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3

## If A+B+C =180^@, prove that cot^2 A + cot^2 B +cot^2 C >= 1 ?

dk_ch
Featured 1 month ago

$A + B + C = {180}^{\circ}$

$\implies \cot \left(A + B\right) = \cot \left({180}^{\circ} - C\right)$

$\implies \frac{\cot A \cot B - 1}{\cot B + \cot A} = - \cot C$

$\implies \left(\cot A \cot B - 1\right) = - \left(\cot B + \cot A\right) \cdot \cot C$

=>(cotAcotB+cotBcotC+cotCcotA=1

Now

${\cot}^{2} A + {\cot}^{2} B + {\cot}^{2} C - 1$

$= {\cot}^{2} A + {\cot}^{2} B + {\cot}^{2} C - \left(\cot A \cot B + \cot B \cot C + \cot C \cot A\right)$

$= \frac{1}{2} \left(2 {\cot}^{2} A + 2 {\cot}^{2} B + 2 {\cot}^{2} C - 2 \cot A \cot B - 2 \cot B \cot C - 2 \cot C \cot A\right)$

$= \frac{1}{2} \left[{\left(\cot A - \cot B\right)}^{2} + {\left(\cot B - \cot C\right)}^{2} + {\left(\cot C - \cot A\right)}^{2}\right]$
This being sum of three squared quantities each of which is difference of two quantities. So this sum is $\ge 0$

Hence

${\cot}^{2} A + {\cot}^{2} B + {\cot}^{2} C - 1 \ge 0$

$\implies {\cot}^{2} A + {\cot}^{2} B + {\cot}^{2} C \ge 1$

3

## How do i show that arcsin ((x-1)/(x+1)) = 2arctan(x^1/2)-π/2?

Steve M
Featured 1 month ago

We wish to show that:

$\arcsin \left(\frac{x - 1}{x + 1}\right) = 2 \arctan \sqrt{x} - \frac{\pi}{2}$

Define a function $f \left(x\right)$ by:

$f \left(x\right) = \arcsin \left(\frac{x - 1}{x + 1}\right) - 2 \arctan \sqrt{x} + \frac{\pi}{2}$ ..... [A]

Put $x = 0$ and we get:

$f \left(0\right) = \arcsin \left(- 1\right) - 2 \arctan 0 + \frac{\pi}{2}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = - \frac{\pi}{2} - 0 + \frac{\pi}{2}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = 0$

We can use the standard derivatives:

 {: (ul("Function"), ul("Derivative"), ul("Notes")), (f(x), f'(x),), (sin^(-1)x, 1/sqrt(1-x^2), ), (tan^(-1)x, 1/(1+x^2), ) :}

So if we differentiate [A] and apply the chain rule, and quotient rule, we have:

$f ' \left(x\right) = \frac{1}{\sqrt{1 - {\left(\frac{x - 1}{x + 1}\right)}^{2}}} \frac{d}{\mathrm{dx}} \left(\frac{x - 1}{x + 1}\right) - \frac{2}{1 + {\left(\sqrt{x}\right)}^{2}} \frac{d}{\mathrm{dx}} \left(\sqrt{x}\right)$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{1}{\sqrt{1 - {\left(x - 1\right)}^{2} / {\left(x + 1\right)}^{2}}} \frac{\left(x + 1\right) \left(1\right) - \left(x - 1\right) \left(1\right)}{x + 1} ^ 2 - \frac{2}{1 + x} \frac{1}{2 \sqrt{x}}$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{1}{\sqrt{\frac{{\left(x + 1\right)}^{2} - {\left(x - 1\right)}^{2}}{x + 1} ^ 2}} \frac{x + 1 - x + 1}{x + 1} ^ 2 - \frac{1}{\left(1 + x\right) \sqrt{x}}$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = \sqrt{{\left(x + 1\right)}^{2} / \left({\left(x + 1\right)}^{2} - {\left(x - 1\right)}^{2}\right)} \cdot \frac{2}{x + 1} ^ 2 - \frac{1}{\left(1 + x\right) \sqrt{x}}$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = \sqrt{{\left(x + 1\right)}^{2} / \left(\left(x + 1 + x - 1\right) \left(x + 1 - x + 1\right)\right)} \cdot \frac{2}{x + 1} ^ 2 - \frac{1}{\left(1 + x\right) \sqrt{x}}$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = \sqrt{{\left(x + 1\right)}^{2} / \left(\left(2 x\right) \left(2\right)\right)} \cdot \frac{2}{x + 1} ^ 2 - \frac{1}{\left(1 + x\right) \sqrt{x}}$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{x + 1}{2 \sqrt{x}} \cdot \frac{2}{x + 1} ^ 2 - \frac{1}{\left(1 + x\right) \sqrt{x}}$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{1}{\sqrt{x}} \cdot \frac{1}{x + 1} - \frac{1}{\left(1 + x\right) \sqrt{x}}$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = 0$

