# 7.1 Solving trigonometric equations with identities  (Page 5/9)

 Page 5 / 9

## Rewriting a trigonometric expression using the difference of squares

Rewrite the trigonometric expression: $\text{\hspace{0.17em}}4\text{\hspace{0.17em}}{\mathrm{cos}}^{2}\theta -1.$

Notice that both the coefficient and the trigonometric expression in the first term are squared, and the square of the number 1 is 1. This is the difference of squares. Thus,

Rewrite the trigonometric expression: $\text{\hspace{0.17em}}25-9\text{\hspace{0.17em}}{\mathrm{sin}}^{2}\text{\hspace{0.17em}}\theta .$

This is a difference of squares formula: $\text{\hspace{0.17em}}25-9\text{\hspace{0.17em}}{\mathrm{sin}}^{2}\text{\hspace{0.17em}}\theta =\left(5-3\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\theta \right)\left(5+3\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\theta \right).$

## Simplify by rewriting and using substitution

Simplify the expression by rewriting and using identities:

${\mathrm{csc}}^{2}\theta -{\mathrm{cot}}^{2}\theta$

$1+{\mathrm{cot}}^{2}\theta ={\mathrm{csc}}^{2}\theta$

Now we can simplify by substituting $\text{\hspace{0.17em}}1+{\mathrm{cot}}^{2}\theta \text{\hspace{0.17em}}$ for $\text{\hspace{0.17em}}{\mathrm{csc}}^{2}\theta .\text{\hspace{0.17em}}$ We have

Use algebraic techniques to verify the identity: $\text{\hspace{0.17em}}\frac{\mathrm{cos}\text{\hspace{0.17em}}\theta }{1+\mathrm{sin}\text{\hspace{0.17em}}\theta }=\frac{1-\mathrm{sin}\text{\hspace{0.17em}}\theta }{\mathrm{cos}\text{\hspace{0.17em}}\theta }.$

(Hint: Multiply the numerator and denominator on the left side by $\text{\hspace{0.17em}}1-\mathrm{sin}\text{\hspace{0.17em}}\theta .\right)$

Access these online resources for additional instruction and practice with the fundamental trigonometric identities.

## Key equations

 Pythagorean identities $\begin{array}{l}{\mathrm{sin}}^{2}\theta +{\mathrm{cos}}^{2}\theta =1\\ 1+{\mathrm{cot}}^{2}\theta ={\mathrm{csc}}^{2}\theta \\ 1+{\mathrm{tan}}^{2}\theta ={\mathrm{sec}}^{2}\theta \end{array}$ Even-odd identities $\begin{array}{l}\mathrm{tan}\left(-\theta \right)=-\mathrm{tan}\text{\hspace{0.17em}}\theta \\ \mathrm{cot}\left(-\theta \right)=-\mathrm{cot}\text{\hspace{0.17em}}\theta \\ \mathrm{sin}\left(-\theta \right)=-\mathrm{sin}\text{\hspace{0.17em}}\theta \\ \mathrm{csc}\left(-\theta \right)=-\mathrm{csc}\text{\hspace{0.17em}}\theta \\ \mathrm{cos}\left(-\theta \right)=\mathrm{cos}\text{\hspace{0.17em}}\theta \\ \mathrm{sec}\left(-\theta \right)=\mathrm{sec}\text{\hspace{0.17em}}\theta \end{array}$ Reciprocal identities $\begin{array}{l}\mathrm{sin}\text{\hspace{0.17em}}\theta =\frac{1}{\mathrm{csc}\text{\hspace{0.17em}}\theta }\\ \mathrm{cos}\text{\hspace{0.17em}}\theta =\frac{1}{\mathrm{sec}\text{\hspace{0.17em}}\theta }\\ \mathrm{tan}\text{\hspace{0.17em}}\theta =\frac{1}{\mathrm{cot}\text{\hspace{0.17em}}\theta }\\ \mathrm{csc}\text{\hspace{0.17em}}\theta =\frac{1}{\mathrm{sin}\text{\hspace{0.17em}}\theta }\\ \mathrm{sec}\text{\hspace{0.17em}}\theta =\frac{1}{\mathrm{cos}\text{\hspace{0.17em}}\theta }\\ \mathrm{cot}\text{\hspace{0.17em}}\theta =\frac{1}{\mathrm{tan}\text{\hspace{0.17em}}\theta }\end{array}$ Quotient identities $\begin{array}{l}\hfill \\ \mathrm{tan}\text{\hspace{0.17em}}\theta =\frac{\mathrm{sin}\text{\hspace{0.17em}}\theta }{\mathrm{cos}\text{\hspace{0.17em}}\theta }\hfill \\ \mathrm{cot}\text{\hspace{0.17em}}\theta =\frac{\mathrm{cos}\text{\hspace{0.17em}}\theta }{\mathrm{sin}\text{\hspace{0.17em}}\theta }\hfill \end{array}$

