin x\)(Hint): \(\sin ( - x) = \sin (0 - x)\)