要三角函數的n倍角公式

sin (nx) = 2 sin [(n – 1)x ] cos x - sin [(n – 2)x] -------------- (1)
cos (nx) = 2 cos [(n – 1)x ] cos x - cos [(n – 2)x] ------------- (2)

2 tan [(n – 1)x ] + tan x
tan (nx) = ---------------------------------------
1 - tan [(n – 1)x] tan x

where n is 2, 3, 4, 5, 6 ……etc

I do one example for cos (5x) using equation (2)

cos (5x) = 2 cos 4x cos x - cos 3x
cos (5x) = 2 (1 – 8 cos^2 x + 8 cos^4 x) cos x - (-3 cos x + 4cos^3x)
cos (5x) = (2 – 16 cos^2 x + 16 cos^4 x) cos x + 3 cos x - 4cos^3x
cos (5x) = 2cos x – 16 cos^3 x + 16 cos^5 x + 3 cos x - 4cos^3x
cos (5x) = 5cos x – 20 cos^3 x + 16 cos^5 x

You have to know what cos 4x and cos 3x in order to derive cos 5x
So does to know cos 3x and cos 2x to derive cos 4x …..etc.

cos 2x = -1 +2 cos^2 x
cos 3x = -3 cos x + 4cos^3x
cos 4x = 1 – 8 cos^2 x + 8 cos^4 x
cos 5x = 5 cos x – 20 cos^3 x + 16 cos^5 x

There are other ways to express multiple angles formula by using summation, series and factorial. I don’t think you understand it. This one is the simplest of all, but rather very tedious,

因為有時的式子很難明，所以最好有一次代入過程比我理解。另外，我不知道cot, sec, csc有沒有。

也太多了吧??????