Question

$$\sin^{\circ3}\frac{2a}{9}-\sin^{2}\frac{2}{9}-\frac{3}{4}(sin\frac{2a}{9}-sin\frac{\sigma}{9})$$

Answer

$$9*sin^32*a-sin^229-306*sin(2*a)+306*IM*g*m*a*sin(s)$$

Solution


Regroup terms.
\[{sin}^{32}a\times 9-{sin}^{229}-34(9\sin{2a}-\sin{s}\imath gma\times 9)\]
Regroup terms.
\[{sin}^{32}a\times 9-{sin}^{229}-34(9\sin{2a}-9\imath gma\sin{s})\]
Regroup terms.
\[9{sin}^{32}a-{sin}^{229}-34(9\sin{2a}-9\imath gma\sin{s})\]
Expand by distributing terms.
\[9{sin}^{32}a-{sin}^{229}-(306\sin{2a}-306\imath gma\sin{s})\]
Remove parentheses.
\[9{sin}^{32}a-{sin}^{229}-306\sin{2a}+306\imath gma\sin{s}\]