## Key concepts

• There are multiple ways to represent a trigonometric expression. Verifying the identities illustrates how expressions can be rewritten to simplify a problem.
• Graphing both sides of an identity will verify it. See [link] .
• Simplifying one side of the equation to equal the other side is another method for verifying an identity. See [link] and [link] .
• The approach to verifying an identity depends on the nature of the identity. It is often useful to begin on the more complex side of the equation. See [link] .
• We can create an identity by simplifying an expression and then verifying it. See [link] .
• Verifying an identity may involve algebra with the fundamental identities. See [link] and [link] .
• Algebraic techniques can be used to simplify trigonometric expressions. We use algebraic techniques throughout this text, as they consist of the fundamental rules of mathematics. See [link] , [link] , and [link] .

## Verbal

We know $\text{\hspace{0.17em}}g\left(x\right)=\mathrm{cos}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ is an even function, and $\text{\hspace{0.17em}}f\left(x\right)=\mathrm{sin}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}h\left(x\right)=\mathrm{tan}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ are odd functions. What about $\text{\hspace{0.17em}}G\left(x\right)={\mathrm{cos}}^{2}x,F\left(x\right)={\mathrm{sin}}^{2}x,$ and $\text{\hspace{0.17em}}H\left(x\right)={\mathrm{tan}}^{2}x?\text{\hspace{0.17em}}$ Are they even, odd, or neither? Why?

All three functions, $\text{\hspace{0.17em}}F,G,$ and $\text{\hspace{0.17em}}H,$ are even.

This is because $\text{\hspace{0.17em}}F\left(-x\right)=\mathrm{sin}\left(-x\right)\mathrm{sin}\left(-x\right)=\left(-\mathrm{sin}\text{\hspace{0.17em}}x\right)\left(-\mathrm{sin}\text{\hspace{0.17em}}x\right)={\mathrm{sin}}^{2}x=F\left(x\right),G\left(-x\right)=\mathrm{cos}\left(-x\right)\mathrm{cos}\left(-x\right)=\mathrm{cos}\text{\hspace{0.17em}}x\mathrm{cos}\text{\hspace{0.17em}}x={\mathrm{cos}}^{2}x=G\left(x\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}H\left(-x\right)=\mathrm{tan}\left(-x\right)\mathrm{tan}\left(-x\right)=\left(-\mathrm{tan}\text{\hspace{0.17em}}x\right)\left(-\mathrm{tan}\text{\hspace{0.17em}}x\right)={\mathrm{tan}}^{2}x=H\left(x\right).$

Examine the graph of $\text{\hspace{0.17em}}f\left(x\right)=\mathrm{sec}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ on the interval $\text{\hspace{0.17em}}\left[-\pi ,\pi \right].\text{\hspace{0.17em}}$ How can we tell whether the function is even or odd by only observing the graph of $\text{\hspace{0.17em}}f\left(x\right)=\mathrm{sec}\text{\hspace{0.17em}}x?$

After examining the reciprocal identity for $\text{\hspace{0.17em}}\mathrm{sec}\text{\hspace{0.17em}}t,$ explain why the function is undefined at certain points.

