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cosine of pi/5

Tuesday 18 February 2014 An interesting exercise for HL students is to prove that the cosine of (36 degrees) is exactly one-half of the golden ratio (phi). Students are aware of 'special' angles, such as and , for which they need to know the sine, cosine and tangent ratios without the aid of a calculator. This is a skill they need to practice and one they will probably need to use on the Paper 1 (non-GDC) exam. But what is special about and in particular the ?  Well, very special indeed since there is no doubt that is a particularly 'special' number that IB maths students would have encountered at some time previously in their mathematical careers - and it can be proven that is exactly or .

One could take different routes to prove that .  It can be a nice exercise for HL students to do on their own (with some helpful hints or prompts) and a nice presention to an SL class during the study of trigonometry.

Best to start with a bit of geometry that illustrates why it's not so unusual for and to have a connection. Some students may have been shown some of the interesting properties of a regular pentagon especially with regard to the golden ratio. There are a few ratios of lengths of line segments in a regular pentagon that are equal to .  For example, in the regular pentagon shown at right the ratio is equal to .

It's easy to show that each interior angle of a regular pentagon must be 108 degress. Thus angle ABF in the figure must be 54 degrees and it follows that angle BAF is 36 degrees. It can be proven by similar triangles that . We won't take that approach here (perhaps in a future blog post) but given this fact we can see from right triangle ABF that .  If we let and since , then . Clearly F is the midpoint of AC; so then . Therefore, with , it follows that .

But let's prove this statement with an analytic approach with no reference to a regular polygon.

Hint 1 ;  that is, cosines of supplementary angles (sum=180) are equal. Students should verify this identity using their knowledge of the unit circle. Given that the cosine of an angle in standard position (vertex at origin) is equal to the x-coordinate of the point where the terminal side intersects the unit circle a diagram such as the one at left gives a good visual confirmation that .

Hint 2:  Rather than , consider ; and and are supplementary .
Now, using using sum and double-angle identities for cosine show that .
Hint 3: Letting , solve for in the equation .
Hint 4:  Using double-angle identity for cosine derive a half-angle formula for cosine. Use this with result for to finally arrive at exact value for , remembering that must be positive since is in the first quadrant.