Download + 2 = 0 sin( x + 30 )

Trigonometric Equations
Further worked examples.
(All angles in degrees)
1. 5sin(x + 30) + 2 = 0
sin( x + 30 )
= -0.4
Reference Angle
x + 30 = arcsine(-0.4) = -23.6 (336.4)
General solutions: Argument values lie in the 3rd and 4th quadrants
3rd Quadrant : x + 30 = 360 – 336.4
+ 360k k IAMO Z
x + 30 = 203.6
+ 360k k IAMO Z
x = 173.6
+ 360k k IAMO Z ------>
Check Sine (173.6 + 30) = Sin (203.6) = -0.4
4th Quadrant x + 30 = 360 – 23.6 + 360k k IAMO Z
x +30 = 336.4
+ 360k k IAMO Z
x = 306.4
+ 360k k IAMO Z ------>
Check Sine (306.4 + 30) = Sin (336.4) = -0.4
What we learnt:
•
•
•
•
•
•
•
The argument is the value or expression on which the function operates. In this
case the function is Sin() and the argument is (x + 30).
The angle(s) corresponding to the argument is/are the reference angle(s), -23.6
degrees in this case. 336.4 is just the positive equivalent.
Once you have your reference angle imagine its position (or, better, make a
sketch). When the angle needs to be placed in other quadrants the sketch will
help you to figure whether to add or subtract it from 180or 360 to accomplish
this.
It is the argument which the function operates on and hence it is the argument
which repeats on each cycle of the function. By setting up correctly as
argument = reference angle + repeat angle e.g
x + 30 = 180 + 23.6 + 360k k IAMO Z
the repetition angle will be correctly transferred to the solution value of x.
The independent variable is exhibited on the x-axis. In his case the
independent variable is “x” and the value(s) of x which satisfies equation in the
original question [5sin(x + 30) + 2 = 0 ] is what we seek.
Trig functions are not 1:1 and so each reference angle may be found in two
quadrants, depending on its sign and the trig function concerned (CAST or
ASTC may be used to find the appropriate quadrants). In this case the reference
angle is negative and so the argument values will fall in the 3rd and 4th quadrants.
The solution values of x need not be in these quadrants.
Solve the original equation for the argument. Set the argument equal to each of
the values for the reference angle in the appropriate quadrants. These values
repeat every 360 degrees for Sine and Cosine and every 180 degrees for Tangent.
e.g.
3rd Quadrant : x + 30 = 180 + 23.6 + 360k k IAMO Z
x + 30 = 203.6
+ 360k k IAMO Z
•
Once we have a value for the argument that solves the original equation, solve
for the independent variable.
x = 173.6
+ 360k k IAMO Z ------>
You can check your work by substituting this value in the original equation
5Sin(173.6 + 30)
+2=0
5sin(203.6)
+2=0
5 x -0.400
+2=0
√
2.
6Sin(x) = Cos(x)
Tan(x) = 0.1667
Reference angle: x = 9.46
Argument values are in the 1st and 3rd quadrants
x = 9.46
+
180k K IAMO Z ----------->
Check Tan (9.46) = 0.1667
x = 180 + 9.46 + 180k K IAMO Z ----------->
Check Tan(189.6) = 0.1667
What we Learnt
•
•
•
•
Much the same as the previous example
We see that Tangent repeats every 180 degrees
Tangent becomes undefined at x = 90 + 180k k IAMO Z
k IAMO (is a member of) Z (implies) that k is any integer, -ve, +ve or zero
3.
