How do we solve trigonometric equations with more than one function?

Aim: How do we solve trigonometric equations with more than one function? 

We can solve trigonometric equations with more than one function by using trigonometric identities or reciprocal identities, so we can have one function and can easily solve by factoring. 

Example #1:
5 sin^2θ-9 cosθ-3=0

1) Replace the sine using the trig identity 1- cos^2θ  which is equivalent to sine. Doing this will allow you to work with one function.  

sin^2θ+cos^2θ=1
         - cos^2θ    -cos^2θ

    sin^2θ= 1-cos^2θ

5 (1- cos^2θ)-9 cosθ-3=0

2) Distribute the 5 into 1- cos^2θ so you can have a quadratic equation. 

5 - 5 cos^2θ-9 cosθ-3=0

3) Simplify the like term and arrange the equation in terms of ax^2+bx+c=0

-5 cos^2θ-9 cosθ+2=0
Multiply by -1 to get rid of the negative sign in front of the coefficient. 
-1(-5 cos^2θ-9 cosθ+2)=0(-1)

5 cos^2θ+9 cosθ-2=0

4)Now factor the equation by factoring them by grouping.

5 cos^2θ+9 cosθ-2=0

Multiply 5 times -2 and get -10. Than look for two numbers that multiply to -10 and add up to 9. Two number that add up to 9 and multiply to -10 are 10 and -1. Replace them with 9 cosθ.

5 cos^2θ+10 cosθ-1 cosθ-2=0

Now, group them into two pairs and factor by using common factors.

5 cos^2θ+10 cosθ                                      -1 cosθ-2
5 cosθ ( cosθ+2)                                    -1(cosθ+2)

Finally, you have (5 cosθ-1)(cosθ+2)=0

5) Solve for the solutions.

5 cosθ-1=0                                                cosθ+2=0
          +1  +1                                                      -2  -2

5/5 cosθ=1/5                                               cosθ= -2

cosθ=1/5

6)Find degrees where cosθ=1/5 and cosθ=-2

cosθ=1/5 in 78.5° and 281.5°                    cosθ= -2 has no solution
 (Do inverse cosine of 1/5 and you will get                                     because the unit circle 
78.5° and later subtract it from 360 to get                                     only goes up to 1. 
281.5°)
               
7) The solutions are 78.5° and 281.5°.

Example #2:

7 cosθ +1= 6 secθ

1)Replace secant with reciprocal of cosine.


7 cosθ +1= 6/cosθ

2) Multiply both side by cosine to get rid of the cosine from the denominator.

cosθ(7 cosθ +1)= (6/cosθ)cosθ 

7 cos^2θ +cosθ= 6

3) subtract 6 from both side so you can create a quadratic equation.

 7 cos^2θ +cosθ= 6

                      -6       -6

7 cos^2θ +cosθ-6=0

4) Now solve the rest of the problem by factoring by grouping and 

solving for the degree solutions as shown in the previous problem.  

How do we solve linear trigonometric equations?

Aim: How do we solve linear trigonometric equations? 

Solving trigonometric equations are easy as long as you know how to do them.  

There are three steps to solving linear equations:
1)  Get the trig function alone.
2)  Use inverse to solve for angle or use the helpful Unit Circle!
3)  Check for other solutions.

Example: 

Solve for all values between 0 to 2:
2 sin θ-√3 = 0
1) To get the trig function alone, add √3 to both sides of the equation.
                             2 sinθ -√3 = 0
                                       +√3   +√3
                             2 sinθ = √3
2) Divide 2 from both sides to get sinθ completely alone.
                              
                                 2 sinθ = √3
                                    2          2
                                 sinθ =  √3
                                              2
3) Now, if you look at the unit circle, you will see that the only places where sin (y-coordinate) is √3 are at π and 2π or in degrees 60° 
                                               2            3         3
and 120°. These are the solutions. You can also achieve these solutions by first using the calculator to solve for inverse sine of √3, 
                                                                                                       2
which will give you 60° and you can subtract it from 180 to get 120° since sine is only positive in the first two quadrants or 180°
Example #2:

tan θ -­ 2 cos θ tan θ = 0

1) Factor to solve for both tanθ and cosθ. Factor out tanθ since it is 

common. 

                                  tan θ -­ 2 cos θ tan θ = 0
                                  
                                   tanθ (1- 2 cosθ) = 0

2) Set each equal to 0 and solve.


                    tanθ = 0               and              1-2 cosθ = 0

                                                                     -1           -1


                                                                    -2 cosθ = -1

                                                                     -2            -2
                                                                      
                                                                      cosθ = 1/2




3) If you look at the unit circle, you will see that tangent (sin/cos) 

equals to 0 at 0° and 180° or 0 and π in radians. The cosine(x-

coordinate) are 60° and 300° or π/3 and 5π/3. These are the 

solutions. 

How do we graph the other trig functions?

Aim: How do we graph the other trig functions ?

To graph secant and cosecant, we fist graph sine and cosine to make our lives easier. After graphing cosine or sine, we can use the characteristics of secant and secant graphs to graph them.

