5.4 Solving Trigonometric Equations

5.4 Solving Trigonometric Equations
[to solve trig equations, we must find x values which satisfy the equation and are within the stated domain - usually 2pi]

So really sorry everyone, but this blog has a very small width, so I wrote my blog on Word. Click images to enlarge them!

Okay, the next example is REALLY important to pay attention to (helpful on # 9,10,12 in hmwk).

This final example was taught in class by Ms. Burchat and I feel it’s really important to go through for future questions (even though it isn’t needed in Part 1 of 5.4 homework).

5.4 Solving Trigonometric Equations

Much like solving any other equation, the goal of solving a trigonometric equation means finding the values of x that satisfy the equation. If there is a stated domain, the goal is to find the values of x within that domain. If not, however, you must find all of the values of x that could possibly satisfy the equation.

For example:

Solve the equation 4 cos x - 1 = 0 within the domain 0 < x < 2.

First, isolate x, as usual.



The ratio is positive, so according to the CAST rule, the angle will be in the first and fourth quadrants.

According to the calculator, to the nearest hundredth, cos^(-1) (1/4) = 1.32 radians. This is the angle from the x-axis.



Because one solution is in the first quadrant, one solution is 1.32. However, the other solution is in the fourth quadrant, and so 1.32 radians from the x-axis. The x-axis represents 2π. Therefore, 2π - 1.32 can be used to find the other solution, 4.96.

The two solutions are x = 1.32 and x = 4.96.


A similar process can be used if the ratio was negative, such as in 4 cos x + 1 = 0 within the domain 0 < x < 2. First isolate x:

According to CAST, the solutions must be in the second and third quadrants because the ratio is negative. Recall that cos^(-1) (1/4) = 1.32 radians, and this represents the distance from the x-axis.



In the second and third quadrants, the x-axis represents π. In the second quadrant, the angle is 1.32 fewer than π, so to solve for x, subtract 1.32 from π. π - 1.32 = 1.82.

In the third quadrant, the angle is 1.32 greater than π. To solve for x, add 1.32 to π. π + 1.32 = 4.45.

The two solutions are x = 1.82 and x = 4.45.


Horizontal Compressions

It won't always be the case that there are only two solutions within the domain. There might, for example, be a horizontal compression, like in 4 cos (2x) - 1 = 0 (domain: 0 < x < 2) Isolate the variable as much as possible.

There is a significant difference between 2.48 and 2π, so it is likely that there are more solutions.

The period of a cosine graph is usually 2π, but since it was horizontally compressed by a factor of 1/2, the new period is just π. This means that π can simply be added to each solution, since the cosine graph will have repeated and be at the same value.

π + 0.66 = 3.80
π + 2.48 = 5.62

The solutions are therefore x = 0.66, x = 2.48, x = 3.80, and x = 5.62.


No Stated Domain

4 cos (2x) - 1 = 0 has been stated above to have a period of π, and π was added to 0.66 and 2.48 to create two new solutions. If there is no domain, π can be added to (or subtracted from) 0.66 or 2.48 ad infinitum.

Trigonometric Identity Tips

So when I was going through the trig. identities homework, I just jotted down some notes that weren't found on the pink sheet :P
I tend to overthink the problems so these were just some reminders as to what i could do next if i got stuck.. It was helpful for me & hopefully for you too ? :)
[Click image to make it larger]

4.5 Trig Identities Part 2

Today, we were given a work period to practice proving trigonometric identites. This gave us a chance to work out the more difficult problems with our classmates. One question was especially difficult, and required several people to solve it! It is as follows:

sin2x + sin2y = 2sin(x+y)cos(x-y)

L.S. :
sin2x + sin2y
= 2sinxcosx + 2sinycosy

R.S. :
2sin(x+y)cos(x-y)
= 2(sinxcosy + cosxsiny)(cosxcosy + sinxsiny)
= 2(sinxcosxcos^2y + sin^2xsinycosy + cos^2xcosysiny + sin^2ysinxcosx)

It appears that this is where many of the student in our class got stuck.

= 2(sinxcosxcos^2y + sin^2ysinxcosx + sin^2xsinycosy + cos^2xcosysiny)
= 2[sinxcosx(cos^2y + sin^2y) + sinycosy(sin^2x + cos^2x)]

Similar terms were collected and 'sinxcosx' and 'sinycosy' were factored out.

= 2[sinxcosx(1) + sinycosy(1)]
= 2sinxcosx + 2sinycosy

Since L.S. = R.S., sin2x + sin2y = 2sin(x+y)cos(x-y).

Remember: when proving trigonometric identities, you must keep an open mind. Don't be afraid to try new things!

Period 1: 4.5 Prove Trigonometric Identities, Part 1

Hello 8)

So... What are trigonometric identities?
Trig. identities are equations that are true for all angles.
A counter example is proving that a trigonometric identity is not true.

Today, we learned about the Double Angle Formula; Used to express double an angle.
So how are these equations formed ?

Recall ;
sin (a+b) = sin a cos b + cos a sin b
|> Set a equal to b. This is similar to saying "double a"
Let x represent our new variables to represent a new equation

sin (x+x) = sin x cos x + cos x sin x

This can be rewritten as:

sin(2x) = 2(sin x cos x)

Recall;
cos (a+b) = cos a cos b + sin a sin b
|> Set a equal to b. This is similar to saying "double a"

cos (x+x) = cos x cos x + sin x sin x

This can rewritten as:

cos(2x) = cos² x - sin² x

Using Pythagorean Identity, this can also be rewritten...
[sin² x = 1 - cos² x ] [cos² x = 1-sin² x]

cos(2x) = cos² x - (1-cos² x)
cos(2x) =2cos² x - 1

-OR-

cos(2x) = (1-sin² x) - sin² x
cos (2x) = 1 - 2sin² x


Example 1: Rewrite using Double Angle Formula
a.
cos (6x)
=cos [2(3x)]
=cos² x (3x) - sin² (3x)
-OR -
=2cos² x (3x) - 1
-OR-
=1-2sin² x (3x)

b.
sin (0.5x) [¼ + ¼ = ½ ]
= sin [2( ¼ x)]
= 2 sin ( ¼ x) cos ( ¼ x)

Example 2: Express as a single sine/cosine function
6 sin θ cos θ [Rewrite as a multiple of 2]

6 sin θ cos θ = 3 · [2 sinθ cos θ ] > [Sine Double Angle Formula! :o]

6 sin θ cos θ = 3 sin 2 θ


Don't forget how to do a Left Side - Right Side check !

Prove secx = tanx cscx

LEFT SIDE: secx	RIGHT SIDE: tanx cscx
=1 / cosx	= sinx /cosx x 1/sinx = 1/cosx