# Make the internet a better place to learn

1

## How do you write the polar form of the equation of the line that passes through the points (4,-1) and (-2,3)?

Douglas K.
Featured 1 month ago

#### Explanation:

The slope, m, of the line is:

$m = \frac{3 - \left(- 1\right)}{- 2 - 4}$

$m = \frac{4}{-} 6$

$m = - \frac{2}{3}$

Use the point-slope form of the equation of a line:

$y = m \left(x - {x}_{1}\right) + {y}_{1}$

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

Here is a graph of that line:

Multiply both sides by 3:

$3 y = - 2 \left(x - 4\right) - 1$

Distribute the -2:

$3 y = - 2 x + 8 - 1$

$2 x + 3 y = 7$

Substitute $r \cos \left(\theta\right)$ for x and $r \sin \left(\theta\right)$ for y:

$2 r \cos \left(\theta\right) + 3 r \sin \left(\theta\right) = 7$

Factor out r:

$r \left(2 \cos \left(\theta\right) + 3 \sin \left(\theta\right)\right) = 7$

Divide both sides by $\left(2 \cos \left(\theta\right) + 3 \sin \left(\theta\right)\right)$

$r = \frac{7}{2 \cos \left(\theta\right) + 3 \sin \left(\theta\right)}$

Here is a graph of that equation:

3

## What is the value of cos(-870)?

Parzival S.
Featured 1 month ago

$\cos \left(- {870}^{o}\right) = \cos \left(- {150}^{o}\right) = - \frac{\sqrt{3}}{2}$

#### Explanation:

Let's first notice that we're looking at $- {870}^{o}$. The negative sign means we'll rotate in a clockwise direction (and so go down through Q4 first).

There are ${360}^{o}$ in a full circle, so when we go around once fully, we've gone 360 of the 870:

$- {870}^{o} + {360}^{o} = - {610}^{o}$

We need to go around once more time:

$- {610}^{o} + {360}^{o} = - {150}^{o}$

Now let's go around a third time, but not all the way. We'll go through Q4 (to do that takes $- {90}^{o}$) and end up in Q3 but not get to Q2 (that would take $- {180}^{o}$). Here's a visual of where we are:

Do you see the point for $- {150}^{o}$? It's in the same place as $+ {210}^{o}$

Now that we know where we are, we can get down to finding the cosine!

The reference angle between $- {150}^{o}$ and ${180}^{o}$ is ${30}^{o}$. This means that the triangle we have is a 30-60-90 triangle and the lengths of the sides are $1 , \sqrt{3} , 2$ for the short side (which is the opposite - the side that drops from the $x$-axis), the medium side (which is the adjacent - that's the $x$-axis), and the long side (which is the hypotenuse (the side that extends from the origin to the red dot on the circle)), respectively.

$\cos = \text{adj"/"hyp} = \frac{\sqrt{3}}{2}$

One last thing to address and we're done - because the adjacent is along the negative $x$-axis, we make the cosine negative. And so:

$\cos \left(- {870}^{o}\right) = \cos \left(- {150}^{o}\right) = - \frac{\sqrt{3}}{2}$

2

## How do I prove this? sin (1/2)(x-y) = +- (1/2) (4-a^2-b^2)^(1/2) Given: sinx+siny=a and cosx+cosy=b

Ratnaker Mehta
Featured 1 month ago

See the Proof in the Explanation.

#### Explanation:

Given that, $\sin x + \sin y = a \Rightarrow 2 \sin \left(\frac{x + y}{2}\right) \cos \left(\frac{x - y}{2}\right) = a \ldots \left(1\right) .$

Similarly, $\cos x + \cos y = b \Rightarrow 2 \cos \left(\frac{x + y}{2}\right) \cos \left(\frac{x - y}{2}\right) = b \ldots \left(2\right) .$

Hence, Squaring and Adding (1) & (2), we get,

$4 {\cos}^{2} \left(\frac{x - y}{2}\right) \left\{{\sin}^{2} \left(\frac{x + y}{2}\right) + {\cos}^{2} \left(\frac{x + y}{2}\right)\right\} = {a}^{2} + {b}^{2} ,$ i.e.,