So the function $f \left(x\right)$ has a zero derivative everywhere, and therefore integrating we have:

$f \left(x\right) = c$

We established earlier that $f \left(0\right) = 0 \implies f \left(0\right) = c = 0$, Hence:

$c = 0$

Hence we have:

$\setminus \setminus \setminus \setminus \arcsin \left(\frac{x - 1}{x + 1}\right) - 2 \arctan \sqrt{x} + \frac{\pi}{2} = 0$

$\therefore \arcsin \left(\frac{x - 1}{x + 1}\right) = 2 \arctan \sqrt{x} - \frac{\pi}{2}$ QED

3

## 6(tanx)^2-4(sinx)^2=1. Solve x?

dk_ch
Featured 1 month ago

$6 {\left(\tan x\right)}^{2} - 4 {\left(\sin x\right)}^{2} = 1$

=>6(tan^2x)xxcos^2x-4sin^2x xxcos^2x=cos^2x

$\implies 6 {\sin}^{2} x - 4 {\sin}^{2} x \times {\cos}^{2} x = {\cos}^{2} x$

$\implies 6 \left(1 - {\cos}^{2} x\right) - 4 \left(1 - {\cos}^{2} x\right) {\cos}^{2} x = {\cos}^{2} x$

$\implies 6 - 6 {\cos}^{2} x - 4 {\cos}^{2} x + 4 {\cos}^{4} x = {\cos}^{2} x$

$\implies 4 {\cos}^{4} x - 11 {\cos}^{2} x + 6 = 0$

$\implies 4 {\cos}^{4} x - 8 {\cos}^{2} x - 3 {\cos}^{2} x + 6 = 0$

$\implies 4 {\cos}^{2} x \left({\cos}^{2} x - 2\right) - 3 \left({\cos}^{2} x - 2\right) = 0$

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

when ${\cos}^{2} x - 2 = 0 \implies \cos x \pm \sqrt{2}$

but $- 1 \le \cos x \le + 1$
so this solution is not acceptable.

when

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

$\cos x = \pm \frac{\sqrt{3}}{2}$

For

$\cos x = + \frac{\sqrt{3}}{2} = \cos \left(\frac{\pi}{6}\right)$

$\implies x = 2 n \pi \pm \frac{\pi}{6} \text{ where } n \in \mathbb{Z}$

For

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

$\implies x = 2 n \pi \pm \frac{5 \pi}{6} \text{ where } n \in \mathbb{Z}$

1

## How do you graph y=cos(x+pi/6)?

Kevin D.
Featured 1 month ago

#### Answer:

Graph $y = \cos \left(x\right)$ and shift everything to the left by $\frac{\pi}{6}$

#### Explanation:

We know that $\sin$ and $\cos$ has a period of $2 \pi$. That is to say that it repeats itself every $2 \pi$ units.
I would assume you know how to graph a $f \left(x\right) = \cos \left(x\right)$ functions, if not, it should look like this:

Now, you need to graph $f \left(x\right) = \cos \left(x + \frac{\pi}{6}\right)$.

Imagine you have a function $f \left(x\right)$ and another function $g \left(x\right) = f \left(x + 1\right)$.

What this means is that for any point $\left(x , y\right)$ on the graph $g \left(x\right)$, it will take $x + 1$ units for $f \left(x\right)$ to reach that same $y$ value.
That is what this $g \left(x\right) = f \left(x + 1\right)$ is saying.

