Arthur Pratt

2021-12-30

Proving $\frac{2\mathrm{sin}x+\mathrm{sin}2x}{2\mathrm{sin}x-\mathrm{sin}2x}={\mathrm{csc}}^{2}x+2\mathrm{csc}x\mathrm{cot}x+{\mathrm{cot}}^{2}x$
Proving right hand side to left hand side:
${\mathrm{csc}}^{2}x+2\mathrm{csc}x\mathrm{cot}x+{\mathrm{cot}}^{2}x=\frac{1}{{\mathrm{sin}}^{2}x}+\frac{2\mathrm{cos}x}{{\mathrm{sin}}^{2}x}+\frac{{\mathrm{cos}}^{2}x}{{\mathrm{sin}}^{2}x}$ (1)
$=\frac{{\mathrm{cos}}^{2}x+2\mathrm{cos}x+1}{{\mathrm{sin}}^{2}x}$ (2)
$=\frac{1-{\mathrm{sin}}^{2}x+1+\frac{\mathrm{sin}2x}{\mathrm{sin}x}}{{\mathrm{sin}}^{2}x}$ (3)
$=\frac{\frac{2\mathrm{sin}x-{\mathrm{sin}}^{3}x+\mathrm{sin}2x}{\mathrm{sin}x}}{{\mathrm{sin}}^{2}x}$ (4)
$=\frac{2\mathrm{sin}x-{\mathrm{sin}}^{3}x+\mathrm{sin}2x}{{\mathrm{sin}}^{3}x}$ (5)
I could not prove further to the left hand side from here.

hysgubwyri3

We have that, using $\mathrm{sin}2x=2\mathrm{sin}x\mathrm{cos}x$ and cancelling out $2\mathrm{sin}x\ne 0$
$\frac{2\mathrm{sin}x+\mathrm{sin}2x}{2\mathrm{sin}x-\mathrm{sin}2x}=\frac{2\mathrm{sin}x+2\mathrm{sin}x\mathrm{cos}x}{2\mathrm{sin}x-2\mathrm{sin}x\mathrm{cos}x}=\frac{1+\mathrm{cos}x}{1-\mathrm{cos}x}$
then, assuming $\mathrm{cos}x\ne -1$ ( which holds since we need $\mathrm{sin}x\ne 0$ for the original expression), multiply by $\frac{1+\mathrm{cos}x}{1+\mathrm{cos}x}$ and simplify to obtain the result using that $1-{\mathrm{cos}}^{2}x={\mathrm{sin}}^{2}x$

Cheryl King

The following relations would be used in the answer:
$1-{\mathrm{cos}}^{2}x={\mathrm{sin}}^{2}x$
$\mathrm{sin}2x=2\mathrm{sin}x\mathrm{cos}x$
Taking LHS,
$\frac{2\mathrm{sin}x+\mathrm{sin}2x}{2\mathrm{sin}x-\mathrm{sin}2x}=\frac{2\mathrm{sin}x\left(1+\mathrm{cos}x\right)}{2\mathrm{sin}x\left(1-\mathrm{cos}x\right)}$
Cancelling $\mathrm{sin}2x$
$=\frac{1+\mathrm{cos}x}{1-\mathrm{cos}x}$
Multiplying $1+\mathrm{cos}x$ to numerator and denominator
$=\frac{{\left(1+\mathrm{cos}x\right)}^{2}}{1-{\mathrm{cos}}^{2}x}$
which is equivalent to
$=\frac{\left(1+{\mathrm{cos}}^{2}x+2\mathrm{cos}x\right)}{{\mathrm{sin}}^{2}x}$
Divinding each term by denominator, we get LHS as
$={\mathrm{csc}}^{2}x+2\mathrm{csc}x\mathrm{cot}x+{\mathrm{cot}}^{2}x$
which is the RHS.
Hence, Proved.

nick1337

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