$4 {\cos}^{2} \left(\frac{x - y}{2}\right) = {a}^{2} + {b}^{2} , \mathmr{and} ,$

$4 \left\{1 - {\sin}^{2} \left(\frac{x - y}{2}\right)\right\} = {a}^{2} + {b}^{2} ,$

$\Rightarrow 4 - 4 {\sin}^{2} \left(\frac{x - y}{2}\right) = {a}^{2} + {b}^{2} ,$

$\Rightarrow 4 {\sin}^{2} \left(\frac{x - y}{2}\right) = \left(4 - {a}^{2} - {b}^{2}\right) ,$ giving,

$\sin \left(\frac{x - y}{2}\right) = \pm \frac{1}{2} {\left(4 - {a}^{2} - {b}^{2}\right)}^{\frac{1}{2}} .$

1

## Cos(x) cos(3x)=0.707 determin value of x ? Thanks

Monzur R.
Featured 1 month ago

$x \approx 0.251$

#### Explanation:

$\cos x \cos 3 x = 0.707$

$f \left(x\right) = \cos x \cos 3 x - 0.707 = 0$

Use Newton method to find a zero of $f$

${x}_{n + 1} = {x}_{n} - \frac{f \left({x}_{n}\right)}{f ' \left({x}_{n}\right)}$

$f ' \left(x\right) = - 2 \sin 2 x - 2 \sin 4 x$

$0.707 = \cos \left(\frac{1}{4} \pi\right)$, so a good ${x}_{0}$ would be $\frac{\frac{1}{4} \pi}{3} = 0.262$

x_1=0.262-(cos(0.262)cos(3(0.262))-0.707)/(2sin(2(0.262)-2sin(4(0.262))=0.253

${x}_{2} = 0.251$

${x}_{3} = 0.251$

${x}_{4} = 0.251$

The iterations approach $0.251$ to $0.251$ is a solution to $\cos x \cos 3 x = 0.707$

1

## A helicopter at A(6, 9, 3) moves with constant velocity. 10 minutes later, it is at B(3, 10, 2.5) Distances are in kilometres. At what angle to the horizontal does is the helicopter flying?

Mr. Raj
Featured 3 weeks ago

$\approx {8.98}^{o}$

#### Explanation:

So first up we need to understand the question completely. A helicopter starts at $A \left(6 , 9 , 3\right)$, and after 10 mins, it is at $B \left(3 , 10 , 2.5\right)$. We need to find the angle between the velocity vector (which would be $\vec{A B}$) and the $X Y$ plane.

So first we need to find the velocity vector which would be $\vec{A B}$:

To find the vector between two points $M \left({x}_{1} , {y}_{1} , {z}_{1}\right)$ and $N \left({x}_{2} , {y}_{2} , {z}_{2}\right)$, use:

$\vec{M N} = \left(\begin{matrix}{x}_{2} - {x}_{1} \\ {y}_{2} - {y}_{1} \\ {z}_{2} - {z}_{1}\end{matrix}\right)$

So from our points, we have:

$A \left(6 , 9 , 3\right)$ and $B \left(3 , 10 , 2.5\right)$

$\implies \vec{A B} = \left(\begin{matrix}3 - 6 \\ 10 - 9 \\ 2.5 - 3\end{matrix}\right)$

$\implies \vec{A B} = \left(\begin{matrix}- 3 \\ 1 \\ - 0.5\end{matrix}\right)$

This vector is the velocity vector of the helicopter in terms of km per 10 minutes (as it travels from $A$ to $B$ in 10 minutes)

This vector would look like:

(the red dot is A and the Blue dot is B)

So we need to find the angle that $\vec{A B}$ makes with the XY plane.

To do this find the angle between $\vec{A B}$ and the vector on the XY plane which is the same as $\vec{A B}$, but the $z$ value is zero. Let's call this vector $a$.

$\vec{A B} = \left(\begin{matrix}- 3 \\ 1 \\ - 0.5\end{matrix}\right)$

$\implies \vec{a} = \left(\begin{matrix}- 3 \\ 1 \\ 0\end{matrix}\right)$

To find the angle between the two vectors $\vec{b}$ and $\vec{b}$, use:

$\cos \theta = \frac{| \vec{b} \cdot \vec{c} |}{| \vec{b} | | \vec{c} |}$

In this equation, the numerator is the modulus of the dot product and the denominator is the length of each vector.

