Layla Velazquez

2022-06-20

How can I express $\frac{\mathrm{sin}3a}{\mathrm{sin}a}$ while using only $\mathrm{cos}a$?

Amy Daniels

We know that
$\mathrm{sin}3a=\mathrm{sin}\left(2a+a\right)=\dots =3\mathrm{sin}a-4{\mathrm{sin}}^{3}a$
So
$\frac{\mathrm{sin}3a}{\mathrm{sin}a}=3-4{\mathrm{sin}}^{2}a=3-4+4{\mathrm{cos}}^{2}a=4{\mathrm{cos}}^{2}a-1$

cazinskup3

$\frac{\mathrm{sin}3a}{\mathrm{sin}a}=\frac{3\mathrm{sin}a-4{\mathrm{sin}}^{3}a}{\mathrm{sin}a}$
$=3-4{\mathrm{sin}}^{2}a$
$=3-4\left(1-{\mathrm{cos}}^{2}a\right)$
$=3-4+4{\mathrm{cos}}^{2}a$
$=4{\mathrm{cos}}^{2}a-1$

Do you have a similar question?