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## How do you simplify #sin(u-v)cosv+cos(u-v)sinv#?

Ratnaker Mehta
Featured 4 months ago

$\sin u$.

#### Explanation:

Letting, $\left(u - v\right) = w ,$ and, knowing that,

$\sin \left(x + y\right) = \sin x \cos y + \cos x \sin y ,$ we have,

The Expression $= \sin w \cos v + \cos w \sin v$

$= \sin \left(w + v\right)$

$= \sin \left(\left(u - v\right) + v\right) \ldots \ldots \ldots \ldots \ldots . . \left[\because , w = u - v\right]$

$= \sin u$

## What is the distance between #(2 , (5 pi)/8 )# and #(-1 , (1 pi )/3 )#?

Douglas K.
Featured 4 months ago

The distance is $\approx 2.7$

#### Explanation:

Change the -1 radius to +1 by adding $\pi$ to the angle:

$\left(1 , \frac{4 \pi}{3}\right)$

Find the difference between the two angles:

$\theta = \frac{4 \pi}{3} - \frac{5 \pi}{8}$

$\theta = \frac{32 \pi}{24} - \frac{15 \pi}{24}$

$\theta = \frac{17 \pi}{24}$

We can find the distance between them using the Law of Cosines where side $a = 2$, side $b = 1$ and the angle between them is $\theta = \frac{17 \pi}{24}$

${c}^{2} = {2}^{2} + {1}^{2} - 2 \left(2\right) \left(1\right) \cos \left(\frac{17 \pi}{24}\right)$

$c \approx 2.7$

## How would you use the sum and difference identities to prove: 2sin(a+b)sin(a-b) = cos(2b) - cos(2a) ?

dk_ch
Featured 4 months ago

Let $a + b = x \mathmr{and} a - b = y$

So $2 a = x + y \mathmr{and} 2 b = x - y$

Now $R H S = \cos 2 b - \cos 2 a$

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

$= \cos x \cos y + \sin x \sin y - \cos x \cos y + \sin x \sin y$

$= 2 \sin x \sin y$

$= 2 \sin \left(a + b\right) \sin \left(a - b\right) = L H S$

## If #A = <2 ,1 ,8 >#, #B = <5 ,7 ,3 ># and #C=A-B#, what is the angle between A and C?

Featured 4 months ago

THe angle is 66.2º

#### Explanation:

The dot product betwwen 2 vectors is
#〈veca〉.〈vecc〉=∥〈veca〉∥∥〈vecc〉∥cos(veca,vecc)#

#cos(veca,vecc)=(〈veca〉 .〈 vecc 〉 ) /(∥〈veca〉∥∥〈vecc〉∥)#

If #〈veca〉=〈a_1,a_2,a_3〉#
and #〈vecc〉=〈c_1,c_2,c_3〉#

The dot product is #〈veca〉.〈vecc〉=a_1c_1+a_2c_2+a_3c_3#

#〈vecc〉=〈veca〉-〈vecb〉 =〈2,1,8〉-〈5,7,3〉 =〈-3,-6,5〉#

#〈veca〉.〈vecc〉=-6+(-6)+40=28#

#∥〈veca〉∥=sqrt(a_1^2+a_2^2+a_3^2)#

#∥〈veca〉∥=sqrt(4+1+64)=sqrt69#

#∥〈vecc〉∥=sqrt(9+36+25)=sqrt70#
let $\cos \left(\vec{a} , \vec{c}\right) = \cos \theta$
so the angle
$\cos \theta = \frac{28}{\left(\sqrt{69}\right) \left(\sqrt{70}\right)} = 0.403$
#theta =66.2º#

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

Douglas K.
Featured 1 month 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.

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