Transcript 42 Harder Trig Equations

“Teach A Level Maths”
Vol. 1: AS Core Modules
42: Harder Trig Equations
© Christine Crisp
Harder Trig Equations
Module C2
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Harder Trig Equations
e.g.1 Solve the equation sin 2  0  5 for the interval
 180    180
Solution: Let x  2 so, sin x  0  5
We can already solve this equation BUT the interval
for x is not the same as for  .
 180    180 
 360  x  360
There will be 4 solutions ( 2 for each cycle ).
1st solution: sin x  0  5  x  30 
( Once we have 2 adjacent solutions we can add or
subtract 360  to get the others. )
Sketch to find the 2nd solution:
Harder Trig Equations
sin x  0  5 for  360  x  360
y  0 5
0
30

180
360 
150
So,
x  2  30 , 150
y  sin x
For  360  x  360, the other solutions are
150  360  210 and 30  360  330
So,
x  2   330 ,  210 , 30 , 150
   165 ,  105 , 15 , 75

N.B. We must get all the solutions for x before we find
 . Alternate solutions for  are NOT 360  apart.
Harder Trig Equations
We can use the same method for any function of
e.g. (a) tan 4  c for
Use x  4
0    180
and 0  x  720
e.g. (b) cos   c for  360    360
2




180

x

180
Use x 
and
2
e.g. (c) sin(  30 )  c for
0     360
Use x    30 and  30  x  330
 .
Harder Trig Equations
SUMMARY
Solving Harder Trig Equations
 Replace the function of
 by x.
 Write down the interval for solutions for x.
 Find all the solutions for x in the required
interval.
 Convert the answers to values of
 .
Harder Trig Equations
Exercise
1. Solve the equation cos 2  0  5 for 0    360
Solution: Let x  2
 cos x  0  5
0    360 
0  x  720
Principal value: cos x  0  5  x  60
1
y  0 5
0
-1
60

180
300

360 
y  cos x
So, x  2  60 , 300 , 60  360 , 300  360
 x  2  60 , 300 , 420 , 660
   30 , 150 , 210 , 330
Harder Trig Equations
We sometimes need to give answers in radians. If
so, we may be asked for exact fractions of .
e.g.
cos x 
1
Principal value is


4
2
Tip: If you don’t remember the fractions of  ,
use your calculator in degrees and then convert to
radians using
 radians  180
So, cos x 
1
2
 x  45 

 x
4
from the calculator
rads.
If an exact value is not required, then switch the
calculator to radian mode and get x  0 785c (3 d.p.)
Harder Trig Equations
e.g. 2 Solve the equation tan 3  1 giving exact
answers in the interval 0     .
The use of

always indicates radians.
Solution: Let x  3
0   
 0  x  3

 
st
( or x  45 
)
tan x  1  1 solution is x 
4
4
 to find
 usually
 keep adding

For “tan” equations we
180
So, x  3  ,   , 2 
more solutions, but working
in
4
4 radians4 we must
remember to add
.
 5 9
 5 3 9
 x  3  ,
,
,
  ,
4 4
4
12 12 4 12
Harder Trig Equations

1


e.g. 3 Solve the equation cos    
for the
4

2
interval 0    2 .
1

Solution: Let x   
 cos x 
4
2


0    2 
 x  2 
4
4

9
 x
4
4
1


rads.
Principal value: cos x 
 x  45 
4
2
Sketch for a 2nd value:
Harder Trig Equations
1 for 
9
cos x 
 x
4
4
2
y
1
0
2nd value:
 07

x  2 
2


4
-1
1
 

 x   
4 


7
y  cos x
4
2
7
 x
4
cos x repeats every 2 , so we add 2 to the
principal value to find the 3rd solution:

9
  7 9
 x   2 
 x    ,
,
4
4
4 4
4
4
36 2 8
3
   0,
,
Ans:   0,
, 2
2
2 4 14
4
Harder Trig Equations
 x
e.g. 4 Solve the equation sin    0  4 for 0  x  4
2
giving the answers correct to 2 decimal places.
x
Solution: We can’t let x 
so we use a capital X
2
( or any another letter ).
x
Let X 
so sin X   0  4
2
2 4
0  x  4  0  X 
21
We need to use radians but don’t need exact
answers, so we switch the calculator to radian mode.
Principal value: X  ( 0  41c )
Sketch for the 1st solution that is in the interval:
Harder Trig Equations
sin X   0  4 for 0  X  2
y
x

X  
2

1
0 412


3 553
5 872
X
2
y   0 4
-1
y  sin X
x
1 solution is X     0  412c  3  553c
2
x
nd
2 solution is
X   2  0  412c  5  872c
2
Multiply by 2: Ans:
x  7  11c , 11  74c ( 2 d.p.)
st
Harder Trig Equations
Exercise
1. Solve the equation tan 2  1 for 0    2
giving the answers as exact fractions of  .
2. Solve the equation sin (  60 )  0  25 for
 180    180 giving answers correct to 1
decimal place.
Harder Trig Equations
Solutions
1. Solve the equation tan 2  1 for
Solution: Let x  2
 tan x  1
0    2 
Principal value:
0  x  4
tan x  1

