How do we solve linear trigonometric equations?

How do we solve linear trigonometric equations?

So when you solve linear trigonometric equations your just solving for any variable or θ.

There are 3 simple steps to solving, but let's do a quick practice to prepare.

Let's find the degrees of 2Sinθ-1 = 0

So you first add 1 to both sides and the divide by 2 to get Sinθ by itself

⬇

So you'd get Sinθ = 1/2

We know that Sin equals y, so then we'd find out where Sin is equal to 1/2. 

                   

According your trigonometric circle, you'd look to see where the point y is equal to 1/2.  *There's always two points, so look closely on the opposite side*

Or another shortcut is plugging the equations above in the calculator. You would have to find the inverse of Sinθ = (1/2) →  Sin⁻¹(1/2) = θ

You'd get 30°, so you have to subtract 180° and make it positive no matter what to get the second answer. Which is 150°

⤤ This method above only works for Sin, so be careful. 

Back on Track !!

Let's talk about those 3 steps then. 

1. Get the Trig function alone

Trig functions are Sin, Cos, Tan, etc. 

2. Use inverse of function to solve for angle

So when we did the inverse of Sin to find the angle of θ, that's what I mean.

3. Check for other solutions

Just like we found a shortcut but we looked for other solution once we got 30°, then we found 150°

Most of the time a problem is going to give you a range in where the answer could be. Let's try this sample problem.

Solve for all values that 0 ≤ θ ≤ 2π :

2Sinθ = √3

So you divide both sides by to to get Sinθ alone.

Sinθ = √3 / 2

Now you do the inverse of Sin ( Sin⁻¹ ) with √3 / 2 to get the angle of θ.

θ = 60°

So we have to check for other solutions, and we get 60° and 120°.

All using the 3 simple steps !! 

Sources:

1. John Schnatterly: johnschnatterly.blogspot.com

How do we evaluate inverse trigonometric relations and functions?

How do we evaluate inverse trigonometric relations and functions? 

Okay, so normal trigonometric functions are like sin(x), cos(x), and tan(x). And some added ones are sec(x), csc(x), and cot(x).

But now there's inverse functions that are just inverse of these regular functions. There's sin⁻¹(x), cos⁻¹(x), and tan⁻¹(x). As for the co-functions there are sec⁻¹(x), csc⁻¹(x), and cot⁻¹(x).

You will used these inverses when looking at arc lines, which are reflected from their original line over line y=x. For example for sin(x)

The red line is sin(x) and the blue line the the inverse sin⁻¹(x).
The green line shows y=x and how both lines reflect
over each other but cross each other's paths.

For arc lines you write them a little differently. For sin⁻¹(x) it's be written as arcsin(x). Here's a chart I made so you'd remember. 

So let's try a practice problem.

Let's find the arcsin (1/2) + arctan(1)

We know that arcsin is sin⁻¹ and arctan is tan⁻¹ , so we can just rewrite this.

sin⁻¹ (1/2) + tan⁻¹ (1)

Plugging that into your calculator you should get sin⁻¹ (1/2) = 30 and tan⁻¹ (1) = 45

So simplifying the problem :

30 + 45 = 75°

Sources:

1. John Schnatterly: johnschnatterly.blogspot.com

How do we solve exponential equations?

How do we solve exponential equations?

Exponential equations is another fancy way of say equations with different types of exponents. From positive to negative exponents, from whole numbers to fractions, variables to non-variables. 

You get the point !

SO to start of with the basics let's introduce the exponents. There are normal positive exponents, which everyone can do (on a calculator of course).

2⁴ = 16

Then there's negative exponents, where the outcome is basically a fraction form. For example if you have 2 to the -4. It's basically the same as 1 divided by 2⁴.

Therefore 2⁴ is equal to 16, so your answer would be   1  

                                                                                       16

Next there's fractioned exponents, where you'd probably get 2 to the 1/4. Here's a trick to that ! When you get 1/2 as an exponent, that's basically the same as square rooting the number you have. 

So if you get any number bigger than 2 as the denominator of the fraction (in the exponent), that would go outside the square root. 

 

Last but not least there are exponents with variables. You'll get a problem that looks like this:

2^x+4 = 8

First you have to make both bases equal, so you'd turn 8 into 2³.

2^x+4 = 2³

Then you take exponents and re-write it as an equation to find x.

x+4 = 3

(x = -1)

To make sure that the answer works you need to plug it back in and see if the equation is correct. When you get problems that have 2 answers, you'll need to plug it back in more often to se if one of the answer don't work. 

Sources:

1. John Schnatterly: johnschnatterly.blogspot.com
2. http://www.purplemath.com/modules/solvexpo2.htm
3. http://www.mathsisfun.com/algebra/exponent-fractional.html

How do we solve more complex rationals?

How do we solve more complex rationals? 

Complex rationals fall in the same category as complex fractions. So here we'll be doing more practice on it. 

When you get a fraction like ant ordinary, but instead it brings in i. 

You remember i don't you? Imaginary number, it SHOULD ring a bell. But anyways ... 

