Winigefx
2022-06-24
Answered

trigonometry equation - ${\mathrm{sin}}^{3}(x)+{\mathrm{sin}}^{3}(2x)+{\mathrm{sin}}^{3}(3x)=(\mathrm{sin}(x)+\mathrm{sin}(2x)+\mathrm{sin}(3x){)}^{3}$

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Lilliana Burton

Answered 2022-06-25
Author has **19** answers

Let $\mathrm{sin}x=a$, $\mathrm{sin}2x=b$, and $\mathrm{sin}3x=c$

Expand out

$(a+b+c{)}^{3}$

Then subtract ${a}^{3},{b}^{3}$ and ${c}^{3}$, which yields

$3{a}^{2}b+3{a}^{2}c+3a{b}^{2}+6abc+3a{c}^{2}+3{b}^{2}c+3b{c}^{2}=0$

The long expression can be factored nicely into :

3(a+b)(b+c)(a+c), which equal to 0.

The factor 3 makes no difference, so solve for:

$\mathrm{sin}(x)+\mathrm{sin}(2x)=0$,

$\mathrm{sin}(2x)+\mathrm{sin}(3x)=0$, and

$\mathrm{sin}(x)+\mathrm{sin}(3x)=0$.

Expand out

$(a+b+c{)}^{3}$

Then subtract ${a}^{3},{b}^{3}$ and ${c}^{3}$, which yields

$3{a}^{2}b+3{a}^{2}c+3a{b}^{2}+6abc+3a{c}^{2}+3{b}^{2}c+3b{c}^{2}=0$

The long expression can be factored nicely into :

3(a+b)(b+c)(a+c), which equal to 0.

The factor 3 makes no difference, so solve for:

$\mathrm{sin}(x)+\mathrm{sin}(2x)=0$,

$\mathrm{sin}(2x)+\mathrm{sin}(3x)=0$, and

$\mathrm{sin}(x)+\mathrm{sin}(3x)=0$.

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