1
$\begingroup$

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?

$\endgroup$
1
1
$\begingroup$

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 $A=\frac\pi4$

$\endgroup$
0
$\begingroup$

Hint:

You only need to know the formula:

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

Also take a look at this link .

$\endgroup$
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.