This means that all points on $g \left(x\right)$ is occurring 1 unit earlier than $f \left(x\right)$ so we shift $f \left(x\right)$ to the left by 1 unit to obtain $g \left(x\right)$.

To generalize:
If $g \left(x\right) = f \left(x + n\right)$ we shift $f \left(x\right)$n$u n i t s \to t h e \ast \le f t \ast \to \ge t$g(x). If g(x)=f(x-n)$w e s h \mathmr{if} t$f(x) $n$ units to the right to get $g \left(x\right)$.

Now, we can apply it to this question:

We have $f \left(x\right) = \cos \left(x + \frac{\pi}{6}\right)$ which is basically saying we should shift $\cos \left(x\right)$ to the left by $\frac{\pi}{6}$ units.

The blue curve is your $y = \cos \left(x + \frac{\pi}{6}\right)$
The red curve is your $y = \cos \left(x\right)$

3

## How cos^2 (pi /11) + cos^2((2 pi )/11 )+ cos^2 ((3 pi)/ 11) + cos^2(( 4 pi)/ 11) + cos^2(( 5 pi )/11) = 9 / 4?

dk_ch
Featured 2 weeks ago

LHS=cos^2 (pi /11) + cos^2((2 pi )/11 )+ cos^2 ((3 pi)/ 11) + cos^2(( 4 pi)/ 11) + cos^2(( 5 pi )/11) )

$= \frac{1}{2} \left(2 {\cos}^{2} \left(\frac{\pi}{11}\right) + 2 {\cos}^{2} \left(\frac{2 \pi}{11}\right) + 2 {\cos}^{2} \left(\frac{3 \pi}{11}\right) + 2 {\cos}^{2} \left(\frac{4 \pi}{11}\right) + 2 {\cos}^{2} \left(\frac{5 \pi}{11}\right)\right)$

$= \frac{1}{2} \left(5 + \cos \left(\frac{2 \pi}{11}\right) + \cos \left(\frac{4 \pi}{11}\right) + \cos \left(\frac{6 \pi}{11}\right) + \cos \left(\frac{8 \pi}{11}\right) + \cos \left(\frac{10 \pi}{11}\right)\right)$ $\textcolor{red}{\left[\text{using } 2 {\cos}^{2} \theta = \left(1 + \cos 2 \theta\right)\right]}$

$= \frac{1}{2} \left(5 + \cos \left(\frac{6 \pi}{11}\right) + \cos \left(\frac{4 \pi}{11}\right) + \cos \left(\frac{2 \pi}{11}\right) + \cos \left(\frac{8 \pi}{11}\right) + \cos \left(\left(\pi - \frac{\pi}{11}\right)\right)\right)$

$= \frac{1}{2} \left(5 + 2 \cos \left(\frac{5 \pi}{11}\right) \cos \left(\frac{\pi}{11}\right) + 2 \cos \left(\frac{5 \pi}{11}\right) \cos \left(\frac{3 \pi}{11}\right) - \cos \left(\frac{\pi}{11}\right)\right)$

$= \frac{1}{2} \left(5 + 2 \cos \left(\frac{5 \pi}{11}\right) \left(\cos \left(\frac{\pi}{11}\right) + \cos \left(\frac{3 \pi}{11}\right)\right) - \cos \left(\frac{\pi}{11}\right)\right)$

$= \frac{1}{2} \left(5 + 2 \cos \left(\frac{5 \pi}{11}\right) \cdot 2 \cos \left(\frac{2 \pi}{11}\right) \cos \left(\frac{\pi}{11}\right) - \cos \left(\frac{\pi}{11}\right)\right)$

$= \frac{1}{2} \left(5 + \cos \frac{\frac{5 \pi}{11}}{\sin} \left(\frac{\pi}{11}\right) \cdot 2 \cos \left(\frac{2 \pi}{11}\right) \cdot 2 \sin \left(\frac{\pi}{11}\right) \cos \left(\frac{\pi}{11}\right) - \cos \left(\frac{\pi}{11}\right)\right)$

