Students may find the following textbook questions difficult.
Refer to the figure below, given that
Show that .
To use the given information , we express and in terms of and .
– – – – – – (*)
Then, many students stop there and don’t know how to find . May be, students find that, there are, at least, two different approaches.
One is geometry. , the direct meaning in the figure is , but, it is not that easy to express explicitly.
Another one is algebra, that is using trigonometric identities to do some algebraic computation so as to figure out the relation between and .
We select the latter.
To find , we need to get rid of , hence all we need is replacing by something involves . Here is the crucial key: , and (*) will be
Well, it is not a must that students can obtain the required results, because, algebraic computation is still nightmares to some students.
Urm, what’s next? Some students may expand further and come up with monsters. But, once we discover that there are similar non-zero terms () on both sides, cancel it and the expression above can be further simplied as
This is actually a past-paper question (1992 paper 2), just highlight one of the parts.
A regular pentagon inscribing in a circle with radius . Prove that
The previous result showing that
According to the marking scheme, we have
But, as I’d mentioned in class, it seems that the presentation above is not clear enough. Something is missing in the marking scheme. Let me explain.
Refer to the following figure.
By cosine law, it is easy to show that
– – – – – – (**)
Imagine that the point is moving along the circumference, like
The formula (**) is still valid.
How about when moves to some position such that ? Like
Is (**) still valid? Well, we need to show it out! Just refer to the figure above, by cosine law
Yes, (**) is still valid for . Hence, we have