# Trigonometric Identities Notes

Standard

Intro to Trigonometry:

There are 6 trigonometric functions, which take angles as their input and output to a real number. An angle is said to be in standard position: x-axis, with it vertex at the origin.  If the angle opens clockwise then the angle is considered to be negative angle. If the angle opens counter-clockwise it is considered to be a positive angle.

Two angles are called coterminal if they share a terminal side in standard position. For example, the angles 150 degrees, 510 degrees, and -210 degrees are all coterminal.

Radians are another unit of measure for angles. There are 360 degrees in a complete angle

so 360 degrees is equal to 2 pi radians; 180 degrees is equal to pi radians; 180 degrees divided by pi is equal to 1 or equal to pi divided by 180 degrees.

The unit circle is the circle centered at the origin with a radius the length of one unit.

The definition of the trig functions in the context of a point on the plane is suppose $\theta$ is in standard position, and p= (x,y) is a point on the terminal side of theta, and r is the distance between P and the origin. Then the six trig functions are defined as follows:

sin $\theta$ = $\frac {y}{r}$

cos $\theta$ = $\frac {x}{r}$

tan $\theta$ = $\frac {y}{x}$

csc $\theta$ = $\frac {r}{y}$

sec $\theta$ = $\frac {r}{x}$

cot $\theta$ = $\frac {x}{y}$

In the context if a point is on the unit circle. Suppose $\theta$ is in standard position , and P= (x,y) is a point on the unit circle. Then the 6 trig functions are defined as follows:

sin $\theta$ =  y

cos $\theta$ = x

tan $\theta$ = $\frac {y}{x}$

csc $\theta$ = $\frac {1}{y}$

sec $\theta$ = $\frac {1}{x}$

cot $\theta$ = $\frac {x}{y}$

Definition of the trig functions in the context of a right triangle. Suppose $\theta$ is an acute angle of a right triangle. Trig functions are defined as follows:

sin $\theta$ = $\frac {opposite}{hypotenuse}$

cos $\theta$ = $\frac {adjacent}{hypotenuse}$

tan $\theta$ = $\frac {opposite}{adjacent}$

csc $\theta$ = $\frac {hypotenuse}{opposite}$

sec $\theta$ = $\frac {hypotenuse}{adjacent}$

cot $\theta$ = $\frac {adjacent}{opposite}$

Basic Identities:

$sin ^2$ x + $cos^2$ x = 1

tanx= $\frac {sinx}{cosx}$

sin(x + y) = sin x cos y + cos x sin y

sin (x – y) = sin x cos y – cos x sin y

cos (x + y) = cos x cos y – sin x sin y

cos (x – y) = cos x cos y + sin x sin y

The derivative of y = sin x also depends on the value of

$\overset {lim}{ h \rightarrow 0}$ $\frac {sin(x+h)-sinx}{h}$

To esitmate the limit, find the qoutient $\frac {sinx}{x}$ for various calues of x close to 0. ( make sure to set your calculator to radian mode not in degree mode)

Derivatives of Trig Functions:

Dx(sin x) = cos x

Dx(cos x) = -sin x

Dx (tan x) = $sec^2$ x

Dx (cot x) = $-csc^2$ x

Dx (sec x) = sec x tan x

Dx (csc x) = -csc x cot x

# Related Rates

Standard

Hey guys we are almost done!! We just took our last test of the semester and now the only thing we have is the final!! Whoooo!! We are so close! We need to not get lazy and keep up the hard work so we can pass this class with ease! I looked ahead and noticed that we are about to learn about related rates in class, so I thought I would give everyone a little insight on what to expect on the next lesson. This lesson should be fairly easy because we are just using information we already know and just adding a little extra twist in it. In these types of problems we will be given a rate of change associated with one value and then be given another rate of change related to a second value. The second value will also be related to the first value. It seems tricky, but it is really not that bad once we perform examples in class. These problems are  very similar to our optimization problems because we will have to figure out what equation we will need that includes both rates of change. When making the equation we will also have to pair up the right rate of change with the right value. These problems take a lot of because starting the problem and making the right equation is key to success.