$= \frac{1}{2} \left(5 + \cos \frac{\frac{5 \pi}{11}}{\sin} \left(\frac{\pi}{11}\right) \cdot 2 \cos \left(\frac{2 \pi}{11}\right) \sin \left(\frac{2 \pi}{11}\right) - \cos \left(\frac{\pi}{11}\right)\right)$

$= \frac{1}{2} \left(5 + \frac{2 \cos \left(\frac{5 \pi}{11}\right) \sin \left(\frac{4 \pi}{11}\right)}{2 \sin \left(\frac{\pi}{11}\right)} - \cos \left(\frac{\pi}{11}\right)\right)$

$= \frac{1}{2} \left(5 + \frac{\sin \left(\frac{9 \pi}{11}\right) - \sin \left(\frac{\pi}{11}\right)}{2 \sin \left(\frac{\pi}{11}\right)} - \cos \left(\frac{\pi}{11}\right)\right)$

$= \frac{1}{2} \left(5 + \sin \frac{\pi - \frac{2 \pi}{11}}{2 \sin \left(\frac{\pi}{11}\right)} - \sin \frac{\frac{\pi}{11}}{2 \sin \left(\frac{\pi}{11}\right)} - \cos \left(\frac{\pi}{11}\right)\right)$

$= \frac{1}{2} \left(5 + \sin \frac{\frac{2 \pi}{11}}{2 \sin \left(\frac{\pi}{11}\right)} - \frac{1}{2} - \cos \left(\frac{\pi}{11}\right)\right)$

$= \frac{1}{2} \left(5 + \frac{2 \sin \left(\frac{\pi}{11}\right) \cos \left(\frac{\pi}{11}\right)}{2 \sin \left(\frac{\pi}{11}\right)} - \frac{1}{2} - \cos \left(\frac{\pi}{11}\right)\right)$

$= \frac{1}{2} \left(5 + \cos \left(\frac{\pi}{11}\right) - \frac{1}{2} - \cos \left(\frac{\pi}{11}\right)\right)$

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

$= \frac{1}{2} \cdot \frac{9}{2} = \frac{9}{4} = R H S$

3

## If [email protected]=(sqrt(5)-1)/4 and [email protected]=(sqrt(5)+1)/4 then prove that (1+cos(pi/10))(1+cos((3pi)/10))(1+cos((7pi)/10))(1+cos((9pi)/10))=(1/16)?

dk_ch
Featured 2 weeks ago

$L H S = \left(1 + \cos \left(\frac{\pi}{10}\right)\right) \left(1 + \cos \left(\frac{3 \pi}{10}\right)\right) \left(1 + \cos \left(\frac{7 \pi}{10}\right)\right) \left(1 + \cos \left(\frac{9 \pi}{10}\right)\right)$

Puttingv$\frac{\pi}{10} = 2 \theta$ we have

$L H S = \left(1 + \cos \left(2 \theta\right)\right) \left(1 + \cos \left(6 \theta\right)\right) \left(1 + \cos \left(14 \theta\right)\right) \left(1 + \cos \left(18 \theta\right)\right)$

$= \left(2 {\cos}^{2} \left(\theta\right)\right) \left(2 {\cos}^{2} \left(3 \theta\right)\right) \left(2 {\cos}^{2} \left(7 \theta\right)\right) \left(2 {\cos}^{2} \left(9 \theta\right)\right)$

$= {\left[\left(2 \cos \left(\theta\right) \cos \left(9 \theta\right)\right) \left(2 \cos \left(3 \theta\right) \cos \left(7 \theta\right)\right)\right]}^{2}$

$= {\left[\left(\cos \left(10 \theta\right) + \cos \left(8 \theta\right)\right) \left(\cos \left(10 \theta\right) + \cos \left(4 \theta\right)\right)\right]}^{2}$

$= {\left[\left(\cos \left(\frac{\pi}{2}\right) + \cos \left(8 \theta\right)\right) \left(\cos \left(\frac{\pi}{2}\right) + \cos \left(4 \theta\right)\right)\right]}^{2}$

$= {\left[\cos \left(8 \theta\right) \cos \left(4 \theta\right)\right]}^{2}$

$= {\left[\frac{4 \cos \left(8 \theta\right) \cos \left(4 \theta\right) \sin \left(4 \theta\right)}{4 \sin \left(4 \theta\right)}\right]}^{2}$