When $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}t=0,$ then $\text{\hspace{0.17em}}\mathrm{sec}\text{\hspace{0.17em}}t=\frac{1}{0},$ which is undefined.

All of the Pythagorean identities are related. Describe how to manipulate the equations to get from $\text{\hspace{0.17em}}{\mathrm{sin}}^{2}t+{\mathrm{cos}}^{2}t=1\text{\hspace{0.17em}}$ to the other forms.

## Algebraic

For the following exercises, use the fundamental identities to fully simplify the expression.

$\mathrm{sin}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}\mathrm{sec}\text{\hspace{0.17em}}x$

$\mathrm{sin}\text{\hspace{0.17em}}x$

$\mathrm{sin}\left(-x\right)\mathrm{cos}\left(-x\right)\mathrm{csc}\left(-x\right)$

$\mathrm{tan}\text{\hspace{0.17em}}x\mathrm{sin}\text{\hspace{0.17em}}x+\mathrm{sec}\text{\hspace{0.17em}}x{\mathrm{cos}}^{2}x$

$\mathrm{sec}\text{\hspace{0.17em}}x$

$\mathrm{csc}\text{\hspace{0.17em}}x+\mathrm{cos}\text{\hspace{0.17em}}x\mathrm{cot}\left(-x\right)$

$\frac{\mathrm{cot}\text{\hspace{0.17em}}t+\mathrm{tan}\text{\hspace{0.17em}}t}{\mathrm{sec}\left(-t\right)}$

$\mathrm{csc}\text{\hspace{0.17em}}t$

$3\text{\hspace{0.17em}}{\mathrm{sin}}^{3}\text{\hspace{0.17em}}t\text{\hspace{0.17em}}\mathrm{csc}\text{\hspace{0.17em}}t+{\mathrm{cos}}^{2}\text{\hspace{0.17em}}t+2\text{\hspace{0.17em}}\mathrm{cos}\left(-t\right)\mathrm{cos}\text{\hspace{0.17em}}t$

$-\mathrm{tan}\left(-x\right)\mathrm{cot}\left(-x\right)$

$-1$

$\frac{-\mathrm{sin}\left(-x\right)\mathrm{cos}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}\mathrm{sec}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}\mathrm{csc}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}x}{\mathrm{cot}\text{\hspace{0.17em}}x}$

$\frac{1+{\mathrm{tan}}^{2}\theta }{{\mathrm{csc}}^{2}\theta }+{\mathrm{sin}}^{2}\theta +\frac{1}{{\mathrm{sec}}^{2}\theta }$

${\mathrm{sec}}^{2}x$

$\left(\frac{\mathrm{tan}\text{\hspace{0.17em}}x}{{\mathrm{csc}}^{2}x}+\frac{\mathrm{tan}\text{\hspace{0.17em}}x}{{\mathrm{sec}}^{2}x}\right)\left(\frac{1+\mathrm{tan}\text{\hspace{0.17em}}x}{1+\mathrm{cot}\text{\hspace{0.17em}}x}\right)-\frac{1}{{\mathrm{cos}}^{2}x}$

$\frac{1-{\mathrm{cos}}^{2}\text{\hspace{0.17em}}x}{{\mathrm{tan}}^{2}\text{\hspace{0.17em}}x}+2\text{\hspace{0.17em}}{\mathrm{sin}}^{2}\text{\hspace{0.17em}}x$

${\mathrm{sin}}^{2}x+1$

For the following exercises, simplify the first trigonometric expression by writing the simplified form in terms of the second expression.