So when we are finding the angle between $\vec{A B}$ and $\vec{a}$, we would use:

$\cos \theta = \frac{| \vec{A B} \cdot \vec{a} |}{| \vec{A B} | | \vec{a} |}$

$\implies \cos \theta = \frac{| \left(\begin{matrix}- 3 \\ 1 \\ - 0.5\end{matrix}\right) \cdot \left(\begin{matrix}- 3 \\ 1 \\ 0\end{matrix}\right) |}{| \left(\begin{matrix}- 3 \\ 1 \\ - 0.5\end{matrix}\right) | | \left(\begin{matrix}- 3 \\ 1 \\ 0\end{matrix}\right) |}$

$\implies \cos \theta = \frac{- 3 \cdot - 3 + 1 \cdot 1 \pm 0.5 \cdot 0}{\sqrt{{\left(- 3\right)}^{2} + {1}^{2} + {\left(- 0.5\right)}^{2}} \sqrt{{\left(- 3\right)}^{2} + {1}^{2} + {\left(0\right)}^{2}}}$

$\implies \cos \theta = \frac{9 + 1 + 0}{\sqrt{9 + 1 + 0.25} \sqrt{9 + 1 + 0}}$

$\implies \cos \theta = \frac{10}{\sqrt{10.25} \sqrt{10}}$

=>costheta=10/(sqrt(10.25*10)

$\implies \cos \theta = \frac{10}{\sqrt{102.5}}$

$\implies \theta = {\cos}^{-} 1 \left(\frac{10}{\sqrt{102.5}}\right)$

$\implies \theta \approx 8.98$ (to two s.f.)

2

## What's cos^3((2pi)/7)+cos^3((4pi)/7)+cos^3((8pi)/7)?

dk_ch
Featured 3 weeks ago

${\cos}^{3} \left(\frac{2 \pi}{7}\right) + {\cos}^{3} \left(\frac{4 \pi}{7}\right) + {\cos}^{3} \left(\frac{8 \pi}{7}\right)$

Using formula

${\cos}^{3} A = \frac{1}{4} \left(\cos 3 A + 3 \cos A\right)$ the given expreession becomes

$= \frac{1}{4} \left(\cos \left(\frac{6 \pi}{7}\right) + 3 \cos \left(\frac{2 \pi}{7}\right) + \cos \left(\frac{12 \pi}{7}\right) + 3 \cos \left(\frac{4 \pi}{7}\right) + \cos \left(\frac{24 \pi}{7}\right) + 3 \cos \left(\frac{8 \pi}{7}\right)\right)$

$= \frac{1}{4} \left(\cos \left(\pi - \frac{\pi}{7}\right) + 3 \cos \left(\frac{2 \pi}{7}\right) + \cos \left(2 \pi - \frac{2 \pi}{7}\right) + 3 \cos \left(\pi - \frac{3 \pi}{7}\right) + \cos \left(3 \pi + \frac{3 \pi}{7}\right) + 3 \cos \left(\pi + \frac{\pi}{7}\right)\right)$

$= \frac{1}{4} \left(- \cos \left(\frac{\pi}{7}\right) + 3 \cos \left(\frac{2 \pi}{7}\right) + \cos \left(\frac{2 \pi}{7}\right) - 3 \cos \left(\frac{3 \pi}{7}\right) - \cos \left(\frac{3 \pi}{7}\right) - 3 \cos \left(\frac{\pi}{7}\right)\right)$

$= \frac{1}{4} \left(4 \cos \left(\frac{2 \pi}{7}\right) - 4 \cos \left(\frac{3 \pi}{7}\right) - 4 \cos \left(\frac{\pi}{7}\right)\right)$

$= \cos \left(\frac{2 \pi}{7}\right) - \cos \left(\frac{3 \pi}{7}\right) - \cos \left(\frac{\pi}{7}\right)$