5Cos(2x)
5(Cos2 (x) – Sin2 (x))
5(1 – Sin2 (x) - Sin2 (x))
5(1 – 2Sin2 (x))
5 – 10 Sin2 (x)
-13Sin2 (x)
13Sin2 (x)
+Sin(x)
+ Sin(x)
+ Sin(x)
+ Sin(x)
+ Sin(x)
+ Sin(x) +3
-Sin(x) -3
Sin(x)
Sin(x)
Sin(x)
Sin(x)
Reference angles:
= 3Sin2 (x)+ 2
= 3Sin2 (x)+ 2
= 3Sin2 (x)+ 2
= 3Sin2 (x)+ 2
= 3Sin2 (x)+ 2
=0
=0
= (1 ± √(1 + 4∙13∙3) /2∙13
= (1 ± √(157)/26
=0.52
=-0.44
= Sin(31.33)
=Sin (-26.1)
x=31.33 (Solutions in Q1 and Q2) for Sine +ve or
x=-26.1 (Solutions in Q3 and Q4) for Sine -ve
Q1 and Q 2
Q3 and Q4
x=31.33
x=180-31.33
x=148.67
x=180 + 26.1
x=206.1
x=360 – 26.1
x=333.9
+360k
+360k
+360k
+360k
+360k
+360k
+360k
k IAMO Z
k IAMO Z
k IAMO Z
k IAMO Z
k IAMO Z
k IAMO Z
k IAMO Z
--->
--->
--->
--->
also 31.33-360= -328.6, 148.67-360=-211.64, 206.1-360=-153.9, 333.9-360=-26.1
What we learnt:
•
The original equation becomes
13Sin2 (x)
-Sin(x) -3
=0
and so there are three 'contributors' to the final functional value.
•
•
•
•
13Sin2 (x) is always positive and ranges from 0 to 13
•
Sin(x) alternates between -1 and +1
•
-3 is constant at -3
We used k=-1 to obtain the values from -360 to 0 degrees
We see that Sin2(x) is always positive.
We see clearly that
Sin(x) = -Sin(-x)
or
Sin(-x)= -Sin(x)
4.
Cos (x + 15)
sin(90 – x - 15)
sin (75 – x)
75 -x
=
=
=
=
sin(2x)
sin(2x)
sin(2x)
2x
=
2x (assuming 2x +ve)
Reference angle:
75 -x
The argument will occur in the 1st and 2nd quadrants
1st Quadrant
75 – x
-3x
-x
x
=
=
=
=
2x
-75
-25
25
Check:
Cos (25 + 15) =
Also 25+120 =145, 25+ 240=265
+ 360k
+360k
+120k
-120k
Sin (2x25)
k IAMO Z
k IAMO Z
k IAMO Z
k IAMO Z ---->
√
2th Quadrant
75 – x
x
x
x
=
=
=
=
180-2x
180 -75
105
105
Check: Cos(105 +15)
=
Sin(210)
+ 360k k IAMO Z
+360k k IAMO Z
+360k k IAMO Z
+360k k IAMO Z ---->
√
Graphs over page /...
What we learnt:
•
If the trig equation cannot be simplified to
Trig-function(argument)
e.g. Sin(3x + 15)
=
=
value
0.7
we may simplify to
Trig-function-1(argument-1)
=
Trig-function-1(argument-2)
Then
argument-1
=
argument-2
e.g. Sin(75 – x) = Sin(2x)
=>
75 – x
=
2x
Now (75 – x) is our argument and and 2x is the reference angle.
Effectively we are solving f(x) = Sin(75 – x) and g(x) = Sin(2x) simultaneously
5.
Cos(2x)
-Sin(x + 30)
Note:
-sin(x)
Sin(-x - 30)
Cos(90 + x +30)
Cos(120 + x)
=
=
=
-Sin(x+30)
cos(2x)
sin(-x)
Cos(2x)
Cos(2x)
Cos(2x)
=
2x
=
=
=
Reference angle:
120 + x
The argument value is assumed positive and so lies in the 1st and 4th quadrants for Cosine.
1st Quadrant:
120 + x
-x
x
x
=
=
=
=
2x
-120
120
120
Cos(2 x 120)
Cos(240)
=
=
-Sin(120 + 30)
-sin(150)
=
=
=
360 – 2x
240
80
+ 360k
+ 360k
-360k
-360k
k
k
k
k
IAMO Z
IAMO Z
IAMO Z
IAMO Z --->
Check:
4th Quadrant:
120 + x
3x
x
Also: 80 + 120 = 200, 80 + 240 =320
Check:
Cos(2 x 80)
Image over page /...