$= \frac{1}{16} {\left[\sin \frac{16 \theta}{\sin} \left(4 \theta\right)\right]}^{2}$

$= \frac{1}{16} {\left[\sin \frac{\frac{8 \pi}{10}}{\sin} \left(\frac{2 \pi}{10}\right)\right]}^{2}$

$= \frac{1}{16} {\left[\sin \frac{\pi - \frac{2 \pi}{10}}{\sin} \left(\frac{2 \pi}{10}\right)\right]}^{2}$

$= \frac{1}{16} {\left[\sin \frac{\frac{2 \pi}{10}}{\sin} \left(\frac{2 \pi}{10}\right)\right]}^{2}$

$= \left(\frac{1}{16}\right) = R H S$

2

## Solve in the interval 0^@, 360^@ 2csc^2x + cotx = 10 How?

George C.
Featured 2 weeks ago

#### Answer:

${\cot}^{- 1} \left(\frac{1 + \sqrt{65}}{4}\right) \approx {23.816}^{\circ}$

${\cot}^{- 1} \left(\frac{1 - \sqrt{65}}{4}\right) \approx {150.473}^{\circ}$

${\cot}^{- 1} \left(\frac{1 + \sqrt{65}}{4}\right) + {180}^{\circ} \approx {203.816}^{\circ}$

${\cot}^{- 1} \left(\frac{1 - \sqrt{65}}{4}\right) + {180}^{\circ} \approx {330.473}^{\circ}$

#### Explanation:

Note that:

${\csc}^{2} x = \frac{1}{\sin} ^ 2 x = \frac{{\cos}^{2} x + {\sin}^{2} x}{\sin} ^ 2 x = 1 + {\cot}^{2} x$

So:

$10 = 2 {\csc}^{2} x + \cot x$

$\textcolor{w h i t e}{10} = 2 \left({\cot}^{2} x + 1\right) + \cot x$

$\textcolor{w h i t e}{10} = 2 {\cot}^{2} x + \cot x + 2$

Subtract $10$ from both ends to get:

$2 {\cot}^{2} x + \cot x - 8 = 0$

This is in the form:

$a {t}^{2} + b t + c = 0$

with $t = \cot x$, $a = 2$, $b = 1$ and $c = - 8$.

So using the quadratic formula, we have:

$\cot x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$\textcolor{w h i t e}{\cot x} = \frac{- \textcolor{b l u e}{1} \pm \sqrt{{\textcolor{b l u e}{1}}^{2} - 4 \left(\textcolor{b l u e}{2}\right) \left(\textcolor{b l u e}{- 8}\right)}}{2 \left(\textcolor{b l u e}{2}\right)}$

$\textcolor{w h i t e}{\cot x} = \frac{- 1 \pm \sqrt{65}}{4}$

Note that (expressed in degrees):

• $\cot x$ is continuous and monotonically decreasing on $\left({0}^{\circ} , {180}^{\circ}\right)$ with range $\left(- \infty , \infty\right)$.

• The range of ${\cot}^{- 1} \left(y\right)$ is $\left({0}^{\circ} , {180}^{\circ}\right)$.

• $\cot x$ has period ${180}^{\circ}$.

Hence all the solutions in $\left({0}^{\circ} , {360}^{\circ}\right)$ are:

${\cot}^{- 1} \left(\frac{1 + \sqrt{65}}{4}\right) \approx {23.816}^{\circ}$

${\cot}^{- 1} \left(\frac{1 - \sqrt{65}}{4}\right) \approx {150.473}^{\circ}$

${\cot}^{- 1} \left(\frac{1 + \sqrt{65}}{4}\right) + {180}^{\circ} \approx {203.816}^{\circ}$

${\cot}^{- 1} \left(\frac{1 - \sqrt{65}}{4}\right) + {180}^{\circ} \approx {330.473}^{\circ}$

Note that your calculator may give a negative value for ${\cot}^{- 1} \left(\frac{1 - \sqrt{65}}{4}\right)$ (namely $\approx - {29.527}^{\circ}$). If it does then just add ${180}^{\circ}$.