$= \frac{1}{2 \sin \left(\frac{\pi}{7}\right)} \left(2 \sin \left(\frac{\pi}{7}\right) \cos \left(\frac{2 \pi}{7}\right) - 2 \sin \left(\frac{\pi}{7}\right) \cos \left(\frac{3 \pi}{7}\right) - 2 \sin \left(\frac{\pi}{7}\right) \cos \left(\frac{\pi}{7}\right)\right)$

$= \frac{1}{2 \sin \left(\frac{\pi}{7}\right)} \left(\sin \left(\frac{3 \pi}{7}\right) - \sin \left(\frac{\pi}{7}\right) - \sin \left(\frac{4 \pi}{7}\right) + \sin \left(\frac{2 \pi}{7}\right) - \sin \left(\frac{2 \pi}{7}\right)\right)$

$= \frac{1}{2 \sin \left(\frac{\pi}{7}\right)} \left(\sin \left(\frac{3 \pi}{7}\right) - \sin \left(\frac{\pi}{7}\right) - \sin \left(\pi - \frac{3 \pi}{7}\right)\right)$

$= \frac{1}{2 \sin \left(\frac{\pi}{7}\right)} \left(\sin \left(\frac{3 \pi}{7}\right) - \sin \left(\frac{\pi}{7}\right) - \sin \left(\frac{3 \pi}{7}\right)\right)$

$= \frac{1}{2 \sin \left(\frac{\pi}{7}\right)} \times \left(- \sin \left(\frac{\pi}{7}\right)\right)$

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

3

## Question on combination of Logarithm and Trigonometry (there are two parts)?

dk_ch
Featured 3 weeks ago

Q-1

Let $x = 4 \sin 9 \cos 9 = 2 \sin 18 = 2 \times \frac{\sqrt{5} - 1}{4} = \frac{1}{2} \left(\sqrt{5} - 1\right)$

Again

${t}_{1} = 4 \sin 63 \cos 63 = 2 \sin 126 = 2 \sin \left(180 - 54\right) = 2 \sin 54 = \frac{1}{2} \left(\sqrt{5} + 1\right)$

So $x \times {t}_{1} = \frac{1}{2} \left(\sqrt{5} - 1\right) \times \frac{1}{2} \left(\sqrt{5} + 1\right) = \frac{1}{4} \times 4 = 1$

Otherwise without using exact values

$x \cdot {t}_{1} = 4 \sin 9 \cos 9 \cdot 4 \sin 63 \cos 63$

$\implies x \cdot {t}_{1} = 2 \sin 18 \cdot 2 \sin 126$

$\implies x \cdot {t}_{1} = 4 \sin 18 \cdot \sin \left(180 - 54\right)$

$\implies x \cdot {t}_{1} = \frac{2}{\cos} 18 \cdot 2 \sin 18 \cdot \cos 18 \cdot \sin 54$

$\implies x \cdot {t}_{1} = \frac{1}{\cos} 18 \cdot 2 \sin 36 \cdot \sin 54$

$\implies x \cdot {t}_{1} = \frac{\cos \left(54 - 36\right) - \cos \left(54 + 36\right)}{\cos} 18$

$\implies x \cdot {t}_{1} = \frac{\cos 18 - \cos 90}{\cos} 18$

$\implies x \cdot {t}_{1} = \frac{\cos 18 - 0}{\cos} 18 = 1$

Hence $x = \frac{1}{t} _ 1$

Now

${\log}_{{t}_{1}} 4 \sin 9 \cos 9$

$= {\log}_{{t}_{1}} x$

$= {\log}_{{t}_{1}} \left(\frac{1}{t} _ 1\right)$

$= {\log}_{{t}_{1}} {\left({t}_{1}\right)}^{- 1}$

$= - {\log}_{{t}_{1}} \left({t}_{1}\right) = - 1$

Q-2

$l = {\left(\frac{{\cot}^{2} x {\cos}^{2} x}{{\cot}^{2} x - {\cos}^{2} x}\right)}^{2}$

$\implies l = {\left(\frac{{\cot}^{2} x {\cos}^{2} x}{\frac{1}{\tan} ^ 2 x - \frac{1}{\sec} ^ 2 x}\right)}^{2}$

$\implies l = {\left(\frac{{\cot}^{2} x \cdot {\tan}^{2} x \cdot {\cos}^{2} x \cdot {\sec}^{2} x}{{\sec}^{2} x - {\tan}^{2} x}\right)}^{2}$