0    2
5
Add  : x  2  ,
,
4 4
 5
  ,
,

8 8

x

9 13
,
4
4
9 13
,
8
8
4
Harder Trig Equations
Solutions
2. Solve the equation sin (  60 )  0  25 for
 180    180 giving answers correct to 1
decimal place.
Solution: Let x    60
 sin x  0  25
 180    180   240  x  120
Principal value: sin x  0  25  x  14 5 
Sketch for the 2nd solution:
Harder Trig Equations
sin x  0  25 for  240  x  120
y
1
y  0 25
180
14 5 
-1
360 
x
( 165  5  )
y  sin x
x    60  14  5  , ( 180  14  5   165  5  )
The 2nd value is too large, so we subtract 360 




x



60

165

5

360


194

5

Add 60  : Ans:   134  5  , 74  5 
Harder Trig Equations
Harder Trig Equations
The following slides contain repeats of
information on earlier slides, shown without
colour, so that they can be printed and
photocopied.
For most purposes the slides can be printed
as “Handouts” with up to 6 slides per sheet.
Harder Trig Equations
SUMMARY
Solving Harder Trig Equations
 Replace the function of
 by x.
 Write down the interval for solutions for x.
 Find all the solutions for x in the required
interval.
 Convert the answers to values of
 .
Harder Trig Equations
e.g. 1 Solve the equation sin 2  0  5 for the interval
 180    180
Solution: Let x  2 so, sin x  0  5
We can already solve this equation BUT the interval
for x is not the same as for  .
 180    180 
 360  x  360
There will be 4 solutions ( 2 for each cycle ).
1st solution: sin x  0  5  x  30 
( Once we have 2 adjacent solutions we can add or
subtract 360  to get the others. )
Sketch to find the 2nd solution:
Harder Trig Equations
sin x  0  5 for  360  x  360
y  0 5
30

So,
x  2  30 , 150

150
y  sin x
For  360  x  360, the other solutions are
150  360  210 and 30  360  330
So,
x  2   330 ,  210 , 30 , 150
   165 ,  105 , 15 , 75

N.B. We must get all the solutions for x before we find
 . Alternate solutions for  are NOT 360  apart.
Harder Trig Equations
We can use the same method for any function of
e.g. (a) tan 4  c for
Use x  4
0    180
and 0  x  720
e.g. (b) cos   c for  360    360
2



Use x 
and  180  x  180
2
e.g. (c) sin(  30 )  c for
0     360
Use x    30 and  30  x  330
 .
Harder Trig Equations
e.g. 2 Solve the equation tan 3  1 giving exact
answers in the interval 0     .
The use of

always indicates radians.
Solution: Let x  3
0   
 0  x  3
 ( or
 
st
)
x  45 
tan x  1  1 solution is x 
4
4
For “tan” equations we usually keep adding 180 to find
more solutions, but working in radians we must
remember to add  .
 5 9
 5 3 9
x  3  ,
,
,
  ,
4 4
4
12 12 4 12
Harder Trig Equations

1


e.g. 3 Solve the equation cos    
for the
4

2
interval 0    2 .
1

Solution: Let x   
 cos x 
4
2


0    2 
 x  2 
4
4

9
 x
4
4
1


rads.
Principal value: cos x 
 x  45 
4
2
Sketch for a 2nd solution:
Harder Trig Equations
1 for 
9
cos x 
 x
4
4
2
y
1
 

 x   
4 

 07
2
So,
7
4
4
y  cos x
7
 x
value:
4
4
x repeats every 2 , so we add 2 to the 1st value:
2nd
cos


x  2 

9
 x   2 
4
4
36 2 8
   0,
,
24
4
  7 9
x    ,
,
4 4
4
4
Ans:   0,
3
2
, 2
Harder Trig Equations
 x
e.g. 4 Solve the equation sin    0  4 for 0  x  4
2
giving the answers correct to 2 decimal places.
x
Solution: We can’t let x 
so we use a capital X
2
( or any another letter ).
x
Let X 
so sin X   0  4
2
2 4
0  x  4  0  X 
21
We need to use radians but don’t need exact
answers, so we switch the calculator to radian mode.
Principal value: X  ( 0  41c )
Sketch for 1st solution that is in the interval:
Harder Trig Equations
sin X   0  4 for 0  X  2
y  sin X
0 412
3 553
5 872
X
y   0 4
x
1 solution is X     0  412c  3  553c
2
x
nd
2 solution is
X   2  0  412c  5  872c
2
Multiply by 2: Ans:
x  7  11c , 11  74c ( 2 d.p.)
st