Instantaneous Rate of Change of a Trigonometric Function

Now, everyone knows the basic method to find the instantaneous rate of change of an equation ; that is, to find the difference between 2 points nearest to the point required a rate of change, divided by the difference in x-values of the respective y-values chosen. SIMPLE. What about trigonometric functions? Well, I have to say that is JUST THE SAME. Like 100% the same.

There is only a few things you need to note:

- Given a trigonometric graph with/out transformation, the turning points of the graph give an instantaneous rate of ZERO!!!
- Given a plain trigonometric function, and asked to find the instantaneous rate when a < x < b, graph the function and use the Tangent Operation to find the instantaneous rate of change  simply subsititute the a and b into the function to find the y-values. Then use the traditional way to find the instantaneous rate of change.
- It really doesnt matter how complicated or how simple the function is, because your only need two distinct points to find the instanatneous rate of change

Take a look at this video to understand some detailed examples of this topic, if you are still unsure or how to find instantaneous rate of change. Remember to practise!

If you come across a quadratic trigonometric equation........hmmmm

Quadratic trigonometric equations are just like any other quadratic fequations. Dont see sinx / cosx / tanx as a BIG thing...if you look carefully, replace the trigonometric identities with a simple x, you will see the nostalgicity of it ; it will look like a regular quadratic equation.

there are generally 4 methods you can use to solve a quadratic trigonometric equation. stay calm and look carefully =)

method 1. common factor (factor out the common trigonometric identity and equate them to zero to find the acute angle,x)

method 2. trinomial factor (factorise the whole equation to get something like (kx + a)(jx + b). equate each polynomial to zero to find the acute angle, x)

method 3. replace 'complex' trigonometric identities with simpler ones (using the trigonometric formulas. then factor them, and again, equate it to zero to find the acute angle, x)

method 4. quadratic formula (use the quadratic formula to determine the roots of an unfactorable trigonometric equation. or simply use the GC)

Sorry to be repetitive, but seriously, it's nothing much of a big deal. Take it as a quadratic equation you are required to solve.

For example,

tanxcos²x = tanx    (uh-ohhhhhhh...what now?!)

Remember the identity cos²x = 1 – sin²x  ? Use it!

Substitute it into the equation and you will get tanx (cos²x – 1)Factor out (cos²x – 1) so you can find the acute angle of x.
See...Told you it isnt THAT tough   =D

Solving Trigonometric Equations

Solving trigonometric equations is all about finding the x value in the equation, or what may be also known as the angle in the equation.
First things first, to be able to solve trigonometric equations, you must first master your special angles and must know it by heart. refer to previous posts if you need kick-start reviews on special angles

Step by Step:
1. Simplify the equation so the left hand side has only the trigonometric identity and right hand side has only the integers
2. Note the positive/negative sign on the right hand side. Then determine which quadrant the angle is in (with reference to the trigonometric identity on the left hand side)
3. Find the value/s of x  (note that the value of x determined from the equation is an acute angle. Thus, the possible values of x must be determined by adding or substracting angles to the acute angle, depending on which quadrant the angle is in)

e.g.
sin x = - s.r3 / 2  ;   0  <  x  <  2pi

Since we know that the acute angle of x is pi / 3 (use special angles to determine the angle in pink), and now after sketching the graph, we know that sin x lies in the 3rd and 4th quadrant (because in the 3rd and 4th quadrant, sin is a negative ; this matches with the answer determined). Hence, there are 2 values of x which lie in the 3rd and 4th quadrant (obeying the restriction that x's maximum is 2pi)

x = pi / 3 + pi     >>>     because to find an angle in the 3rd quadrant is pi + acute angle
   = 4pi / 3

AND

x = 2pi - pi / 3
   = 5pi / 3

Solving trigonometric identities is really about practising by doing more exercises and familiarizing yourself with the special angles and methods to solve it.

Definitely not a one-time-go math topic.

Check this video out. It'll probably help you understand the above much, much better.
Good Luck!

Sinusoidal Functions of the form f(x) = a sin/cos [k (x-d)] + c

SINUSOIDALLLLLL! (big bombastic word here) ; but it is not self-explanatory. the things under it is not as bombastic as it sounds. hehehehehe...

As usual, when you get a function  f(x) = a sin/cos [k (x-d)] + c,

|a| = amplitude of the graph (refer to the previous 2 posts for a clear definition of an amplitude)
k = horizontal stretch / compression
d = horizontal shift
c = vertical shift

The only thing that's gonna be new here is the method to find the graph's period, given its equation. Using the formula
2π / k = Period
remember, k is the horizontal stretch / compression you can determine by the equation given. Hence, substitute k into the formula, and you will be able to find the period of your trigonometric graph.

