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How do you plot the polar coordinate #(2, (5pi)/6)#?

Alan P.
Featured 6 months ago

Polar coordinate $\left(2 , \frac{5 \pi}{6}\right)$ is a distance of $2$ from the origin
and at an angle of $\frac{5 \pi}{6}$

Explanation:

Note that the angle $\frac{5 \pi}{6}$ has a reference angle of $\frac{\pi}{6}$ above the $\pi$ axis.
$\frac{\pi}{6}$ is a common angle with hypotenuse $= 2$ and opposite side $= 1$

The inverse of the function sin(x) + cos(x) is...?

Douglas K.
Featured 5 months ago

Explanation:

Here is the graph:

This looks like a cosine function that has been multiplied by an amplitude and phase shift to the right:

$f \left(x\right) = A \cos \left(x - \phi\right) = \sin \left(x\right) + \cos \left(x\right)$

My graphing tool allows me to obtain values of points and I can tell you that $A = \sqrt{2}$

$\cos \left(x - \phi\right) = \frac{1}{\sqrt{2}} \sin \left(x\right) + \frac{1}{\sqrt{2}} \cos \left(x\right)$

This fits the trigonometric identity:

$\cos \left(x - \phi\right) = \sin \left(x\right) \sin \left(\phi\right) + \cos \left(x\right) \cos \left(\phi\right)$

where $\cos \left(\phi\right) = \sin \left(\phi\right) = \frac{1}{\sqrt{2}}$

This happens at $\phi = \frac{\pi}{4}$

The graphing tool confirms that the x coordinates are shift by $\frac{\pi}{4}$.

$f \left(x\right) = \sqrt{2} \cos \left(x - \frac{\pi}{4}\right)$

Substitute ${f}^{-} 1 \left(x\right)$ for every x:

$f \left({f}^{-} 1 \left(x\right)\right) = \sqrt{2} \cos \left({f}^{-} 1 \left(x\right) - \frac{\pi}{4}\right)$

The left side becomes x by definition:

$x = \sqrt{2} \cos \left({f}^{-} 1 \left(x\right) - \frac{\pi}{4}\right)$

Make the $\sqrt{2}$ on the right disappear by multiplying both sides by $\frac{\sqrt{2}}{2}$:

$\frac{\sqrt{2}}{2} x = \cos \left({f}^{-} 1 \left(x\right) - \frac{\pi}{4}\right)$

Use the inverse cosine on both sides:

${\cos}^{-} 1 \left(\frac{\sqrt{2}}{2} x\right) = {f}^{-} 1 \left(x\right) - \frac{\pi}{4}$

Solve for ${f}^{-} 1 \left(x\right)$:

${f}^{-} 1 \left(x\right) = {\cos}^{-} 1 \left(\frac{\sqrt{2}}{2} x\right) + \frac{\pi}{4}$

To confirm that this is truly an inverse, verify that $f \left({f}^{-} 1 \left(x\right)\right) = {f}^{-} 1 \left(f \left(x\right)\right) = x$

How do you sketch a right triangle corresponding to #sectheta=2# and find the third side, then find the other five trigonometric functions?

Parzival S.
Featured 4 months ago

See below:

Explanation:

With a triangle with $\sec \theta = 2$, let's first remember that:

$\sec \theta = \text{hypotenuse"/"adjacent} = \frac{2}{1}$

We could run through the pythagorean theorem for the third side, but we can also remember that these two measurements are part of a 30-60-90 triangle, and so the third side is $\sqrt{3}$. To prove it, let's go ahead and do Pythagorean Theorem:

${a}^{2} + {b}^{2} = {c}^{2}$

${1}^{2} + {\left(\sqrt{3}\right)}^{2} = {2}^{2}$

$1 + 3 = 4$

This gives us the 6 trig ratios:

$\sin \theta = \frac{\sqrt{3}}{2}$

$\cos \theta = \frac{1}{2}$

$\tan \theta = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \left(\frac{\sqrt{3}}{2}\right) \left(\frac{2}{1}\right) = \sqrt{3}$

$\csc \theta = \frac{2}{\sqrt{3}} = \frac{2 \sqrt{3}}{3}$

$\sec \theta = \frac{2}{1} = 2$

$\cot \theta = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$

The angle of the triangle we're looking at is $60 = \frac{\pi}{3}$, with the opposite being the middle length of $\sqrt{3}$, the adjacent length of 1, and hypotenuse of 2.

How do you divide #( i-3) / (5i +2)# in trigonometric form?

