Evaluate.

$\mathrm{tan}1/2x=\mathrm{csc}x-\mathrm{cot}x$

$\mathrm{tan}1/2x=\mathrm{csc}x-\mathrm{cot}x$

PoentWeptgj
2022-08-01
Answered

Evaluate.

$\mathrm{tan}1/2x=\mathrm{csc}x-\mathrm{cot}x$

$\mathrm{tan}1/2x=\mathrm{csc}x-\mathrm{cot}x$

You can still ask an expert for help

Quenchingof

Answered 2022-08-02
Author has **14** answers

$\mathrm{tan}1/2x=\mathrm{csc}x-\mathrm{cot}x$

$\mathrm{sin}(A+B)=\mathrm{sin}A\mathrm{cos}B+\mathrm{cos}A\mathrm{sin}B\text{}and\text{}\mathrm{cos}(A+B)=\mathrm{cos}A\mathrm{cos}B-\mathrm{sin}A\mathrm{sin}B$

let B = A then $\mathrm{sin}(2A)=2\mathrm{sin}A\mathrm{cos}A$ ..... (1) and

$\mathrm{cos}(2A)={\mathrm{cos}}^{2}A-{\mathrm{sin}}^{2}A=(1-{\mathrm{sin}}^{2}A)-{\mathrm{sin}}^{2}A=1-2{\mathrm{sin}}^{2}A\therefore 2{\mathrm{sin}}^{2}A=1-\mathrm{cos}(2A)$,,,,,,(2)

$\mathrm{tan}(x/2)=[\mathrm{sin}(x/2)]/[\mathrm{cos}(x/2)]=[2\mathrm{sin}(x/2)\mathrm{sin}(x/2)]/[2\mathrm{sin}(x/2)\mathrm{cos}(x/2)]$

multiply numerator and denominator by $2\mathrm{sin}(x/2)$

$=[1-\mathrm{cos}x]/[\mathrm{sin}x]$ use (1) and (2)

$=[1/\mathrm{sin}x]-[\mathrm{cos}x/\mathrm{sin}x]=\mathrm{csc}x-\mathrm{cot}x$

$\mathrm{sin}(A+B)=\mathrm{sin}A\mathrm{cos}B+\mathrm{cos}A\mathrm{sin}B\text{}and\text{}\mathrm{cos}(A+B)=\mathrm{cos}A\mathrm{cos}B-\mathrm{sin}A\mathrm{sin}B$

let B = A then $\mathrm{sin}(2A)=2\mathrm{sin}A\mathrm{cos}A$ ..... (1) and

$\mathrm{cos}(2A)={\mathrm{cos}}^{2}A-{\mathrm{sin}}^{2}A=(1-{\mathrm{sin}}^{2}A)-{\mathrm{sin}}^{2}A=1-2{\mathrm{sin}}^{2}A\therefore 2{\mathrm{sin}}^{2}A=1-\mathrm{cos}(2A)$,,,,,,(2)

$\mathrm{tan}(x/2)=[\mathrm{sin}(x/2)]/[\mathrm{cos}(x/2)]=[2\mathrm{sin}(x/2)\mathrm{sin}(x/2)]/[2\mathrm{sin}(x/2)\mathrm{cos}(x/2)]$

multiply numerator and denominator by $2\mathrm{sin}(x/2)$

$=[1-\mathrm{cos}x]/[\mathrm{sin}x]$ use (1) and (2)

$=[1/\mathrm{sin}x]-[\mathrm{cos}x/\mathrm{sin}x]=\mathrm{csc}x-\mathrm{cot}x$

asked 2021-08-20

Let P(x, y) be the terminal point on the unit circle determined by t. Then

asked 2020-12-05

Add the polynomials:
(6.1y + 3.2x) + (4.8y – 3.2x)

asked 2022-06-03

Prove $\mathrm{tan}{84}^{\circ}=\mathrm{tan}{78}^{\circ}+\mathrm{tan}{72}^{\circ}+\mathrm{tan}{60}^{\circ}$

asked 2022-05-22

Prove: ${\mathrm{cos}}^{2}x(\mathrm{sec}x-1)(\mathrm{sec}x+1)=(1-\mathrm{cos}x)(1+\mathrm{cos}x)$

asked 2022-04-11

Understanding equality $\frac{1}{2}{\int}_{0}^{2\pi}\frac{\mathrm{cos}\left(\theta \right)}{2+\mathrm{cos}\left(\theta \right)}d\theta =\pi -{\int}_{0}^{2\pi}\frac{d\theta}{2+\mathrm{cos}\left(\theta \right)}$

asked 2022-05-03

Factor the polynomial $${s}^{2}-9s+14$$.

asked 2022-04-10

Translate $2xy={x}^{2}-{y}^{2}$ to polar coordinates

Why it confuses me:

$x=r\mathrm{cos}\varphi$

$y=r\mathrm{sin}\varphi$

Then:

$2{r}^{2}\mathrm{cos}\left\{\varphi \right\}\mathrm{sin}\left\{\varphi \right\}={r}^{2}\mathrm{cos}\left\{\varphi \right\}-{r}^{2}\mathrm{sin}\left\{\varphi \right\}={r}^{2}({\mathrm{cos}}^{2}\left\{\varphi \right\}-{\mathrm{sin}}^{2}\left\{\varphi \right\})$

$r}^{2}\mathrm{sin}\left\{2\varphi \right\}={r}^{2}\mathrm{cos}\left\{2\varphi \right\$

$\mathrm{sin}\left\{2\varphi \right\}=\mathrm{cos}\left\{2\varphi \right\}$

Which gives:

$\varphi =\frac{\pi}{8}+\frac{\pi n}{2},\text{}\text{}\text{}\text{}\text{}n\in \mathbb{Z}$

But it's just a set of$\varphi$ values. How do i plot such a function? In polar coordinates r is a function of some $\varphi$ , isnt that?

Why it confuses me:

Then:

Which gives:

But it's just a set of