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3

## What is \cos ^{-1}(.60)?

EZ as pi
Featured 1 month ago

cos^-1 (0.6)= 53.1°

#### Explanation:

Note that ${\cos}^{-} 1$ does not mean $\frac{1}{\cos}$ as we are used to in algebra.

${\cos}^{-} 1$ is the notation used for arc-cos.

Cos 30° = 0.866 " "hArr" "cos^-1(0.866) = 30°

In this case ${\cos}^{-} 1 \left(0.60\right)$ is asking the question..

"Which angle has a $C o s$ value of 0.60?"

The only way to determine this is with a calculator or tables.
Using a graph is possible, but not accurate enough.

Depending on which calculator you have, the following are possible key presses:

D.A.L. calculators:

shift cos^-1 (0.6) =" larr ( ) might not be needed on some models

The answer will be 53.1°

Or $0.6 \text{ shift" cos^-1" } \leftarrow =$ sign not needed, the answer $\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . .}$appears immediately

The answer will be 53.1°

Note that this is only the acute angle, the 1st quadrant angle.
In the 4th quadrant, Cos 306.9° = 0.6

2

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

Alan P.
Featured 1 month 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)
$\textcolor{w h i t e}{\text{XXX}} {\theta}_{C} = \arctan \left(\frac{0.5862 \ldots}{- 0.03448 \ldots}\right) + \pi \approx 1.62955215$

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

1

## How do you find the exact value cos(x+y) if tanx=5/3,siny=1/3?

mason m
Featured 4 days ago

$\cos \left(x + y\right) = \frac{6 \sqrt{2} - 5}{3 \sqrt{34}}$

#### Explanation:

Use the cosine angle addition formula:

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

We need to determine $\sin x$, $\cos x$, and $\cos y$ from whatever we have.

Determining $m a t h b f \left(\sin x , \cos x :\right)$

${\tan}^{2} x + 1 = {\sec}^{2} x \text{ "" "" }$use $\tan x = \frac{5}{3}$

$\frac{25}{9} + 1 = {\sec}^{2} x$

$\sec x = \sqrt{\frac{34}{9}} = \frac{\sqrt{34}}{3}$

Then:

color(blue)(cosx=1/secx=3/sqrt34

Now using:

${\sin}^{2} x + {\cos}^{2} x = 1 \text{ "" "" }$where $\cos x = \frac{3}{\sqrt{34}}$

${\sin}^{2} x + \frac{9}{34} = 1$

color(blue)(sinx=sqrt(25/34)=5/sqrt34

Determining $m a t h b f \left(\cos y :\right)$

${\sin}^{2} y + {\cos}^{2} y = 1 \text{ "" "" }$use $\sin y = \frac{1}{3}$

$\frac{1}{9} + {\cos}^{2} y = 1$

color(blue)(cosy=sqrt(8/9)=(2sqrt2)/3

Returning to $m a t h b f \left(\cos \left(x + y\right) :\right)$

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

$\textcolor{w h i t e}{\cos \left(x + y\right)} = \frac{3}{\sqrt{34}} \left(\frac{2 \sqrt{2}}{3}\right) - \frac{5}{\sqrt{34}} \left(\frac{1}{3}\right)$

$\textcolor{w h i t e}{\cos \left(x + y\right)} = \frac{6 \sqrt{2} - 5}{3 \sqrt{34}}$

Note these are all assuming that $x$ and $y$ are in the first quadrant (everything is positive).

2

## How do you convert r(2 - cosx) = 2 to rectangular form?

Douglas K.
Featured 4 days ago

The explanation is written with the assumption that you meant $\theta$ for the argument of the cosine function.

#### Explanation:

Use the distributive property :

$2 r - r \cos \left(\theta\right) = 2$

$2 r = r \cos \left(\theta\right) + 2$

Substitute x for $r \cos \left(\theta\right)$ and $\sqrt{{x}^{2} + {y}^{2}}$ for r:

$2 \sqrt{{x}^{2} + {y}^{2}} = x + 2$

Square both sides:

$4 \left({x}^{2} + {y}^{2}\right) = {x}^{2} + 4 x + 4$

$4 {x}^{2} + 4 {y}^{2} = {x}^{2} + 4 x + 4$

$3 {x}^{2} - 4 x + 4 {y}^{2} = 4$

Add $3 {h}^{2}$ to both sides:

$3 {x}^{2} - 4 x + 3 {h}^{2} + 4 {y}^{2} = 3 {h}^{2} + 4$

Remove a factor of the 3 from the first 3 terms:

$3 \left({x}^{2} - \frac{4}{3} x + {h}^{2}\right) + 4 {y}^{2} = 3 {h}^{2} + 4$

Use the middle in the right side of the pattern ${\left(x - h\right)}^{2} = {x}^{2} - 2 h x + {h}^{2}$ and middle term of the equation to find the value of h:

$- 2 h x = - \frac{4}{3} x$

$h = \frac{2}{3}$

Substitute the left side of the pattern into the equation:

$3 {\left(x - h\right)}^{2} + 4 {y}^{2} = 3 {h}^{2} + 4$

Substitute $\frac{2}{3}$ for h and insert a -0 into the y term:

$3 {\left(x - \frac{2}{3}\right)}^{2} + 4 {\left(y - 0\right)}^{2} = 3 {\left(\frac{2}{3}\right)}^{2} + 4$

$3 {\left(x - \frac{2}{3}\right)}^{2} + 4 {\left(y - 0\right)}^{2} = \frac{16}{3}$

Divide both sides by $\frac{16}{3}$

$3 {\left(x - \frac{2}{3}\right)}^{2} / \left(\frac{16}{3}\right) + 4 {\left(y - 0\right)}^{2} / \left(\frac{16}{3}\right) = 1$

${\left(x - \frac{2}{3}\right)}^{2} / \left(\frac{16}{9}\right) + {\left(y - 0\right)}^{2} / \left(\frac{16}{12}\right) = 1$

Write the denominators as squares:

${\left(x - \frac{2}{3}\right)}^{2} / {\left(\frac{4}{3}\right)}^{2} + {\left(y - 0\right)}^{2} / {\left(\frac{4}{\sqrt{12}}\right)}^{2} = 1$

The is standard Cartesian form of the equation of an ellipse with a center at $\left(\frac{2}{3} , 0\right)$; its semi-major axis is $\frac{4}{3}$ units long and is parallel to the x axis and its semi-minor axis is $\frac{4}{\sqrt{12}}$ units long and is parallel to the y axis.

2

## What is the exact value of cos(tan^-1(2)+tan^-1(3)) ?

mason m
Featured 1 week ago

$\cos \left({\tan}^{-} 1 \left(2\right) + {\tan}^{-} 1 \left(3\right)\right) = - \frac{1}{\sqrt{2}}$

#### Explanation:

First use the cosine angle-addition formula:

$\cos \left(a + b\right) = \cos \left(a\right) \cos \left(b\right) - \sin \left(a\right) \sin \left(b\right)$

Then the original expression equals:

$= \cos \left({\tan}^{-} 1 \left(2\right)\right) \cos \left({\tan}^{-} 1 \left(3\right)\right) - \sin \left({\tan}^{-} 1 \left(2\right)\right) \sin \left({\tan}^{-} 1 \left(3\right)\right)$

Note that if $\theta = {\tan}^{-} 1 \left(2\right)$, then $\tan \left(\theta\right)$. That is, a right triangle with angle $\theta$ has $\tan \left(\theta\right) = 2$, which is the triangle with the leg opposite $\theta$ being $2$ and the leg adjacent to $\theta$ being $1$. The Pythagorean theorem tells us that the hypotenuse of this triangle is $\sqrt{5}$.

So, when $\theta = {\tan}^{-} 1 \left(2\right)$, we see that:

$\cos \left({\tan}^{-} 1 \left(2\right)\right) = \cos \left(\theta\right) = \text{adjacent"/"hypotenuse} = \frac{1}{\sqrt{5}}$

$\sin \left({\tan}^{-} 1 \left(2\right)\right) = \sin \left(\theta\right) = \text{opposite"/"hypotenuse} = \frac{2}{\sqrt{5}}$

Now let angle $\phi$ be defined by $\phi = {\tan}^{-} 1 \left(3\right)$, such that $\tan \left(\phi\right) = 3$. This is the right triangle with:

$\left\{\begin{matrix}\text{opposite"=3 \\ "adjacent"=1 \\ "hypotenuse} = \sqrt{10}\end{matrix}\right.$

Then:

$\cos \left({\tan}^{-} 1 \left(3\right)\right) = \cos \left(\phi\right) = \text{adjacent"/"hypotenuse} = \frac{1}{\sqrt{10}}$