Example,
Transform a sine function such that g(x) has amplitude 4, period π, phase shift π/6rad and 2 units up.

Using the formula given above,
the period of the function = 2π/k = π. Therefore, k is 2.
Hence, g(x) = 4 sin [2 (x + π/6)] + 2

I found this exercise quite beneficial especially if ur new to these. After mastering a few questions, you're set and ready to go! (of course, only for this subtopic...hehehe) EXERCISES!

Sayonara!

Graphs of Reciprocal Trigonometric Functions

Now, the R-E-C-I-P-R-O-C-A-L!!! you really cant deny that mathematicians from the past have genuinely creative and curious minds. =]

This part of math has nothing new to learn, but only new things to remember...
Since trigonometric function graphs are periodic, they are continuous and will definitely have x-intercepts.

To remember:

1. The x-intercepts of a trigonometric function become the vertical asymptotes of the respective trigonnometric function's reciprocal.

2. The reciprocal of a trigonometric function IS NOT THE SAME as the inverse of the trigonometric function.

3. Practise, practise, practise with the graphing calculator!

Examples:

If y = sin x is graphed, its x-intercept lies at π.
If y = 1/sin x is graphed, its x-intercept of  π becomes its vertical asymptote. Hence, y = 1/sin x has a vertical asymptote at x =  π

The reciprocal of a trigonometric function is 1 divided by the trigonometric function. However, Trigo^-1  is asking for the angle that has the trigonometric ratio equal to x (the inverse).

For further help of this, this video can probably help you understand a little better as it provides detailed examples. Good luckie!!!~

Graphs of Sine, Cosine and Tangent Functions

FIRST OF ALL, you must remember that trigonometric functions like sine, cosine and tangent are periodic. In other words, they are continuous with the same pattern  =D  think of eternal~
I would say this part of math is fairly easy to understand. But it's just the matter of understanding and knowing it well without memorizing.

Remember how polynomial functions could have their transformations? Like the vertical stretch, horizontal shift, etc? Trigonometric functions can too. But there are 'special ways' to understand their transformation. Not that much of a difference, its just the way to find the value of each transformation.

Before that, note there are 3 graphs at the top. Now, these 3 graphs are all trigonometric graphs over ONE CYCLE.

What is a PERIOD?
A period is basically the x value at the end of a cycle (from the beginning of the first cycle to the end of the first cycle). In this case, the period for sinx graph is 2pi, for cosx graph is 2pi, and pi for the tanx graph. It means how 'long' it takes for one cycle to be completed.

What is an AMPLITUDE?
An amplitude is like the 'height' of the graph from the x axis. Ask yourself "How high and how low does the graph go, in terms of the y values?" and you will get the amplitude of the graphs. In this case, the amplitude for the sinx graph is 1unit, for the cosx graph is 1unit, and 1 unit as well for the tanx graph.
To refresh,

y = a sin k (x - d) + c, where

a = vertical stretch / vertical compression
k = horizontal stretch / horizontal compression
d = horizontal shift (left if +, and right if -)
c = vertical shift (upwards or downwards)

Normally, the questions asked in this topic will ask for a new equation based on the information they provide you with, such as the vertical shift, horizontal compression and more. All you need to do is just to fill in the value they provide at the correct space in the general equation. For example,

A sine function has a vertical stretch of 2.
Hence, the equation of the function is  y = 2sinx

However, sometimes they give you the period of the graph and expect you to find its horizontal stretch/compression. Then comes the formula  

Period = 2pi / k, where k is the horizontal stretch/compression.

For example,
Find the value of k if the period is 2pi
Hence, 2pi = 2pi / k   >>>   k = 4.
y = sin4x

Simple as that! If you need more help, Click Here for a YouTube Tutorial

Bye!

By the way. . .

After learning about speedometers and secants yesterday, I became mathcrazee while driving my way home.

I noted the mileage on my dashboard and the time I left college. Drove home as usual and noted my end mileage and the time I arrived home. Just for the fun of it, I calculated my average rate of speed throughout my journey. 

I subtracted the difference in my beginning and end mileage and divided it by the time I reached home. As a result, I found out that I was driving at 80km/h. 

Ok I know...sounds crazy. But why not apply these stuffs to real life?? Wouldn't it be much easier to remember or understand it this way, instead of blindly memorizing from the book?

Try it yourself. It really works!