Write the trigonometric expression as an algebraic expression. $\mathrm{cos}(\mathrm{arcsin}x-\mathrm{arctan}2x)$

DofotheroU
2021-09-16
Answered

Write the trigonometric expression as an algebraic expression. $\mathrm{cos}(\mathrm{arcsin}x-\mathrm{arctan}2x)$

You can still ask an expert for help

odgovoreh

Answered 2021-09-17
Author has **107** answers

1. Let $u=\mathrm{arctan}2x$ , then $\mathrm{tan}u=\frac{2x}{1}=\frac{adj}{opp}.$

2.$hyp=\sqrt{1+4{x}^{2}}$

3. Then$\mathrm{cos}u=\frac{1}{\sqrt{1-4{x}^{2}}}=\mathrm{cos}\left(\mathrm{arctan}2x\right)$

4. And$\mathrm{sin}u=\frac{2x}{\sqrt{1-4{x}^{2}}}=\mathrm{sin}\left(\mathrm{arctan}2x\right)$

5. Let$v=\mathrm{arccos}x$ , then $\mathrm{cos}v=x.$

6.$opp=\sqrt{1-{x}^{2}}$

7. Then$\mathrm{sin}v=\frac{\sqrt{1-{x}^{2}}}{1}=\mathrm{sin}\left(\mathrm{arccos}x\right)$

8.$\mathrm{cos}(\mathrm{arcsin}x-\mathrm{arctan}2x)-\mathrm{cos}\left(\mathrm{arcsin}x\right)\mathrm{cos}\left(\mathrm{arctan}2x\right)+\mathrm{sin}\left(\mathrm{arcsin}x\right)\mathrm{sin}\left(\mathrm{arctan}2x\right)=$

$=\mathrm{cos}\left(\mathrm{arcsin}x\right)\mathrm{cos}\left(\mathrm{arctan}2x\right)+x\mathrm{sin}\left(\mathrm{arctan}2x\right)=\frac{2{x}^{2}+\sqrt{1-{x}^{2}}}{\sqrt{1-4{x}^{2}}}$

2.

3. Then

4. And

5. Let

6.

7. Then

8.

asked 2022-07-25

Show that f is continuous on $(-\mathrm{\infty},\mathrm{\infty})$.

$f(x)=\{\begin{array}{l}\mathrm{sin}(x)\text{}if\text{}x\pi /4\\ \mathrm{cos}(x)\text{}if\text{}x\ge \pi /4\end{array}$

$f(x)=\{\begin{array}{l}\mathrm{sin}(x)\text{}if\text{}x\pi /4\\ \mathrm{cos}(x)\text{}if\text{}x\ge \pi /4\end{array}$

asked 2022-11-03

Evaluate ${\int}_{0}^{\pi}\mathrm{log}(5-4\mathrm{cos}x)\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x$

asked 2021-09-12

h is related to one of the parent functions described in this chapter. Describe the sequence of transformations from f to h.

h(x)=|x+3|−5.

h(x)=|x+3|−5.

asked 2021-05-31

asked 2022-11-03

Solving $\frac{4}{\pi}\mathrm{arctan}(1+x)\mathrm{cosh}x=1$

asked 2022-11-05

Solve $\mathrm{sin}(z)=\frac{3+i}{4}$

What i did so far was this:

$$\mathrm{sin}(z)=\frac{3+i}{4}\Rightarrow \frac{{e}^{iz}-{e}^{-iz}}{2i}=\frac{3+i}{4}\Rightarrow {e}^{2iz}-1=\left(\frac{-1+3i}{2}\right){e}^{iz}$$

setting $u={e}^{iz}$ we'll have

$${u}^{2}+\left(\frac{1-3i}{2}\right)u-1=0$$

completing squares

$${(u+\frac{1-3i}{4})}^{2}=\frac{3}{8}(1+i)$$

seting $w=u+\frac{1-3i}{4}$

$${w}^{2}=\frac{3}{8}(1+i)$$

solving

$$w=\pm {2}^{\frac{-5}{4}}\sqrt{3}(\mathrm{cos}\left(\frac{\sqrt{2}}{4}\right)+i\mathrm{sin}\left(\frac{\sqrt{2}}{4}\right))$$

now i just have to substitute this in $w=u+\frac{1-3i}{4}$ and then in $u={e}^{iz}$, but the solution looks really big . So what did i do wrong?

What i did so far was this:

$$\mathrm{sin}(z)=\frac{3+i}{4}\Rightarrow \frac{{e}^{iz}-{e}^{-iz}}{2i}=\frac{3+i}{4}\Rightarrow {e}^{2iz}-1=\left(\frac{-1+3i}{2}\right){e}^{iz}$$

setting $u={e}^{iz}$ we'll have

$${u}^{2}+\left(\frac{1-3i}{2}\right)u-1=0$$

completing squares

$${(u+\frac{1-3i}{4})}^{2}=\frac{3}{8}(1+i)$$

seting $w=u+\frac{1-3i}{4}$

$${w}^{2}=\frac{3}{8}(1+i)$$

solving

$$w=\pm {2}^{\frac{-5}{4}}\sqrt{3}(\mathrm{cos}\left(\frac{\sqrt{2}}{4}\right)+i\mathrm{sin}\left(\frac{\sqrt{2}}{4}\right))$$

now i just have to substitute this in $w=u+\frac{1-3i}{4}$ and then in $u={e}^{iz}$, but the solution looks really big . So what did i do wrong?

asked 2021-12-15

How do you simplify $\mathrm{cos}\left(\mathrm{arctan}\left(x\right)\right)$ ?