$\implies l = {\left(\frac{1 \cdot 1}{1}\right)}^{2} = 1$

Now

m=a^(log_(sqrta)abs(2cos(y/2))

=>m=a^(log_(sqrta)abs(2cos((4pi)/2))

=>m=a^(log_(sqrta)abs(2cos(2pi))

=>m=a^(log_(sqrta)2

=>m=((sqrta)^2)^(log_(sqrta)2

=>m=(sqrta)^(2log_(sqrta)2

$\implies m = {\left(\sqrt{a}\right)}^{{\log}_{\sqrt{a}} {2}^{2}} = 4$

So

${l}^{2} + {m}^{2} = {1}^{2} + {4}^{2} = 17$

5

## What is phi, how was it discovered and are its uses?

George C.
Featured 2 weeks ago

A few thoughts...

#### Explanation:

$\phi = \frac{1}{2} + \frac{\sqrt{5}}{2} \approx 1.6180339887$ is known as the Golden Ratio.

It was known and studied by Euclid (approx 3rd or 4th century BCE), basically for many geometric properties...

It has many interesting properties, of which here are a few...

The Fibonacci sequence can be defined recursively as:

${F}_{0} = 0$

${F}_{1} = 1$

${F}_{n + 2} = {F}_{n} + {F}_{n + 1}$

It starts:

$0 , 1 , 1 , 2 , 3 , 5 , 8 , 13 , 21 , 34 , 55 , 89 , 144 , 233 , 377 , 610 , 987 , \ldots$

The ratio between successive terms tends to $\phi$. That is:

${\lim}_{n \to \infty} {F}_{n + 1} / {F}_{n} = \phi$

In fact the general term of the Fibonacci sequence is given by the formula:

${F}_{n} = \frac{{\phi}^{n} - {\left(- \phi\right)}^{- n}}{\sqrt{5}}$

A rectangle with sides in ratio $\phi : 1$ is called a Golden Rectangle. If a square of maximal size is removed from one end of a golden rectangle then the remaining rectangle is a golden rectangle.

This is related to both the limiting ratio of the Fibonacci sequence and the fact that:

phi = [1;bar(1)] = 1+1/(1+1/(1+1/(1+1/(1+1/(1+...)))))

which is the most slowly converging standard continued fraction.

If you place three golden rectangles symmetrically perpendicular to one another in three dimensional space, then the twelve corners form the vertices of a regular icosahedron. Hence we can calculate the surface area and volume of a regular icosahedron of given radius. See https://socratic.org/s/aFZyTQfn

An isosceles triangle with sides in ratio $\phi : \phi : 1$ has base angles $\frac{2 \pi}{5}$ and apex angle $\frac{\pi}{5}$. This allows us to calculate exact algebraic formulae for $\sin \left(\frac{\pi}{10}\right)$, $\cos \left(\frac{\pi}{10}\right)$ and ultimately for any multiple of $\frac{\pi}{60}$ (${3}^{\circ}$). See https://socratic.org/s/aFZztx8s

2

dk_ch
Featured 2 weeks ago

As per problem the movement of the London Eye is periodic one and its height at $t$ th instant $h \left(t\right)$ is given by the following equation

$h \left(t\right) = a + b \cos \left(c t\right) \ldots \ldots \left[1\right]$, where $h \left(t\right)$ in meter and t in sec.

Here $c$ should be the angular velocity associated with the periodic motion. Again it is also given that time taken by the London Eye to complete one cycle is 3o min. This means time period $T = 30 \min$

Hence $\text{Angular velocity} = c = \omega = \frac{2 \pi}{T} = \frac{2 \pi}{30} = \frac{\pi}{15}$ rad/min.

