# the double angle identities-sin2A

I have a question that asks: Express each of the following in the form $a\sin bA$. The first part of the question asks me to do this for

$a) 6\sin A\cos A$

The answer they give is $3\sin 2A$, but I though it would be $\sin 6A$. I don't understand why I am wrong, could someone please explain?

Also the two other parts of the question was:

$b)4\sin 2A\cos 2A$

$c)\sin \frac A2 \cos \frac A2$

And I just get those ones plain wrong. Is there something I am missing about the concept?

You know that $2\sin A \cos A = sin (2 A)$. You learnt that in school, most probably. Now you can write $6\sin A \cos A = 3(2\sin A \cos A) = 3\sin(2A)$.

On the other hand, there is no rule where you could equate $$6\sin A\cos A = \sin 6A$$

This argument is clearly wrong for $A=\frac\pi4$

Hint:

You only need to know the formula:

$$\sin 2a = 2 \sin a \cos a$$

Also take a look at this link .

$\sin2A=2\sin A\cos A$. So $3\sin 2A=6\sin A\cos A$. Look at the list of trigonometric identites here