Alan P.
Featured 4 months ago

$\frac{i - 3}{5 i + 2} \approx 0.587 \cdot \left(\cos \left(1.630\right) + i \cdot \sin \left(- 1.630\right)\right)$

Explanation:

$\textcolor{red}{\text{~~~ Complex Conversion: Rectangular " harr " Tigonometric~~~}}$
$\textcolor{w h i t e}{\text{XX }} \textcolor{b l u e}{a + b i \leftrightarrow r \cdot \left[\cos \left(\theta\right) + i \cdot \sin \left(\theta\right)\right]}$
$\textcolor{w h i t e}{\text{XXX")color(blue)("where } r = \sqrt{{a}^{2} + {b}^{2}}}$
#color(white)("XXX")color(blue)("and "theta={("arctan"(b/a),"if "(a,bi) in "Q I or Q IV"),("arctan"(b/a)+pi,"if " (a,bi) in "Q II or Q III):})#

$\textcolor{red}{\text{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~}}$

$\textcolor{red}{\text{~~~ Trigonometric Division ~~~}}$
#color(white)("XX ")color(blue)((cos(theta)+i * sin(theta))/(cos(phi)+i * sin(phi)))#

#color(white)("XXX")color(blue)(= [color(green)((cos(theta) * cos(phi) + sin(theta) * sin(phi))] +i * [color(magenta)(sin(theta) * cos(phi)-cos(theta) * sin(phi))]#

$\textcolor{red}{\text{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~}}$

If $A = i - 3 = - 3 + 1 i$
then
$\textcolor{w h i t e}{\text{XXX}} {r}_{A} = \sqrt{{\left(- 3\right)}^{2} + {1}^{2}} = \sqrt{10}$
and
$\textcolor{w h i t e}{\text{XXX}} {\theta}_{A} = \arctan \left(- \frac{1}{3}\right) + \pi \approx 2.819842099$

If $B = 5 i + 2 = 2 + 5 i$
then
$\textcolor{w h i t e}{\text{XXX}} {r}_{B} = \sqrt{{2}^{2} + {5}^{2}} = \sqrt{29}$
and
$\textcolor{w h i t e}{\text{XXX}} {\theta}_{B} = \arctan \left(\frac{5}{2}\right) \approx 1.19028995$

$\frac{A}{B} = \frac{i - 3}{5 + 2 i}$

$\textcolor{w h i t e}{\text{XXX}} = \frac{\sqrt{10}}{\sqrt{29}} \cdot \left(\frac{\cos \left({\theta}_{A}\right) + i \sin \left({\theta}_{A}\right)}{\cos \left({\theta}_{B}\right) + i \sin \left({\theta}_{B}\right)}\right)$

#color(white)("XXX")=sqrt(10)/sqrt(29) *([color(green)(cos(theta_A)*cos(theta_B)+sin(theta_A) * sin(theta_B))+i[color(magenta)(sin(theta_A) * cos(theta_B) - cos(theta_A) * sin(theta_B))])#

Plugging in the values for ${\theta}_{A}$ and ${\theta}_{B}$ and using a calculator/spreadsheet:
$\textcolor{w h i t e}{\text{XXX}} \approx - 0.034482759 + 0.586206897 i$

If we call this value $C$
then
$\textcolor{w h i t e}{\text{XXX}} {r}_{C} = \sqrt{{\left(- 0.03448 \ldots\right)}^{2} + {\left(0.5862 \ldots\right)}^{2}} \approx 0.58722022$
and (since $C$ is in Quadrant 2)
$\textcolor{w h i t e}{\text{XXX}} {\theta}_{C} = \arctan \left(\frac{0.5862 \ldots}{- 0.03448 \ldots}\right) + \pi \approx 1.62955215$

$\frac{A}{B} = C = {r}_{C} \left(\cos \left({\theta}_{C}\right) + i \cdot \sin \left({\theta}_{C}\right)\right)$

#Cos2x+3cosx=-2# How do I solve this?

Monzur R.
Featured 1 month ago

${x}_{1} = n \pi , n = 2 k \pm 1 , k \in \mathbb{Z}$

${x}_{2} = \frac{n}{3} \pi , \mod \left(n , 2\right) = 0$

Explanation:

$\cos 2 x + 3 \cos x = - 2$

Use the double angle formula for cosine to expand $\cos 2 x$ and rewrite the equation in standard form

$2 {\cos}^{2} x - 1 + 3 \cos x + 2 = 0$

$2 {\cos}^{2} x + 3 \cos x + 1 = 0$

Let $\cos x = y$

$y = \frac{- 3 \pm \sqrt{9 - 4 \left(2\right) \left(1\right)}}{2 \left(2\right)} = \frac{- 3 \pm 1}{4}$

$y = - 1$ or $y = - \frac{1}{2}$

$\cos x = - 1$

$x = n \pi , n = 2 k \pm 1 , k \in \mathbb{Z}$

$\cos x = - \frac{1}{2}$

${x}_{2} = \frac{n}{3} \pi , \mod \left(n , 2\right) = 0$

How do you graph #y=sin^-1x# over the interval #-1<=x<=1#?

Wataru
Featured 1 month ago

Here is the graph of $y = {\sin}^{- 1} x$.

Explanation:

Step 1: Sketch the graph of $y = \sin x$ on $\left[- \frac{\pi}{2} , \frac{\pi}{2}\right]$.

Step 2: Sketch the line $y = x$.

Step 3: Reflect the graph of $y = \sin x$ about the line $y = x$.

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