=
-sin (80 + 30)
+ 360k k IAMO Z
+ 360k k IAMO Z
+ 120k k IAMO Z --->
What we learned:
•
•
•
This example is similar to the previous one.
Since the reference is nominally positive (2x) and since it came from cosine we assume that
the argument values will fall in the first and fourth quadrants and go through the appropriate
motions even though the values don’t turn out as expected. With no initial value for the
reference angle we really have no idea where it is until the algebra has been simplified.
Since we are solving two different functions simultaneously the solutions can be anywhere
the graphs intersect.
Since k IAMO Z -120k becomes +120 by choosing k = -1
6.
Sin(x)Sin(20) + Cos(x)Cos(20)
Cos(x)Cos(20) + Sin(x)Sin(20)
Cos(x – 20)
Sin(90 -x +20)
Sin(110 – x)
=
=
Sin(50)
Sin(50)
=
Sin(50)
=
Sin(50)
=
Sin(50)
=
50
Reference angle:
110 – x
Reference angles should be in the first and second quadrants
First Quadrant
110 – x
-x
x
=
Check:
Sin(60)Sin(20) + Cos(60)Cos(20)
0.296 + 0.470
=
0.776
Second Quadrant
110 – x
110 – x
-x
x
x
50 + 360k k IAMO Z
=
-60 + 360k k IAMO Z
=
+60 – 360k k IAMO Z --->
=
=
=
=
Sin(50)
0.766
180 – 50 + 360k k IAMO Z
130 + 360k k IAMO Z
=
20 + 360k k IAMO Z
=
-20 - 360k k IAMO Z
=
340 - 360k k IAMO Z --->
Check:
Sin(340)Sin(20) + Cos(340)Cos(20)
-0.117 + 0.883
=
0.766
=
=
Sin(50)
0.776
What we learned:
•
When the RHS has a numeric value the equation is easier to solve
7.
√(1 + Sin(x) )
1 + Sin(x)
Sin2(x) + Sin(x)
Sin(x)(Sin(x) + 1)
=
=
=
=
Cos(x)
1 – Sin2(x)
0
0
Solution set 1:
--------------------------------Sin(x)
=
0
Reference angle:
x
x
=
=
0
0 + 360k k IAMO Z
=
1
=
Cos(0)
=
=
0
-1
=
Sin(270)
=
=
270
270 + 360k k IAMO Z --->
=
0
= Cos2(x)
Two solution sets:
Check:
√(1 + Sin(0) )
--------------------------------Solution set 2:
--------------------------------Sin(x + 1)
Sin(x)
Reference angle:
x
x
check √(1 + Sin(270) )
---------------------------------
=
=
What we learned:
• The real argument to sqrt() is always positive
• Sin(x) is zero twice on (0 ; 360) but only x=0 solves the equation
•
Sin(0)
Cos(270)
8.
√(2)/Cos(x) + 2
√(2)/Cos(x)
√(2)
Cos(x)
Cos(x)
x
Reference angle:
x
=
=
=
=
=
=
0
-2
-2Cos(x)
√(2)/-2
-1/√(2)
135
=
135
=
Cos(135)
Since Cos(x) is negative, the reference angle lies in the 2nd and 3rd quadrants
2nd Quadrant:
x
=
135
+ 360k k IAMO Z --->
Check:
√(2)/Cos(135) + 2
=
-2 + 2
=
0
3rd Quadrant
x
x
Check:
√(2)/Cos(225) + 2
=
=
360 - 135
225
=
0
+ 360k k IAMO Z
+ 360k k IAMO Z --->
What we learnt:
• for the 3rd quadrant angle, 180 + 135 does not work. 360 – 135 does. Why? The second
quadrant value 135 is 45 degrees 'behind 180' and we just need to 'flip' this angle into the 3rd
quadrant. Hence 360 -135. Always check your answers.