Download 8.1 Simple Trig Equations

8.1 Simple Trig Equations
There are often multiple (infinite) solutions to
trigonometric equations. For example take the
equation sin(x)=.5. Find the solutions.
Understand that this graph repeats itself every interval of 2pi, so there will also be
𝝅
𝝅
𝝅
solutions at + 𝟐𝝅 and then at + 𝟒𝝅 and then at + 𝟔𝝅.
𝟔
𝟔
So we would re-write these infinite answers as
𝝅
𝟔
𝟔
+ 𝟐𝒌𝝅 where k is an integer.
• With the sine and cosine, because the period is 2pi,
or the distance that it takes to repeat itself is 2pi,
every time you find a solution if you add a multiple
of 2pi to that solution you will find another
solution.
1
± ,
2
• When the equation involves sin(x)=
±
±1, 0 then you do not need a calculator.
2
,
2
±
3
,
2
• However if the sin(x)=.348 then you need to use
the sin-1 on your calculator to locate the solutions.
• The calculator gives back .355437. This is the
reference angle. Knowing that sin(x) is positive an
angle in the second quadrant also satisfies the
equation. So take π − .355437 and this gives you
both the 1st and 2nd quadrant angles. Now add 2kπ
to both of those to get the entire solution set.
4 cos   12  8
• Solve this equation just like you normally would,
trying to isolate the variable, understand that the
variable is stuck to cosine so you are actually going
to isolate cos ϴ.
4 cos   4
cos   1
Now ask yourself, where is cos ϴ equal to -1.
Its at π, and then at 3π, and then 5π
So solution is π+2πk
3cos   6  4
3cos   2
2
cos  
3
This is a little trickier, we know that the cosine is negative so it exists in the 2nd and 3rd
quadrants. But first we must find a reference angle that exists in the 1st quadrant, so
neglect the negative and find cos-1(2/3).
That will yield .841069. This is the 1st quadrant reference angle. Now take that
reference angle and place it in the 2nd and 3rd quadrants.
π
.841069 radians
or 48.1897⁰
Now looking for these 2 angles
In order to do this, take
.841069 and add it to pi to
find the yellow angle, and
subtract it from pi to find
the red one.
The same idea can be thought of using
the graph of cos(ϴ)
2
3
2

3
48.2⁰
Slopes of lines with the use of an angle of inclination.
An angle of inclination is the angle
formed by a line and the horizontal
(or in this case the x axis).
If we look at slope as
rise/run we end up with
the idea of a right triangle.
Rise is opposite alpha.
Run is adjacent to alpha.
So if we talk about the angle alpha we recognize the
relationship of tangent.
α
run
rise
Thus tan(α)=rise/run
Thus tan(α) = m
And α = tan-1(m)
m = tan(α) and α = tan-1(m)
• Line l passes through the point (-1,3) and
makes an angle of 70⁰ with the x-axis, find its
slope to the nearest hundredths.
• Then find the equation for line l.
• Line l has an equation of 9x+4y=108. First
where does the line cross the x-axis? What is
the angle of inclination for the line?
•
•
•
•
Consider the two lines.
l1  5x+3y=30
l2  5x-2y=-10
Find the measure of the acute angle alpha that they form at their
intersection.
α
Homework pg. 299 1-18 evens 19-24,
27, 28