$\frac{\mathrm{tan}\text{\hspace{0.17em}}x+\mathrm{cot}\text{\hspace{0.17em}}x}{\mathrm{csc}\text{\hspace{0.17em}}x};\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}x$

$\frac{\mathrm{sec}\text{\hspace{0.17em}}x+\mathrm{csc}\text{\hspace{0.17em}}x}{1+\mathrm{tan}\text{\hspace{0.17em}}x};\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}x$

$\frac{1}{\mathrm{sin}\text{\hspace{0.17em}}x}$

$\frac{\mathrm{cos}\text{\hspace{0.17em}}x}{1+\mathrm{sin}\text{\hspace{0.17em}}x}+\mathrm{tan}\text{\hspace{0.17em}}x;\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}x$

$\frac{1}{\mathrm{sin}\text{\hspace{0.17em}}x\mathrm{cos}\text{\hspace{0.17em}}x}-\mathrm{cot}\text{\hspace{0.17em}}x;\text{\hspace{0.17em}}\mathrm{cot}\text{\hspace{0.17em}}x$

$\frac{1}{\mathrm{cot}\text{\hspace{0.17em}}x}$

$\frac{1}{1-\mathrm{cos}\text{\hspace{0.17em}}x}-\frac{\mathrm{cos}\text{\hspace{0.17em}}x}{1+\mathrm{cos}\text{\hspace{0.17em}}x};\text{\hspace{0.17em}}\mathrm{csc}\text{\hspace{0.17em}}x$

$\left(\mathrm{sec}\text{\hspace{0.17em}}x+\mathrm{csc}\text{\hspace{0.17em}}x\right)\left(\mathrm{sin}\text{\hspace{0.17em}}x+\mathrm{cos}\text{\hspace{0.17em}}x\right)-2-\mathrm{cot}\text{\hspace{0.17em}}x;\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}x$

$\mathrm{tan}\text{\hspace{0.17em}}x$

$-4\mathrm{sec}\text{\hspace{0.17em}}x\mathrm{tan}\text{\hspace{0.17em}}x$

$\mathrm{tan}\text{\hspace{0.17em}}x;\text{\hspace{0.17em}}\mathrm{sec}\text{\hspace{0.17em}}x$

$\mathrm{sec}\text{\hspace{0.17em}}x;\text{\hspace{0.17em}}\mathrm{cot}\text{\hspace{0.17em}}x$

$±\sqrt{\frac{1}{{\mathrm{cot}}^{2}x}+1}$

$\mathrm{sec}\text{\hspace{0.17em}}x;\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}x$

$\mathrm{cot}\text{\hspace{0.17em}}x;\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}x$

$\frac{±\sqrt{1-{\mathrm{sin}}^{2}x}}{\mathrm{sin}\text{\hspace{0.17em}}x}$

$\mathrm{cot}\text{\hspace{0.17em}}x;\text{\hspace{0.17em}}\mathrm{csc}\text{\hspace{0.17em}}x$

For the following exercises, verify the identity.

$\mathrm{cos}\text{\hspace{0.17em}}x-{\mathrm{cos}}^{3}x=\mathrm{cos}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}{\mathrm{sin}}^{2}\text{\hspace{0.17em}}x$

$\mathrm{cos}\text{\hspace{0.17em}}x-{\mathrm{cos}}^{3}x=\mathrm{cos}\text{\hspace{0.17em}}x\left(1-{\mathrm{cos}}^{2}x\right)$
$=\mathrm{cos}\phantom{\rule{0.2em}{0ex}}x{\mathrm{sin}}^{2}x$

$\mathrm{cos}\text{\hspace{0.17em}}x\left(\mathrm{tan}\text{\hspace{0.17em}}x-\mathrm{sec}\left(-x\right)\right)=\mathrm{sin}\text{\hspace{0.17em}}x-1$

$\frac{1+{\mathrm{sin}}^{2}x}{{\mathrm{cos}}^{2}x}=\frac{1}{{\mathrm{cos}}^{2}x}+\frac{{\mathrm{sin}}^{2}x}{{\mathrm{cos}}^{2}x}=1+2\text{\hspace{0.17em}}{\mathrm{tan}}^{2}x$