Where i :

And i goes in a cycle:

So i² = -1, i³ = -i, and i⁴ = 1

For example now, let's say you have a problem including i like this

You'd usually do the numerator and the denominator separate, and then re-write it. 

But here it's different because it's including i and your product won't come out the same. 

So instead you'd multiply the top and the bottom by something to turn the denominator into a whole number. In this case you'd multiply top and bottom by 7 - 4i.

Then you'd cancel out the denominators and multiply the numerators. 

That then simplifies to :

So 43-6i divided by 65 would be your final answer. 

That's about it for the i segment, but here's some more practice on normal complex rationals.

So your problem is this:

Remember you need to have it all simplified before you can do anything. So the numerator would simplify to 3y( y + 7).

Then you'd cross out common factors / multiples. Which would be (y + 7), and you would get your final, simplified answer. 

So that's mostly more practice on complex rationals, fractions, etc. Here's a problem i'll leave if you want to do. (It's not mandatory)

Problem :

Sources:

1. John Schnatterly: johnschnatterly.blogspot.com
2. http://www.mathwarehouse.com/algebra/complex-number/divide/how-to-divide-complex-numbers.php
3. http://www.mathsisfun.com/numbers/imaginary-numbers.html

How do we simplify complex fractions?

How do we simplify complex fractions?

Complex fractions are really solved the same way rational expressions. 

When you add or subtract you have to have a common denominator, when you divide you use the KCF method, and when you multiply you just need to remember to simplify. 

But with complex fractions you'll get fractions like these:

With a problem like this you'd just re-write the problem so that it looks normal, and then do the same as division. Use the KCF method. 

But just remember that before you "flip over", re-write the fraction out. Incase of any mistakes or anything. 

That's it for that part, but if you get a real complex fractions with things you don't know ... Here's and example:

You're just like -_- " what is this? "

Well let's break it down. Let's being with simplifying the numerator.

You'll get:

because you have to have a
common denominator
when adding fractions.

Which simplifies to:

Now for the denominator it's look like:

common denominator !

And simplify to:

So now that we've simplified the numerator and denominator, you'll plug it back in and re-write the problem. 

⬇

KCF method

Cross out common factors

Then your answer would simply be:

Because it wouldn't be simplified more than that.

So that's basically it !

Sources:

1. John Schnatterly: johnschnatterly.blogspot.com
2. http://www.purplemath.com/modules/compfrac.htm
3. http://www.basic-mathematics.com/examples-of-complex-fractions.html

How do we solve problems with the rationals expression operations?

How do we solve problems with the rationals expression operations?

Let me show you.

When you get a normal problem with multiplication, you just multiply across.

When you get a division problem you have to use the KCF method. 

You Keep the first fraction, then you Change the sign, and finally you Flip the fraction. 

When you add or subtract fractions you normally just add the numerator and keep the denominator, but that's only if you have the same denominator.


When you don't have the same denominator, you have to convert them.

1	+	1
3		6

One third would be equal to 2/6 and one sixth would still stay the same.

Then you add both fraction and get your answer. 

So when you get a expression like this :

You have to do the KCF method. You keep the first fraction, change the sign to multiply, and flip the 2nd fraction. While you're at it, simplify it. 

You'd get    x    times (x-3)(x+3)

                                                                         x+3                3 * x

That would simplify to   x-3  

                                      x

That's how you do a radical expression operation, that simple.

Sources: 

1. John Schnatterly: johnschnatterly.blogspot.com
2. http://www.mathsisfun.com/algebra/rational-numbers-operations.html

How do we simplify radical expressions?

How do we simplify radical expressions?

Radical expressions look like this:

√108y³

Now when you see that your just like o.O

But it's not that hard as it's made to look, you just have to know how to solve it. 

Just like you'd do when you just have a normal radical like √108 , you first separate the biggest square and number that could multiply to that number under the radical. 

So for √108:

√36 * √3

36 is a perfect square and is the biggest one that fits into 108.

So in this case since you still have variables, don't forget to simplify those before you jump to let's say √36 * √3 = 6√3

We have y³, so we'd have to look for the greatest squared number.

Which in this case would be y².

That leads to √y² * √y

So lets put everything together.

√36 * √3 * √y² * √y

Now is when we can simplify and put numbers and variables outside of the radical. 

36 and y² are both prefect squares, which would go outside of the radical. 3 and y would stay inside the radical since they're not prefect squares. It would look something like this:

Since √36 = 6

and

√y² = y

↓

The outside would be 6y.

And since the 3 and y are simplified as they're gonna get, they stay the same. 

Your result would be:

6y√3y

Note:  The examples shown in these lessons on radicals show ALL of the steps in the process.  It mayNOT be necessary for you to list EVERY step.  As long as you understand the process and can arrive at the correct answer, you are ALL SET!!

Sources:

1. John Schnatterly: johnschnatterly.blogspot.com
2. http://www.regentsprep.org/Regents/math/ALGEBRA/AO1/Lsimplify.htm