Tuesday, April 7, 2015

[AM_FTDA20150402] Proof of Triple Angle Sine Identity



     We are given a triple angle, and we want an expression in terms of  sin q  only, without any double or triple angle.  First, we split up 3q  into  2q +q.  see [1].  This allows us to use the compound angle formula, and then double angle formulas.  At every step, it is a good tactic is to compare what you have with what you want.  This problem solving heuristic is not in the official list, but from my experience it is a useful one.  So at [2], we have three choices for the cos 2q :  cos 2q  = cos2q  – sin2q ,  cos 2q  = 2cos2q  – 1,  and  cos 2q  = 1– 2sin2q.  Which one shall we choose?  Since everything needs to be expressed in terms of  sin q  only,  the best choice is the third formula.  At [3], we have a  cos2q  appearing in the first term.  Again we express that in terms of  sin q  via  cos2q  = 1 – sin2q.  After that, we simplify to get to the RHS.

H10. Simplify the problem
H11. Solve part of the problem
Hxx. compare what you have with what you want

Suitable Levels
GCE ‘O’ Level Additional Mathematics
* revision for GCE ‘A’ Level H2 Mathematics
* revision for IB Mathematics HL / SL
* other syllabuses that teach further trigonometry

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