So the equation[1] takes the following form

$h \left(t\right) = a + b \cos \left(\frac{2 \pi}{30} t\right) \ldots \ldots \left[2\right]$

Now at $t = 0$ the London Eye is at the bottom i.e.$h \left(0\right) = 0$

Applying this condition on [2] we get

$h \left(0\right) = a + b \cos \left(\frac{2 \pi}{30} \times 0\right)$

$0 = a + b \implies a + b = 0. \ldots . \left[3\right]$

Since at $t = 0$ it is at the bottom and it completes the cycle returning at the bottom at $t = 30$ min,
then we can say that at $t = 15$ min it will reach at maximum height 135 m i.e. $h \left(15\right) = 135$m
So we can write

$h \left(15\right) = 135 = a + b \cos \left(\frac{2 \pi}{30} \times 15\right)$

$\implies a + b \cos \left(\pi\right) = 135$

$\implies a - b = 135. \ldots . \left[4\right]$

Now adding [3] and [4] we have $2 a = 135 \implies a = 67.5$ m

Subtracting [4] from [3] we get $2 b = - 135 \implies b = - 67.5$m

So finally the given equation takes the following form

$\textcolor{red}{h \left(t\right) = 67.5 - 67.5 \cos \left(\frac{2 \pi}{30} t\right)}$

The variation of height with time can be represented by following graph.

1

## Use any method to find the EXACT VALUE of tan 15°.?

George C.
Featured 18 hours ago

$\tan {15}^{\circ} = 2 - \sqrt{3}$

#### Explanation:

Use:

$\sin {30}^{\circ} = \frac{1}{2}$

$\cos {30}^{\circ} = \frac{\sqrt{3}}{2}$

$\sin {45}^{\circ} = \cos {45}^{\circ} = \frac{\sqrt{2}}{2}$

$\sin \left(\alpha - \beta\right) = \sin \alpha \cos \beta - \sin \beta \cos \alpha$

$\cos \left(\alpha - \beta\right) = \cos \alpha \cos \beta + \sin \alpha \sin \beta$

$\tan \theta = \sin \frac{\theta}{\cos} \theta$

Then:

$\sin {15}^{\circ} = \sin \left({45}^{\circ} - {30}^{\circ}\right)$

$\textcolor{w h i t e}{\sin {15}^{\circ}} = \sin {45}^{\circ} \cos {30}^{\circ} - \sin {30}^{\circ} \cos {45}^{\circ}$

$\textcolor{w h i t e}{\sin {15}^{\circ}} = \frac{\sqrt{2}}{2} \frac{\sqrt{3}}{2} - \frac{1}{2} \frac{\sqrt{2}}{2}$

$\textcolor{w h i t e}{\sin {15}^{\circ}} = \frac{\sqrt{2}}{4} \left(\sqrt{3} - 1\right)$

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

$\cos {15}^{\circ} = \cos \left({45}^{\circ} - {30}^{\circ}\right)$

$\textcolor{w h i t e}{\cos {15}^{\circ}} = \cos {45}^{\circ} \cos {30}^{\circ} + \sin {45}^{\circ} \sin {30}^{\circ}$

$\textcolor{w h i t e}{\cos {15}^{\circ}} = \frac{\sqrt{2}}{2} \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \frac{1}{2}$

$\textcolor{w h i t e}{\cos {15}^{\circ}} = \frac{\sqrt{2}}{4} \left(\sqrt{3} + 1\right)$

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

$\tan {15}^{\circ} = \sin {15}^{\circ} / \cos {15}^{\circ}$

$\textcolor{w h i t e}{\tan {15}^{\circ}} = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{\frac{\sqrt{2}}{4}}}} \left(\sqrt{3} - 1\right)}{\textcolor{red}{\cancel{\textcolor{b l a c k}{\frac{\sqrt{2}}{4}}}} \left(\sqrt{3} + 1\right)}$

$\textcolor{w h i t e}{\tan {15}^{\circ}} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}$

$\textcolor{w h i t e}{\tan {15}^{\circ}} = \frac{{\left(\sqrt{3} - 1\right)}^{2}}{\left(\sqrt{3} + 1\right) \left(\sqrt{3} - 1\right)}$

$\textcolor{w h i t e}{\tan {15}^{\circ}} = \frac{3 - 2 \sqrt{3} + 1}{3 - 1}$

$\textcolor{w h i t e}{\tan {15}^{\circ}} = 2 - \sqrt{3}$

$\textcolor{w h i t e}{}$
Footnote

We can see the original values of $\sin {30}^{\circ}$, etc used above by considering the right angled triangles: