`sin^4(2x)` Use the power reducing formulas to rewrite the expression in terms of the first power of the cosine.

Friday, October 25, 2013

$\displaystyle{{\sin}}^{{4}}{\left({2}{x}\right)}$ Use the power reducing formulas to rewrite the expression in terms of the first power of the cosine.

According to the power reducing formulas, you may re-wrute the expression such that:

$\displaystyle{{\sin}}^{{4}}{\left({2}{x}\right)}={{\sin}}^{{2}}{\left({2}{x}\right)}\cdot{{\sin}}^{{2}}{\left({2}{x}\right)}=\frac{{{1}-{\cos{{2}}}\cdot{\left({2}{x}\right)}}}{{2}}\cdot\frac{{{1}-{\cos{{2}}}\cdot{\left({2}{x}\right)}}}{{2}}$

$\displaystyle{{\sin}}^{{4}}{\left({2}{x}\right)}=\frac{{{{\left({1}-{\cos{{4}}}{x}\right)}}^{{2}}}}{{4}}$

$\displaystyle{{\sin}}^{{4}}{\left({2}{x}\right)}=\frac{{{1}-{2}{\cos{{4}}}{x}+{{\cos}}^{{2}}{4}{x}}}{{4}}$

$\displaystyle{{\sin}}^{{4}}{\left({2}{x}\right)}=\frac{{{1}-{2}{\cos{{4}}}{x}+\frac{{{1}+{\cos{{8}}}{x}}}{{2}}}}{{4}}$

$\displaystyle{{\sin}}^{{4}}{\left({2}{x}\right)}=\frac{{{2}-{4}{\cos{{4}}}{x}+{1}+{\cos{{8}}}{x}}}{{8}}$

$\displaystyle{{\sin}}^{{4}}{\left({2}{x}\right)}=\frac{{{3}-{4}{\cos{{4}}}{x}+{\cos{{8}}}{x}}}{{8}}$

Hence, eusing the power reducing formulas yields $\displaystyle{{\sin}}^{{4}}{\left({2}{x}\right)}=\frac{{{3}-{4}{\cos{{4}}}{x}+{\cos{{8}}}{x}}}{{8}}.$

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Friday, October 25, 2013

$\displaystyle{{\sin}}^{{4}}{\left({2}{x}\right)}$ Use the power reducing formulas to rewrite the expression in terms of the first power of the cosine.

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