$\frac{1+{\mathrm{sin}}^{2}x}{{\mathrm{cos}}^{2}x}=\frac{1}{{\mathrm{cos}}^{2}x}+\frac{{\mathrm{sin}}^{2}x}{{\mathrm{cos}}^{2}x}={\mathrm{sec}}^{2}x+{\mathrm{tan}}^{2}x={\mathrm{tan}}^{2}x+1+{\mathrm{tan}}^{2}x=1+2{\mathrm{tan}}^{2}x$

${\left(\mathrm{sin}\text{\hspace{0.17em}}x+\mathrm{cos}\text{\hspace{0.17em}}x\right)}^{2}=1+2\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}x\mathrm{cos}\text{\hspace{0.17em}}x$

${\mathrm{cos}}^{2}x-{\mathrm{tan}}^{2}x=2-{\mathrm{sin}}^{2}x-{\mathrm{sec}}^{2}x$

${\mathrm{cos}}^{2}x-{\mathrm{tan}}^{2}x=1-{\mathrm{sin}}^{2}x-\left({\mathrm{sec}}^{2}x-1\right)=1-{\mathrm{sin}}^{2}x-{\mathrm{sec}}^{2}x+1=2-{\mathrm{sin}}^{2}x-{\mathrm{sec}}^{2}x$

## Extensions

For the following exercises, prove or disprove the identity.

$\frac{1}{1+\mathrm{cos}\text{\hspace{0.17em}}x}-\frac{1}{1-\mathrm{cos}\left(-x\right)}=-2\text{\hspace{0.17em}}\mathrm{cot}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}\mathrm{csc}\text{\hspace{0.17em}}x$

${\mathrm{csc}}^{2}x\left(1+{\mathrm{sin}}^{2}x\right)={\mathrm{cot}}^{2}x$

False

$\left(\frac{{\mathrm{sec}}^{2}\left(-x\right)-{\mathrm{tan}}^{2}x}{\mathrm{tan}\text{\hspace{0.17em}}x}\right)\left(\frac{2+2\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}x}{2+2\text{\hspace{0.17em}}\mathrm{cot}\text{\hspace{0.17em}}x}\right)-2\text{\hspace{0.17em}}{\mathrm{sin}}^{2}x=\mathrm{cos}\text{\hspace{0.17em}}2x$

$\frac{\mathrm{tan}\text{\hspace{0.17em}}x}{\mathrm{sec}\text{\hspace{0.17em}}x}\mathrm{sin}\left(-x\right)={\mathrm{cos}}^{2}x$

False

$\frac{\mathrm{sec}\left(-x\right)}{\mathrm{tan}\text{\hspace{0.17em}}x+\mathrm{cot}\text{\hspace{0.17em}}x}=-\mathrm{sin}\left(-x\right)$

$\frac{1+\mathrm{sin}\text{\hspace{0.17em}}x}{\mathrm{cos}\text{\hspace{0.17em}}x}=\frac{\mathrm{cos}\text{\hspace{0.17em}}x}{1+\mathrm{sin}\left(-x\right)}$

Proved with negative and Pythagorean identities

For the following exercises, determine whether the identity is true or false. If false, find an appropriate equivalent expression.

$\frac{{\mathrm{cos}}^{2}\theta -{\mathrm{sin}}^{2}\theta }{1-{\mathrm{tan}}^{2}\theta }={\mathrm{sin}}^{2}\theta$

$3\text{\hspace{0.17em}}{\mathrm{sin}}^{2}\theta +4\text{\hspace{0.17em}}{\mathrm{cos}}^{2}\theta =3+{\mathrm{cos}}^{2}\theta$

True $\text{\hspace{0.17em}}3\text{\hspace{0.17em}}{\mathrm{sin}}^{2}\theta +4\text{\hspace{0.17em}}{\mathrm{cos}}^{2}\theta =3\text{\hspace{0.17em}}{\mathrm{sin}}^{2}\theta +3\text{\hspace{0.17em}}{\mathrm{cos}}^{2}\theta +{\mathrm{cos}}^{2}\theta =3\left({\mathrm{sin}}^{2}\theta +{\mathrm{cos}}^{2}\theta \right)+{\mathrm{cos}}^{2}\theta =3+{\mathrm{cos}}^{2}\theta$