$\sin \left({\tan}^{-} 1 \left(3\right)\right) = \sin \left(\phi\right) = \text{opposite"/"hypotenuse} = \frac{3}{\sqrt{10}}$

Plugging these into the expression from earlier we get:

$= \frac{1}{\sqrt{5}} \left(\frac{1}{\sqrt{10}}\right) - \frac{2}{\sqrt{5}} \left(\frac{3}{\sqrt{10}}\right)$

Note that $\sqrt{5} \left(\sqrt{10}\right) = \sqrt{50} = 5 \sqrt{2}$:

$= \frac{1 - 6}{5 \sqrt{2}} = - \frac{1}{\sqrt{2}}$

1

## Given f(x)=2sin(x-pi/4)? a) State all x-intercepts over the interval (0,4pi) b) State the point(s) in (0,2pi) where the graph of this function reaches a maximum. Give your answer as a point (x,y).

dk_ch
Featured 4 days ago

Given
$y = f \left(x\right) = 2 \sin \left(x - \frac{\pi}{4}\right)$

To get x-intercepts we can put y=0 and solve for $x \in \left[0 , 4 \pi\right]$

So $2 \sin \left(x - \frac{\pi}{4}\right) = 0$

$\implies x - \frac{\pi}{4} = n \pi \text{ where } n \in \mathbb{Z}$

$\implies x = n \pi + \frac{\pi}{4} \text{ where } n \in \mathbb{Z}$

Putting $n = 0$ we get

$x = \frac{\pi}{4}$

Putting $n = 1$ we get

$x = \frac{5 \pi}{4}$

Putting $n = 2$ we get

$x = \frac{9 \pi}{4}$

Putting $n = 3$ we get

$x = \frac{13 \pi}{4}$

(a) So points of x-intercepts over $\left[0 , 4 \pi\right]$ are

(pi/4,0);((5pi)/4,0);((9pi)/4,0);((13pi)/4,0)

b) For maximum value of y the value of $\sin \left(x - \frac{\pi}{4}\right) = 1$

So $x - \frac{\pi}{4} = \frac{\pi}{2}$
$\implies x = \frac{\pi}{4} + \frac{\pi}{2} = \frac{3 \pi}{4}$

For $x = \frac{3 \pi}{4} \text{ } y = 2$

The point(in (0,2pi) where the graph of this function reaches
maximum is $\left(\frac{3 \pi}{4} , 2\right)$

1

## How do you graph y=5+tan(x+pi)?

Alan P.
Featured 4 days ago

Start from a "basic cycle" for the $\tan$ function to obtain the results below.

#### Explanation:

Starting with the "basic cycle" for $\tan \left(\theta\right)$ i.e. for $\theta \in \left[- \frac{\pi}{2} , + \frac{\pi}{2}\right]$

Then consider what values of $x$ would place $\left(x + \pi\right)$ in this same range, i.e. $\left(x + \pi\right) \in \left[- \frac{\pi}{2} , + \frac{\pi}{2}\right]$
(Yes; I know: because $\tan$ has a cycle length of $\pi$ we could recognize that the cycle will repeat at exactly the same place, but let's do this for the more general case.)
$\left(x + \pi\right) \in \left[- \frac{\pi}{2} , \frac{\pi}{2}\right] \textcolor{w h i t e}{\text{X"rarrcolor(white)("X}} x \in \left[- \frac{3 \pi}{2} , - \frac{\pi}{2}\right]$
Giving us the "basic cycle" for $\tan \left(x + \pi\right)$:

Adding $5$ to this to get $y = 5 + \tan \left(x + \pi\right)$ simply shifts the points upward $5$ units:

In the image below, I have added the "non-basic cycles" as well as the "basic cycle" used for analysis:

2

## Can anyone Prove (sin theta)^2+(cos theta)^2=1 ?

George C.
Featured 4 days ago

See explanation...

