Another valid approach, which was posted but deleted earlier, is to use the rule that
the sine of any angle is the cosine of its complement and vice-versa
This is easy to prove... draw a right angled triangle, call the two non-right angles
and
. We know that the hypoteneuse (with length
) is opposite the right angle. Lets arbitrarily call the side opposite
as having length
, so that the third side (opposite
and adjacent to
) has length
. Now:
and
and, since
we can re-write these as
} \qquad \qquad \qquad \cos\alpha = \sin{\left(90^\circ - \alpha\right)})
or, in radians
} \qquad \qquad \qquad \cos\alpha = \sin{\left(\frac{\pi}{2} - \alpha\right)})
Now, I admit that this only establishes these results for acute angles, but with quadrants diagrams and angles of any magnitude, I assure you that they do generalise. And, unlike the double angle formulae / expansions above (which are true but only covered in MX1 and MX2), knowing this result is definitely within the Advanced Maths syllabus.
So, how does this help?
We want
)
. This deals with the supplementary angle (angles adding to 180 degrees) but need a form with the complementary angle (angles adding to 90 degrees), so we rewrite:
 - \frac{\pi}{2} \qquad \text{adding and substracting $\frac{\pi}{2}$ as we need one to make a complementary angle} \\ &= \frac{\pi}{2} - \left(-\pi + x + \frac{\pi}{2}\right) \qquad \text{as a complementary angle will be of the form $\frac{\pi}{2} - \theta$} \\ &= \frac{\pi}{2} - \left(x - \frac{\pi}{2}\right) \qquad \text{on simplifying} \\ &= \cfrac{\pi}{2} - \theta \qquad \text{letting $\theta = x - \frac{\pi}{2}$} \end{align*})
Alternatively, we can start with the complementary form and rearrange to find
, as in:
Either way, having found a form involving a complementary angle, we can use the above rule to conclude that:
} = \sin{\left(\frac{\pi}{2} - \theta\right) = \cos\theta)
And we know need only simplify
by using the definition of
:
} \\ &= \cos{\left[-\left(\frac{\pi}{2} - x\right)\right]} \\ &= \cos{\left(\frac{\pi}{2} - x\right)} \qquad \text{as the cosine function is even} \\ &= \sin{x} \qquad \text{as $x$ and $\frac{\pi}{2} - x$ are complementary angles} \\ \text{Thus,} \quad \sin{(\pi - x)} &= \sin x \qquad \text{as both are equal to $\cos\theta$} \end{align*})
