Arthur Pratt

Answered

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.

Proving right hand side to left hand side:

I could not prove further to the left hand side from here.

Answer & Explanation

hysgubwyri3

Expert

2021-12-31Added 43 answers

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 $1-{\mathrm{cos}}^{2}x={\mathrm{sin}}^{2}x$

then, assuming

Cheryl King

Expert

2022-01-01Added 36 answers

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,

$\frac{2\mathrm{sin}x+\mathrm{sin}2x}{2\mathrm{sin}x-\mathrm{sin}2x}=\frac{2\mathrm{sin}x(1+\mathrm{cos}x)}{2\mathrm{sin}x(1-\mathrm{cos}x)}$

Cancelling$\mathrm{sin}2x$

$=\frac{1+\mathrm{cos}x}{1-\mathrm{cos}x}$

Multiplying$1+\mathrm{cos}x$ to numerator and denominator

$=\frac{{(1+\mathrm{cos}x)}^{2}}{1-{\mathrm{cos}}^{2}x}$

which is equivalent to

$=\frac{(1+{\mathrm{cos}}^{2}x+2\mathrm{cos}x)}{{\mathrm{sin}}^{2}x}$

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.

Taking LHS,

Cancelling

Multiplying

which is equivalent to

Divinding each term by denominator, we get LHS as

which is the RHS.

Hence, Proved.

nick1337

Expert

2022-01-08Added 573 answers

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