If your calculator does not have ${\cot}^{- 1}$ then take the reciprocal and use ${\tan}^{- 1}$, but note that this will give values in the range $\left(- {90}^{\circ} , {90}^{\circ}\right)$, so you will need to add an initial ${180}^{\circ}$ to the $\approx - {29.527}^{\circ}$ value to get a solution in the requested range.

1

## A pole leans away from the sun at an angle of 7° to the vertical, as shown in Figure 27. When the elevation of the sun is 55°, the pole casts a shadow 42 feet long on the level ground. How long is the pole? Round the answer to the nearest tenth.

Jason K.
Featured 3 weeks ago

#### Answer:

$p \approx 51.4 \text{ft}$

#### Explanation:

See the diagram below.

In the diagram, $\alpha$ represents the angle the pole makes with the ground, while point $P$ marks the edge of the shadow on the ground.

From inspection, we can see that the angles $\alpha$, $\beta$, and ${55}^{o}$ must all sum to ${180}^{o}$. Furthermore, since the pole makes an angle of ${7}^{o}$ to the vertical, we know that $\alpha + {7}^{o} = {90}^{o}$, and thus $\alpha = {83}^{o}$.

This means that we can calculate $\beta$:

$\alpha + \beta + {55}^{o} = {180}^{o}$

${83}^{o} + \beta + {55}^{o} = {180}^{o}$

$\beta = {42}^{o}$

Since we now have $\angle P$, $\beta$, and $b$, and we are looking for $p$, the Law of Sines can be helpful here:

$\sin \frac{P}{p} = \sin \frac{\beta}{b}$

$\sin {55}^{o} / p = \sin {42}^{o} / 42$

$\frac{42 \cdot \sin {55}^{o}}{\sin {42}^{o}} = p$

$p \approx 51.4 \text{ft}$

2

## Please help?

Jason K.
Featured 3 weeks ago

#### Answer:

$\approx 24.0 \text{km}$

#### Explanation:

In the picture, the unknown you are looking for (the distance between the two cities) is unlabeled, but is opposite of the angle of ${2.1}^{o}$ that is shown at the satellite. For convenience sake I will label the angle of ${2.1}^{o}$ as $\angle C$, and the unknown distance between the cities as $c$. Furthermore, I will arbitrarily label the "top" distance of 370km as $a$ and the "bottom" distance of 350km as $b$.

In this triangle problem, we have 2 sides ($a$ and $b$) and the angle between those sides ($\angle C$), and we are solving for the opposite side ($c$), which means the Law of Cosines is best suited for this problem:

${c}^{2} = {a}^{2} + {b}^{2} - 2 a b \cos C$

Filling in the knowns as listed above, we have to solve for $c$, which should be a simple matter so long as care is taken to calculate all values properly. (This is particularly true when using a calculator when some questions may be in radians and others in degrees!)

${c}^{2} = {a}^{2} + {b}^{2} - 2 a b \cos C$

${c}^{2} = {\left(370\right)}^{2} + {\left(350\right)}^{2} - 2 \left(370\right) \left(350\right) \left(\cos 2.1\right)$

${c}^{2} \approx 136900 + 122500 - 259000 \left(0.999\right)$

${c}^{2} \approx 259400 - 258826.054$

${c}^{2} \approx 573.946$

$c \approx \sqrt{573.946} \approx 24.0 \text{km}$

1

## How do you graph  y=3-cosx#?

Alan P.
Featured 1 week ago

#### Answer:

Plot $\cos \left(x\right)$
then reflect in the X-axis to get $- \cos \left(x\right)$
and finally shift all point up $3$ units to get $3 - \cos \left(x\right)$

#### Explanation:

I have assumed that you are familiar with the graph for $\textcolor{red}{y = \cos \left(x\right)}$:

Reflecting this in the X-axis causes every $y$ coordinate to be come the negative of what it was for $\textcolor{g r e y}{y = \cos \left(x\right)}$;
that is we get $\textcolor{red}{y = - \cos \left(x\right)}$

Then adding $3$ to every $y$ coordinate increases each $y$ value of $\textcolor{g r e y}{y = - \cos \left(x\right)}$ by $3$ to give $\textcolor{red}{y = 3 - \cos \left(x\right)}$