Once you find the equations that associates the rates of change correctly, we will use the familiar chain rule to find the derivative of the equation. We will then be able to find the differentials of the equation.  We can also use the differential approximation to different rates of changes of variables, if the rate of change is not given for a certain value.

I found some helpful strategies on the internet  that might help you, when approaching these types of problems:

1. Draw a diagram. This is the most helpful step in related rates problems. It allows us to visualize the problem.
2. Assign variables to each quantity in  the problem that is a function of time. Each of these values will have some rate of change over time.
3. List all information that is given in the problem and the rate of change that we are trying to find.
4. Write an equation that associates the variables with one another. If there are variables for which we are not given the rates of change (except for the rate of change that we are trying to determine), we must find some relation form the nature of the question that allows us to write these variables in terms of variables for which the rates of change are given. We mist then substitute these relations into the main equation.
5. Using the chain rule, differentiate each side of the equation with respect to time
6. Substitute all given information into the equation and solve for the required rate of change.

Like I said earlier, rates of change problems are really easy. The hardest part is getting starting and organizing your information. No need to worry! :]

# Differentials

Standard

In almost all of the examples and applications so far, all functions have been defined in the form y=f (x) with y given explicitly in terms of x, or as an explicit function of x, for example y = 3x – 2 is an all explicit functions of x. The equation 4xy – 3x = 6 can be expressed as an explicit function of x by solving for y. This is:

4xy – 3x= 6

4xy = 3x + 6

y = $\frac {3x+6}{4x}$

On the other hand, some equations in x and y cannot be readily solved for y, and some equations cannot be solved for y at all. For example, while it would be possible (but tedious) to use the formula to solve for y in the equation $y^2+2yx+4x^2=0$, it is not possible to solve for y in the equation $y^5+by^3+6y^2x^2+2yx^3+6=0.$ In equations such as these last two, y is said to be given implicitly in terms of x.

In such cases, it may still be possible to find the derivative $\frac {dy}{dx}$  by a process called implicit differentiation. In doing so, we assume that there exists some function or functions f, which we may not be able to find, such that y= f(x) and $\frac {dy}{dx}$ exsist. It is useful to use $\frac {dy}{dx}$  here rather than f'(x) to make it clear which variable is independent and which is dependent.

One notation for the derivative of y is with respect to x is $\frac {dy}{dx}$ Notice that this similar to one way of starting slope which is $\frac {\Delta y}{\Delta x}$. $\Delta$ is used to represent change. When considering $\frac {dy}{dx}$ it is often useful to tink of the x as an infinitesimally small change. Then dy is the corresponding change in y.

REVIEW: Earlier in the year we pointed out that when y is given as a function of x, x is the independent variable. We later defined the derivative $\frac {dy}{dx}$ when y is a function of x. in an equation such as  3xy + $4y^2$ = 10 , either variable can be considered the independent variable; if it asks for $\frac {dy}{dx}$, consider x the independent variable ; if it ask for $\frac {dx}{dy}$, consider y the independent variable. A similar rule holds when other variables are used.

If y= f(x) then $\frac {dy}{dx}$ = f”(x); dy= f'(x)dx

Steps to using Implicit Differentiation:

To find $\frac {dy}{dx}$ for an equation containing x and y:

1. Differentiate on both sides of the equation with respect to x, keeping in mind that y is assumed to be a function of x.

2. Place all terms with $\frac {dy}{dx}$ on one side of the equal sign, and al terms without $\frac {dy}{dx}$ on the other side.

3. Factor out $\frac {dy}{dx}$, and then solve for $\frac {dy}{dx}$ .

HINT: When an applied problem involves an equation that is not given in explicit form, implicit differentiation can be used to locate maxima and minima or to find rates of change.

GOOD LUCK ON THE TEST MONDAY! 😀