**Question**

**Introduction**

When
proving trigonometrical identities, one usually starts from the “more difficult”
side and try to work towards the other side. The above identity looks like a tough nut to
crack. Both sides look equally
complicated. Where do we even begin?

Here is one
way to “cheat”. Starting from, say, the
LHS, we multiply the RHS expression and also multiply by its reciprocal i.e. dividing
by the same amount. It is like using a

*magic wand*to*create something out of nothing*(无中生有 in Chinese), but doing so*does not change*the value of the LHS expression. Now we do not touch the part that is equal to the RHS (highlighted in yellow below), but we try to find a way to cancel away the other stuff, as shown here.**Solution**

**Remarks**

In
the third step, I had replaced cos

^{2}*A*with 1 – sin^{2}*A*and sin^{2}*B*with 1 – cos^{2}*B*. This leads to the required cancellation and we are left with the yellow patch, which was never touched since the first step and it is the RHS. This completes the proof.
Just
as in martial arts where deadly opponents require deadly strokes to counter
them, evil questions require “unorthodox” techniques. Even though most people would not have
thought of it, all the steps presented above are actually legitimate. This is because at every step, the equality
is preserved. All the “=” are really
equal, and it’s legit (either work hard or you might as well quit), although my “can’t touch this” tactic
seems a bit clairvoyant.

.

H04. Look for pattern(s)

H05. Work backwards

H08. Make suppositions

H09. Restate the problem in
another way

H10. Simplify the problem

H11. Solve part of the problem

H13* Use Equation / write a
Mathematical Sentence

**Suitable Levels**

*****GCE ‘O’ Level Additional Mathematics

* other syllabuses that involve proof of trigonometric
identities

* anyone who loves mathematical challenges

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