1) Graph y=sinx
(The graph starts out at zero, goes up to positive 1 and than comes back to zero, later it goes down to negative 1 and again comes back to zero and continues in this pattern).  

2) Since cosecant is reciprocal of sine, every time sin of x equals to zero, you have an asymptote. Draw the asymtotes. 

y=cscx
 = 1/sinx
So, if 1/0, it is undefined. (Cannot have zero in the denominator).


- The asymptote occurs at pi and repeats every pi units. (That is where sine is zero).

3) The graph of csc comes on where the graph of sine does not equal to zero and stays in between the asymptotes. 

 -Remember it does not touch the asymptotes. It does not have an amplitude because the graph continues vertically and has a period of 2pi. 

The images keep on rotating for some weird reason. 

The same steps apply for the graph of secant, except you start out with graphing the cosine of x. (The graph of cosine starts from one)

Why the name Pythagorean Identity is appropriate?

Why the name Pythagorean Identity is appropriate? 

The Pythagorean Identities are trigonometric functions that are made using the Pythagorean Theorem.   The name Pythagorean Theorem is appropriate because the function is based on the Pythagorean Theorem.

The Pythagorean Theorem is the equation a^2+b^2=c^2, where a and b are the legs of a right triangle and c is the hypotenuse. 

Similarly, in a unit circle, there are right triangles. The radius of a unit circle is always 1 and it is the hypotenuse of the right triangle created by an angle. The legs of the triangle are the  length of the y and x axis. The length of the y axis is called sine and the length of the x axis is called cosine.

Now, the Pythagorean Theorem can be used to make an equation for the right triangle in the unit circle. 

cos^2θ+sin^2θ=1^2

or 

cos^2θ+sin^2θ=1

This equation if called Pythagorean Identity number 1.When this 

equation is solved in terms of one of the legs, it leads to 

Pythagorean Identity #2 and  #3. 

#2.cos^2θ+sin^2θ = 1        

cos^2θ  cos^2θ  cos^2θ
1+tan^2θ=sec^2θ

#3.cos^2θ+sin^2θ = 1        

sin^2θ   sin^2θ     sin^2θ
cot^2θ+1=csc^2θ

Aim: How do we solve complex fraction?

Aim: How do we solve complex fraction?

Complex Fraction: An expression in which the denominator or numerator, or they both contain fractions. 

To solve complex fraction, you should always solve the numerator and denominator expressions one at a time to make the complex fraction into one single fraction expression.  

Example: 
(1/x)-1
x-(1/x)

1)Simplify the numerator expression.
1-x
  x    

2) Simplify the denominator expression.
x^2-1
   x         

3) You have the single fraction

1-x             x^2-1
   x       ÷     x 

4) Do keep change flip.

1-x           x        
  x   *    x^2-1

5) Factor 1-x and x^2-1.
-1(x-1)   *        x        
    x            (x+1)(x-1)


6)Simplify

-1(x-1)   *        x   1         
    x           (x+1)(x-1)  1

7) Answer
     -1     

 (x+1)     



Source:
http://www.www-mathtutor.com/simplifying-complex-fractions-1.html

How do we use inverse variation?

Aim: How do we use inverse variation?

Inverse Variation is a relationship between two variables x and y, where x varies inversely with y and their product is constant. In other words, when x increases, y decreases by the same factor to 
make sure the product remains the same. 

x*y=k  ( K is constant)

We can find K by multiplying x and y.

Example:

1. If x=4 and y=5, write an equation that describing this inverse variation?
Multiply x and y
4*5=20
xy=20 or y=20/x

2. Find y using inverse variation?

x	y
3	20
6	10
12	y

First, multiply 3 and 20 to find the constant product.
20*3=60
xy=60 (The product of x and y will always be 60)
Then, substitute 12 for x.
12y=60
Now, divide 12 from both side to get the y alone. 
12y=60
12    12
The answer is y=5.

Aim: How do we multiply and divide rational expressions?
       To multiply or divide a rational expression, you have to factor out any expression that can be factored and then multiply or divide. 

Example
3x-9/x-2*x^2-2x/x-3

1) Factor 3x-9 
3(x-3)/x-2

2) Factor x^2-2x
x(x-2)/x-3

4) Divide x-3/x-3 and x-2/x-2 
*You can cross them out as they both equal to 1

5) Multiply the remaining 3 and x on the numerator 
3x

 Excluded Value:
Find any problem that would make the denominator equal to zero.
x-2=0                           x-3=0

  +2  +2                          +3  +3

x=2                               x=3
x canont be eqaul to 2 or 3

Example
(2b^2-12b/b+5)/(b-6/b+5)

1) Keep the the fraction on the left but change the division sign and fraction 
on the right
(2b^2-12b/b+5)*(b+5/b-6)

2) Factor 2b^2- 12b
2b(b-6)

3) Divide b-6/b-6 and b+5/b+5
* Cross them out because they equal to one

4) This only leaves 2b on the numerator 
2b 
Excluded value:

b+5=0            b-6=0         

  -5  -5              +6  +6

b= 5                b=6

b cannot be equal to 5 or 6.

* Note you also use the denominator you                                                                                                      get after flipping