Section 8-5 Solving More Difficult Trig. Functions

sec^2 x = 9
These trig functions are solved very similar to way they were done in Section 8-1
This time you would take the square root of 9 and get 3.
Then take the inverse of sin3 in you calculator and get no solution because Sin can not be greater than 1.
That would make your final answer no solution.

- Parrish Masters

9-1

This week I'm going to explain to you how to use sinx, cosx, and tanx to find side lengths of a triangle.

To remember how to do this you are going to use the word SOHCAHTOA. This is an easy way to remember the formulas to do this.

• Sinx=Opp/Hyp
• Cosx=Adj/Hyp
• Tanx=Opp/Adj

These formulas are only to be used of RIGHT triangles. Also, when finding these side lengths never use the 90 degree angle. Always use the other angle given.

Okay, so now I'm going to do an example.

Example 1: Find d and f

Okay, so we are going to find the side length f first.

Since you are give angle E, which is 12 degrees and you need to find f. You are going to use tanx=opp/adj. Which is going to give you tan12=9/f. When you solve for f you are going to get f=42.342 degrees.

Now we are going to find the side length d.

Your going to use angle E again, but this time you are going to use sinx=opp/hyp. This is going to give you sin12=9/d. When you solve for d you are going to get d=43.288 degrees.

So you answers are going to be f=42.342 degrees and d=43.288 degrees.

--Halie!

8-5

8-5 is almost exactly like 8-1.
***You can not divide by a trig function to cancel it on both sides in an equation. You must move it to the other side and factor it out.***
I will tell/show you the steps as I work the problem.

Ex.1: 8sin^2x-4=0

• First, you have to get the trig function by itself. You subtract 4 on both sides leaving you with 8sin^2x=4.
• Now you have to divide by 8 on both sides leaving you with sin^2x=4/8 which equals to sin^2x=1/2.
• After you do that, you have to take the square root on both sides. That gives you sinx=the square root of 1/2 which equals to sinx=+-square root of 2/2.
• Now you solve for x which gives you x=sin-1(+-square root of 2/2). When you are solving for x, you take the inverse of sin.
• Now you have to find what quadrant it is in. Sin-1(+-square root of 2/2)=45. So 45 is in quadrant one and since it is a positive and negative answer, you have to find all of the quadrants.
• Quadrant 2: -45+180=135. Quadrant 3: 45+180=225. Quadrant 4: -45+360=315. Your final answer: X=45 degrees, 135 degrees, 225 degrees, 315 degrees.

Ex.2: 6sinx=cosx for 0 degrees less than or equal to x less than or equal to 360 degrees

• First, you have to divide by cosx on both sides leaving you with 6sinx/cosx=1.
• Now you can change 6sinx/cosx to 6tanx so that gives you 6tanx=1.
• Then you have to divide by 6 on both sides which gives you tanx=1/6.
• Now solve for x: x=tan-1(1/6). tan-1(1/6)=9.462 which is in quadrant one.
• Now you have to find the quadrants where tan is positive. Tan is positive in one and three. We already found quadrant one so now we have to find quadrant three. Quadrant three: 9.462+180=189.462.
• Now you have to put those answers into degrees minutes and seconds. Quadrant one: .462x60=27.72, .72x60=43.2. Quadrant three: .462x60=27.72, .72x60=43.2.
• Final answer: X= 9 degrees 27 minutes 43 seconds, 189 degrees 27 minutes 43 seconds

Amber :)

8-4

8-4 Relationships Among the Functions

Today I'm doing my blog on 8-4 because I can guarantee you that I bombed this stuff on my test, and I need all the practice I can get. The following are all the relationship equations you will use in solving the problems in this section.

-sin x/cos x= tan x cos x/sin x= cot x

-Reciprocal Relationships:
csc(theta)=1/sin(theta) sec(theta)=1/cos(theta) cot(theta)=1/tan(theta)

-Pythagorean Relationships:
sin^2(theta)+cos^2(theta)=1 1+tan^2(theta)=sec^2(theta) 1+cot^2(theta)=csc^2(theta)

-Cofunction Relationships:
sin(theta)=cos(90-theta) & cos(theta)=sin(90-theta)
tan(theta)=csc(90-theta) & cot(theta)=tan(90-theta)
sec(theta)=csc(90-theta) & csc(theta)=sec(90-theta)

Example 1:
sec x - sin x tan x
=1/cos x - sin x (sin x/cos x)
=1-sin^2 x/cos x
=cos^2 x/cos x
=cos x
answer= cos x

Example 2:
Prove cot A(1+tan^2 A)/tan A= csc^2 A
=(cot A) (sec^2 A)/tan A
=(cos^2 A/sin^2 A) (1/cos^2 A)
=1/sin^2 A
=csc^2 A
answer= csc^2 A

*Note that when you're dealing with negatives in these problems, treat them just as they'd be positive, using all of your formulas.

-Jordan Duhon

review on 8-5

today i am reviewing 8-5 because this week was mostly review. rememebr that the same rules in 8-1 apply to the rules of 8-5 except these problems are harder. also the major rule is that you cannot divide by a trig function to cancel it on both sides in an equation. you must nove it to the other side and factor it out.



EX: 2 tan^2 x = 3 tan x - 1
 first move 3 tan x - 1 to the other side because you cannot divide by a trig function to cancel.
2 tan^2 x - 3 tan x + 1 = 0
factor: 2 tan^2 x - 2 tan x - 1 tan x + 1
group: (2 tan^2 x - 2 tan x) (-1 tan x + 1)
simplify: 2 tan x(tan x - 1)   -1(tan x - 1)
set to zero: 2 tan x - 1 = 0      tan x - 1 = 0
do the simple equation: tan x = 1/2     tan x = 1
inverse: x = inverse of tan (1/2)        x = inverse of tan (1)

Pythagorean Identites

1+tan^2x=secx

1+cot^2x=cosx

sin^2x+cos^2x=1

These pythagorean identites are the gist of how to figure out if there are trig identities in your problem. Knowing these identities is essential to one's success in getting to the right answer.

For example in verifying identites:

1. cos^2x-sin^2x =1-2sin^2x

Start with left side using the identity sin^2x+cos^2x=1 or cos^2x= -sin^2x+1

2. cos^2x-sin^2x= (1-sin^2x)- sin^2x

3. It equals 1-2sin^2x

8-4 (and no bad algebra)

This week we learned about relationships among trig functions…how riveting. The first thing you will need to know is your Pythagorean Identities.

Sin^2Ɵ+cos^2Ɵ=1, tan^2Ɵ+1=sec^2Ɵ, cot^2Ɵ=csc^2Ɵ

Now, the following steps are to be loosely followed.

1. Algebra (In this step you should factor, FOIL, etc. oh and don’t do it badly.)

2. Identities (first you will want to try one of the Pythagoreans I listed earlier. If none of these work move everything to sin and cos if that helps)

3. ALGEBRA :D but don’t do it badly.

4. Continue with steps 1-3, but no bad algebra.

It may also help you to look back over some algebra to save yourself some time while doing these sorts of problems.

Now, time for the example :D

Solve.

(sinƟ-1)(sinƟ+1)

The first thing you want to do here is FOIL (without bad algebra).

Sin^2Ɵ-1

Next, you’ll want to check your Pythagorean identities and, lucky for me, this is one! In the margin, write that sin^2Ɵ+cos^2Ɵ=1

Subtract sin^2Ɵ from both sides and you will discover your answer is cos^2Ɵ.

Oh and don't forget, no bad algebra.

--Sarah