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Examples
1) Solve for is 2 cos = 0,646cos = 0,323
Now: = cos1(0,323) + k360 = 108,8 + k360,k
= 108,8 + k360, k
2) Solve for if cos 3= 0,632. Get the angle 3= cos1( 0,632) + k360: k
3= 129,2 + k360.
We dont want 3we want , so = 43,06 + k120k. Notice here that the
period of cos 3is no longer 360, but 360_3
= 120
3) Solve for if cos( 40) = tan 22.
cos( 40) = [0,4040262258]
Leave that answer in the calculator and press shift cos to get the angle.
40= 66,2 + k360= 4066,2+ k360
= (106,2 or 26,2)+ k360, k
4) Solve for if cos ( 50) = cos 2. You have the angle. Spread out
cos ( 50) = cos 2
So: 50= 2+ k360
= 2= 50+ k360
Thus 3= 50+ k360 or = 50+ k360
= 16,7+ k120 or = 50+ k360, k
5) Solve for
if cos
2
2
=
1
_4
cos 2= 1_2
2= cos1(1_2)+ k360 or 2= cos1( 1_2)+ k360
2= 60+ k360 or 2= 120+ k360
= 30+ k180 or = 60+ k180, k
6) cos 3= cos
We need to remove the negative by reading from the left to the right.
cos 3= cos and is thus negative.
So cos 3
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+ = 450+ k90 (1) = 90+ k120 (2), k
Since [ 180; 90]
For 1: = 45; 135
From 2: = 90; 30; 150
Co-ratio equations
7. Solve for if cos ( 10) = sin 2 .
If this was cos ( 10) = cos 2 as in a similar example (4) above, then the
ratios are already balanced. So we would only focus on the angles.
Here our duty is to get those ratios the same:
Remember sine becomes cosine in vertical reduction, so cos ( 10) = cos 2
and the sin 2 has a positive sign in front of it. So we ask where is
( 10) positive?
According to the CAST rule,+
+ , it will be positive in the first and fourthquadrants.
90
2
2
So in the first quadrant sin2= cos (90 2) and in the fourth quadrant sin 2
= cos (2 90) quadrant
So: cos ( 10) = cos (90 2) or cos ( 10) = cos (2 90)
10= 90 2+ k360 or 10= 2 90+ k360
3= 100+ 360 = 80+ k360
and= (100_3 )
+ k120 = 80 + k360,k
Workbook: Lesson 16 Activity
For conclusion
This is the graph ofy= tanx(you learnt to draw it in Grade 10)
Lesson 1718
180 180
63,4
45
(45; 1)
(45; 1)
45
We will use the graph to solve tanx= 1_
2
Draw the liney= 1_2
. (Press shift tan.5, and the calculator gives us 26,6).
We need to include all the solutions. We simply add k180if kbecause the period
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of the tan graph is 180
General solution:x= 26,6+ k180 k.
Now lets solve tanx= 2_3
(Draw the line and see what the calculator gives you,
then + k180 k)
sox= 63,4+
k180
k
General solution for tan =p
= tan1(p) + k180k
k
Worked examples
1. Solve for
tan 3= 2,7
Press Shift tan (2,7) to get the angle and write down the general solution. 3= 69,7 + k180 Divide by 3
= 23,2 + k60, k
2. Solve for if [180; 180]
tan2= 0,81
tan = 0,9 Split them
tan = 0,9 or tan = 0,9 Find angles for the general solution
= 42+ k180 or = 42+ k180, k
Substitute for kto find angles between [180; 180]
{138; 42; 42; 138}
3. Solve for if tan 3= tan(+ 20)
tan 3= tan (+ 20)
tan 3
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2x= 33,7+ k180
x= 16,85+ k90
General solution for sin =p
sinis +ve in quadrants 1 and 2
So if sin =p
then:
= {sin1(p)
180 sin1(p)}+ k360; k
Example 1
Find the general solution for insin= 1_2
:
sin = 1_2
=
{sin1
(
1
_
2
)
180 sin1(1_2)}+ k360;
= {30150} + k360, k
Example 2
sin 2= 3_4
2=
{
sin1( 3_4)
180 sin1(
3_
4)}
+ k360
2= { 48,6180( 48,6)}+ k360
2= { 48,6228,6}+ k360
= { 24,3114,3}+ k180, kExample 3
Solve for if sin 2= 0,8 2= 53,13+ k360 or 2= 180 (53,13) + k360
= 26,56+ k180 or 2= 233,13+ k360
= 116,56+ k180, k
Co-ratio equations
Solve forxif sin 2x= cos 3x
In front of the cos, there is a + sign
So sin 2x>0 and according to the CAST rule,+
+ , this is in the first andsecond quadrants.