$\frac{\mathrm{sec}\text{\hspace{0.17em}}\theta +\mathrm{tan}\text{\hspace{0.17em}}\theta }{\mathrm{cot}\text{\hspace{0.17em}}\theta +\mathrm{cos}\text{\hspace{0.17em}}\theta }={\mathrm{sec}}^{2}\theta$

The center is at (3,4) a focus is at (3,-1), and the lenght of the major axis is 26
The center is at (3,4) a focus is at (3,-1) and the lenght of the major axis is 26 what will be the answer?
Rima
I done know
Joe
What kind of answer is that😑?
Rima
I had just woken up when i got this message
Joe
Rima
i have a question.
Abdul
how do you find the real and complex roots of a polynomial?
Abdul
@abdul with delta maybe which is b(square)-4ac=result then the 1st root -b-radical delta over 2a and the 2nd root -b+radical delta over 2a. I am not sure if this was your question but check it up
Nare
This is the actual question: Find all roots(real and complex) of the polynomial f(x)=6x^3 + x^2 - 4x + 1
Abdul
@Nare please let me know if you can solve it.
Abdul
I have a question
juweeriya
hello guys I'm new here? will you happy with me
mustapha
The average annual population increase of a pack of wolves is 25.
how do you find the period of a sine graph
Period =2π if there is a coefficient (b), just divide the coefficient by 2π to get the new period
Am
if not then how would I find it from a graph
Imani
by looking at the graph, find the distance between two consecutive maximum points (the highest points of the wave). so if the top of one wave is at point A (1,2) and the next top of the wave is at point B (6,2), then the period is 5, the difference of the x-coordinates.
Am
you could also do it with two consecutive minimum points or x-intercepts
Am
I will try that thank u
Imani
Case of Equilateral Hyperbola
ok
Zander
ok
Shella
f(x)=4x+2, find f(3)
Benetta
f(3)=4(3)+2 f(3)=14
lamoussa
14
Vedant
pre calc teacher: "Plug in Plug in...smell's good" f(x)=14
Devante
8x=40
Chris
Explain why log a x is not defined for a < 0
the sum of any two linear polynomial is what
Momo
how can are find the domain and range of a relations
the range is twice of the natural number which is the domain
Morolake
A cell phone company offers two plans for minutes. Plan A: $15 per month and$2 for every 300 texts. Plan B: $25 per month and$0.50 for every 100 texts. How many texts would you need to send per month for plan B to save you money?
6000
Robert
more than 6000
Robert
can I see the picture
How would you find if a radical function is one to one?
how to understand calculus?
with doing calculus
SLIMANE
Thanks po.
Jenica
Hey I am new to precalculus, and wanted clarification please on what sine is as I am floored by the terms in this app? I don't mean to sound stupid but I have only completed up to college algebra.
I don't know if you are looking for a deeper answer or not, but the sine of an angle in a right triangle is the length of the opposite side to the angle in question divided by the length of the hypotenuse of said triangle.
Marco
can you give me sir tips to quickly understand precalculus. Im new too in that topic. Thanks
Jenica
if you remember sine, cosine, and tangent from geometry, all the relationships are the same but they use x y and r instead (x is adjacent, y is opposite, and r is hypotenuse).
Natalie
it is better to use unit circle than triangle .triangle is only used for acute angles but you can begin with. Download any application named"unit circle" you find in it all you need. unit circle is a circle centred at origine (0;0) with radius r= 1.
SLIMANE
What is domain
johnphilip
the standard equation of the ellipse that has vertices (0,-4)&(0,4) and foci (0, -15)&(0,15) it's standard equation is x^2 + y^2/16 =1 tell my why is it only x^2? why is there no a^2?
what is foci?
This term is plural for a focus, it is used for conic sections. For more detail or other math questions. I recommend researching on "Khan academy" or watching "The Organic Chemistry Tutor" YouTube channel.
Chris