- #1

- 219

- 0

sin4A and sinA?

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

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- Thread starter Celluhh
- Start date

- #1

- 219

- 0

sin4A and sinA?

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

- #2

- 5,439

- 9

\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}[/tex]

Where a is the angle and n an integer.

- #3

- 20

- 0

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.

- #4

- 219

- 0

Oh ok thank you !!

- #5

- 219

- 0

What about for fractions ? For example sin1/3 x?

- #6

- 13,046

- 594

For fractions it's essentially not doable, except for n=2,3,4, because of the algebra involved.

- #7

- 5,439

- 9

Did you have a problem with my general formulae?

- #8

- 219

- 0

Now , so I can't exactly use it in my exam ! Thanks a lot though !!

- #9

- 219

- 0

Um wait what is C1 ,C2 etc...

- #10

- 20

- 0

If you have not studied permutations, combinations, factorial yet, then you wont understand them.

- #11

- 219

- 0

- #12

- 20

- 0

- #13

- 5,439

- 9

[tex]\left( {\begin{array}{*{20}{c}}

n \\

r \\

\end{array}} \right)[/tex]

They are normally studied before trigonometry.

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