Question

\[\sin ^{3}\alpha=3\sin \alpha-4\sin ^{3}\alpha\]

Answer 1

identities

Answer 2

you just need to find the value for \(\alpha\) - correct?

Answer 3

no, i need to prove they are equal

Answer 4

to simplify, just set \(x=\sin\alpha\) to get:\[x^3=3x-4x^3\]solve for x and then you can find \(\alpha\)

Answer 5

It's not an equation asnaseer. It's and identity

Answer 6

how can it be an identity if it is only true for a certain set of values for \(\alpha\)?

Answer 7

It's true for all values of alpha.

Answer 8

\[\sin (\alpha )=\frac{e^{i \alpha }-e^{-i \alpha }}{2 i}\]
\[\left(\frac{e^{i \alpha }-e^{-i \alpha }}{2 i}\right)^3\]=
\[\frac{3}{4} \frac{\left(i e^{-i \alpha }- i e^{i \alpha }\right)}{2}-\frac{1}{4}\frac{\left(i e^{-3 i \alpha }- i e^{3 i \alpha }\right)}{2}\]

\[\frac{3}{4}(\text{Sin} \alpha )-\frac{1}{4}\text{Sin} 3\alpha\]

Answer 9

That is the problem. To show that the two sides are the same thing.

Answer 10

take \(\alpha=90\), we get:
1 = 3 - 4
which is not true.

Answer 11

@tomaasc - maybe you have stated the identity incorrectly? - please double check.

Answer 12

asnaseer is right, it's not an identity.

Answer 13

It is \[\sin[x]^3=\frac{3}{4}(\text{Sin} \alpha )-\frac{1}{4}\text{Sin} 3\alpha\]

not what you wrote

Answer 14

Then he copied the problem wrong because he is trying to prove identities.

Answer 15

@imranmeah91 has stated a correct identity - is that the one you wanted to prove @tomaasc?

Answer 16

maybe the left side is 3sin

Answer 17

Well, it's your problem. Don't you know what it is/

Answer 18

yes 3 Sin[x]^3 is what you wrote