Random Maths...

Dear bloggie,

Randomly flip a page from my old Form 6 Math book and randomly selected one math question. Decided since nothing to blog about, might as well post my solution instead.

Prove that cos 3Ø + cos 5Ø + cos 7Ø = cos 5Ø (4 cos²Ø - 1)

Click to see my solution

I noticed the right side got cos 5Ø rather than normally all cos Ø. Besides I noticed 3Ø, 5Ø and 7Ø has a pattern each is +2Ø. May mean something or not, but my instinct tell me is worth investigation. Also, the power of 2 on the cos also bug me, I realized the right side can further be factorize to

cos 5Ø (4 cos²Ø - 1)
= cos 5Ø (2 cosØ - 1) (2 cosØ + 1)

Not sure if will be useful or not but just put a mental note, incase if I hit this value later when manipulating the left side. Also recalled a formula which involves cos²Ø;

cos² Ø + sin² Ø = 1

I decided to focus on 5Ø since it's the common point between the left and right. I came up with the idea of:

cos 3Ø + cos 5Ø + cos 7Ø
= cos (5Ø - 2Ø) + cos 5Ø + cos (5Ø + 2Ø)

Now I recall the formula

cos(a + b) = cos a cos b - sin a sin b

But problem is I don't recall any formula on cos(x - y). So I gotta see if there is a formula for cos(x - y). On immediate Math instinct, I get the urge to replace b = -y to get the formula but this is an angle rather than a integer value, I'm not clearly sure if I can do such replacement but I decided to follow the trail to see where replacing b = -y will lead me to first

cos(a - b)
= cos (a + (-b))
= cos a cos (-b) - sin a sin (-b)

Now I recalled seeing some formula bout transforming -vef values angle to +vef angle for cos and sin but I couldn't 100% recall the formula, so time to go back to definitions. After some experiments (drew some triangles on a Cartesian chart and calc values for -Ø), I found out that

sin (-Ø) = - sin Ø
cos (-Ø) = cos Ø

So I continue where I left of in the cos(a - b) problem.

cos(a - b)
= cos (a + (-b))
= cos a cos (-b) - sin a sin (-b)
= cos a cos b - sin a (- sin b)
= cos a cos b + sin a sin b

Ok, time to use the formula on the main problem,

cos 3Ø + cos 5Ø + cos 7Ø
= cos (5Ø - 2Ø) + cos 5Ø + cos (5Ø + 2Ø)
= cos 5Ø cos 2Ø + sin 5Ø cos 2Ø + cos 5Ø + cos 5Ø cos 2Ø - sin 5Ø sin 2Ø
= 2 cos 5Ø cos 2Ø + cos 5Ø
= cos 5Ø (2 cos 2Ø + 1)

Close, onli left the cos 2Ø, since cos 2Ø = cos² Ø - sin ² Ø

= cos 5Ø (2 (cos² Ø - sin ² Ø ) + 1)

Need convert sin ² Ø into cos² Ø

cos² Ø + sin² Ø = 1
sin² Ø = 1 - cos² Ø

Ok, from main one...

= cos 5Ø (2 (cos² Ø - sin ² Ø ) + 1)
= cos 5Ø (2 (cos² Ø - (1 - cos² Ø)) + 1)
= cos 5Ø (2 (cos² Ø - 1 + cos² Ø) + 1)
= cos 5Ø (2 (2 cos² Ø - 1) + 1)
= cos 5Ø (4 cos² Ø - 2 + 1)
= cos 5Ø (4 cos² Ø - 1) Q.E.D

And ya, been watching the tv series Numb3rs lately...