#### Explanation:

Consider a right angled triangle with an internal angle $\theta$:

Then:

$\sin \theta = \frac{a}{c}$

$\cos \theta = \frac{b}{c}$

So:

${\sin}^{2} \theta + {\cos}^{\theta} = {a}^{2} / {c}^{2} + {b}^{2} / {c}^{2} = \frac{{a}^{2} + {b}^{2}}{c} ^ 2$

By Pythagoras ${a}^{2} + {b}^{2} = {c}^{2}$, so $\frac{{a}^{2} + {b}^{2}}{c} ^ 2 = 1$

So given Pythagoras, that proves the identity for $\theta \in \left(0 , \frac{\pi}{2}\right)$

For angles outside that range we can use:

$\sin \left(\theta + \pi\right) = - \sin \left(\theta\right)$

$\cos \left(\theta + \pi\right) = - \cos \left(\theta\right)$

$\sin \left(- \theta\right) = - \sin \left(\theta\right)$

$\cos \left(- \theta\right) = \cos \left(\theta\right)$

So for example:

${\sin}^{2} \left(\theta + \pi\right) + {\cos}^{2} \left(\theta + \pi\right) = {\left(- \sin \theta\right)}^{2} + {\left(- \cos \theta\right)}^{2} = {\sin}^{2} \theta + {\cos}^{2} \theta = 1$

$\textcolor{w h i t e}{}$
Pythagoras theorem

Given a right angled triangle with sides $a$, $b$ and $c$ consider the following diagram:

The area of the large square is ${\left(a + b\right)}^{2}$

The area of the small, tilted square is ${c}^{2}$

The area of each triangle is $\frac{1}{2} a b$

So we have:

${\left(a + b\right)}^{2} = {c}^{2} + 4 \cdot \frac{1}{2} a b$

That is:

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

Subtract $2 a b$ from both sides to get:

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

2

## Can anyone Prove (sin theta)^2+(cos theta)^2=1 ?

Geoff K.
Featured 4 days ago

Use the formula for a circle $\left({x}^{2} + {y}^{2} = {r}^{2}\right)$, and substitute $x = r \cos \theta$ and $y = r \sin \theta$.

#### Explanation:

The formula for a circle centred at the origin is

${x}^{2} + {y}^{2} = {r}^{2}$

That is, the distance from the origin to any point $\left(x , y\right)$ on the circle is the radius $r$ of the circle.

Picture a circle of radius $r$ centred at the origin, and pick a point $\left(x , y\right)$ on the circle:

graph{(x^2+y^2-1)((x-sqrt(3)/2)^2+(y-0.5)^2-0.003)=0 [-2.5, 2.5, -1.25, 1.25]}

If we draw a line from that point to the origin, its length is $r$. We can also draw a triangle for that point as follows:

graph{(x^2+y^2-1)(y-sqrt(3)x/3)((y-0.25)^4/0.18+(x-sqrt(3)/2)^4/0.000001-0.02)(y^4/0.00001+(x-sqrt(3)/4)^4/2.7-0.01)=0 [-2.5, 2.5, -1.25, 1.25]}

Let the angle at the origin be theta ($\theta$).

Now for the trigonometry.

For an angle $\theta$ in a right triangle, the trig function $\sin \theta$ is the ratio $\text{opposite side"/"hypotenuse}$. In our case, the length of the side opposite of $\theta$ is the $y$-coordinate of our point $\left(x , y\right)$, and the hypotenuse is our radius $r$. So:

$\sin \theta = \text{opp"/"hyp" = y/r" "<=>" } y = r \sin \theta$

Similarly, $\cos \theta$ is the ratio of the $x$-coordinate in $\left(x , y\right)$ to the radius $r$:

$\cos \theta = \text{adj"/"hyp"=x/r" "<=>" } x = r \cos \theta$

So we have $x = r \cos \theta$ and $y = r \sin \theta$. Substituting these into the circle formula gives

$\text{ "x^2" "+" "y^2" } = {r}^{2}$
${\left(r \cos \theta\right)}^{2} + {\left(r \sin \theta\right)}^{2} = {r}^{2}$
${r}^{2} {\cos}^{2} \theta + {r}^{2} {\sin}^{2} \theta = {r}^{2}$

The ${r}^{2}$'s all cancel, leaving us with

${\cos}^{2} \theta + {\sin}^{2} \theta = 1$

This is often rewritten with the ${\sin}^{2}$ term in front, like this:

${\sin}^{2} \theta + {\cos}^{2} \theta = 1$

And that's it. That's really all there is to it. Just as the distance between the origin and any point $\left(x , y\right)$ on a circle must be the circle's radius, the sum of the squared values for $\sin \theta$ and $\cos \theta$ must be 1 for any angle $\theta$.