So we need to make the cosine a sine, and we can only do this through verticalreduction. That iscos3xbecomessin(90 3x)
Solution
Example
Example
Example
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So:sin2x= sin(90 3x)
2x= 90 3x + k360
2x 3x= 90 + k360 or 2x+ 3x = 90 + k360
x= 90 + k360 5x= 90 + k360
x= 90 + k360x= 18 + k72;k
Trigonometric equations involving factorisation.
a) Common factor problems are those that usually have two terms.
Example
Solve forx:3cosxsinx= 2cosx
Do not divide bycosxbecause you will lose solutions
3cosxsinx 2cosx= 0. Write in the form where one side is zero.
cosx(3sinx 2)= 0. Take out a common factor
cosx= 0 or sinx= 2_3
x= 90 + k360 orx= 41,8 + k360 orx= 138,2 + k360 k
b) Trinomials(3 terms)
a2 a 2 is a quadratic trinomial that we can factorise.
So isa2 ab b2
In trigonometry we sometimes need to create trinomials.
Example
1. Solve for if 2 sin2+ 5 cos + 1 = 0
Look at the nonsquared term.
We need a trinomial in terms of cos that is with only cos and no sine terms.
Usesin2= 1 cos2
2(1 cos2)+ 5cos+ 1= 0 Simplify
2 2cos2+ 5cos+ 1= 0
2cos2+ 5cos+ 3= 0 Change signs
2cos2 5cos 3= 0Factorise. (This is similar to2x2 5x3= 0
wherex= cos.)
(2cos+ 1)(cos 3)= 0
2cos= 1 or cos= 3
2cos= 1 or cos= 3
cos= 1_2
= cos1(1_2)+ k360 = 120 + k360
This is a quadratic trinomial with middle term 3sincos. The algebraicequivalent is:a2+ 2ab + b2= 0. So we havecos2,and3sincos and we now need
Example
Example
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asin2to complete the trinomial. The only way we can bring that in is by using the1,since= sin2+ cos2.
Socos2 3sincos+ sin2+ cos2= 0
2. Solveforif[180 ;360] and
cos2 3sincos+ 1= 0
cos2 3sincos+ sin2+ cos2= 0
2cos2 3sincos+ sin2= 0
(2cos sin)(cos sin)= 0 Factors
2cos= sin or cos= sin Create tan
2= sin_cos
or sin_cos
= 1
tan= 2 or tan= 1
= 63,4 + k80 or = 45 + k180,k
{116,6;135;45;63,4;225;243,4}
Equations with four terms: Group in twos
Example
Solve forxif2sin2x+ 2sinxcosx+ sinx+ cosx= 0
2sinx(sinx+ cosx)+ (sinx+ cosx)= 0
(sinx+ cosx)(2sinx+ 1)= 0
sinx= cosx or sinx= 1_2
tanx= 1x= 45 + k180 orx= 30 + k360 orx= 210 + k360, k
Your Fact File
If sin =p and 1 p 1,
then = {sin1(p)
180 sin1(p)}+ k360 ,k
If cos =p and 1 p1,
then = cos1(p) + k360 ,k
If tan =p andp,
then = tan1(p) + k180 ,k
1 = sin + cos2
If equations have 2 terms
You look for a common factor
You may have to use co-ratios
You may have to form the tan ratio if each side has a cos and sine of the same
angle
If equations have three terms, it is usually a quadratic trinomial
If there are 4 terms, you have to group them in pairs
Example
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