## trigonometry-ratios of angles

is there a relation between the numerical answers of cos5A and cosA?

sin4A and sinA?

i want to work backwards, if it is possible. tried deriving a formula by myself, but couldnt.:(

 $$\begin{array}{l} \sin na = {}^n{C_1}{\cos ^{n - 1}}\sin a - {}^n{C_3}{\cos ^{n - 3}}a{\sin ^3}a + {}^n{C_5}{\cos ^{n - 5}}a{\sin ^5}a...... \\ \cos na = {\cos ^n}a - {}^n{C_2}{\cos ^{n - 2}}a{\sin ^2}a + {}^n{C_4}{\cos ^{n - 4}}a{\sin ^4}a............ \\ \end{array}$$ Where a is the angle and n an integer.
 I think, you can use the sum of two angles approach Sin4A = 2 Sin2A Cos2A = 2 (2 SinA CosA) Cos2A = 4 SinA CosA (Cos²A - Sin²A) = 4 SinA CosA (1 - 2Sin²A) = 4 CosA (SinA - 2 Sin³A) = 4 √(1 - Sin²A)(SinA - 2 Sin³A) Similar approach can be taken for other one.

## trigonometry-ratios of angles

Oh ok thank you !!

 What about for fractions ? For example sin1/3 x?
 Blog Entries: 9 Recognitions: Homework Help Science Advisor For fractions it's essentially not doable, except for n=2,3,4, because of the algebra involved.
 Did you have a problem with my general formulae?
 @studiot, no that's not it but it's hard to memorise it and it's not one of the formulas learnt in school for Now , so I can't exactly use it in my exam ! Thanks a lot though !! 😊😊😊😊
 Um wait what is C1 ,C2 etc...
 They are symbols for combination. Also written as C(n,1). If you have not studied permutations, combinations, factorial yet, then you wont understand them.
 Oh I see yep I'm only at the double angle formulae level 😓... And having problems with expressing cos4a or others in the form of simple trigo ratio eg. Cosa. Does anyone have any online website to recommend that solves this kind of problems ?
 Have you ever heard of wolframalpha? I am not sure if I should post links in this forum, but you can google it.
 These are the binomial coefficients also written $$\left( {\begin{array}{*{20}{c}} n \\ r \\ \end{array}} \right)$$ They are normally studied before trigonometry.