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