Transcript Powerpoint Presentation

“Teach A Level Maths”
Vol. 1: AS Core Modules
42: Harder Trig Equations
Harder Trig Equations
e.g.1 Solve the equation sin 2  0  5 for the interval
0    360
Solution: Let x  2 so, sin x  0  5
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
y  0 5
0
30

180
360 
150
So,
x  2  30 , 150
y  sin x
The other solutions are
360  30  390 and 360  150  510
x  2  30 , 150 ,390,510
So,
  15 , 75 , 195, 255

N.B. We must get all the solutions for x before we find
 . Alternate solutions for  are NOT 360  apart.
Harder Trig Equations
SUMMARY
Solving Harder Trig Equations
 Replace the function of
 by x.
 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
0    360
 cos x  0  5
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
e.g. 3 Solve the equation
cos   45  
1
2
interval 0    360 .
Solution: Let
 cos x 
x    45
2
0    360
Principal value: cos x 
1
2
Sketch for a 2nd value:
1

x  45
for the
Harder Trig Equations
1 for
45  x  405
cos x 
2
y
1
2nd value:
 07
2
0
-1
1
x    45 
45
180
x  360  45
315 360
 x  315
y  cos x
cos x repeats every 360, so we add 360 to the
principal value to find the 3rd solution:
 x  45  360  405
   45  45, 315, 405
   0, 270, 360
Harder Trig Equations
 x
e.g. 4 Solve the equation sin    0  4 for 0  x  720
2
giving the answers correct to 2 decimal places.
x
Solution: We can’t let x 
so we use a capital A
2
( or any another letter ).
Let
x
A 
so sin A   0  4
2
0  x  720
Principal value: A  sin 1  0.4  23.60
Sketch for the 1st solution that is in the interval:
Harder Trig Equations
sin A   0  4 for 0  A  360
y
x

A  
2

1
23.6
203.6
336.4
180
180
X
360
y   0 4
-1
y  sin X
x
A 
 180  23 .6  203.6
2
x
2nd solution is A 
 360  23 .6  336.4
2
Multiply by 2: Ans: x  7  11c , 11  74c ( 2 d.p.)
1st solution is
Harder Trig Equations
Exercise
1. Solve the equation sin (  60 )  0  25 for
 180    180 giving answers correct to 1
decimal place.
Harder Trig Equations
Solutions
2. Solve the equation sin (  60 )  0  25 for
 180    180 giving answers correct to 1
decimal place.
 sin x  0  25
Solution: Let x    60
 180    180


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
2sin(2x + 45°) = 1
Solution: Let y  2 x  45
0<x<360

2 sin y  1
0  x  360
2 sin y  1 
sin y  0.5
Principal value: sin y  0  5  x  30
Harder Trig Equations
sin x  0  5
y
y 0  5
1
180
30
-1
360 
x
150
y  sin x
y  2 x  45  30 , (180  30  150 )
Add 360 to find further values : 390° , 510° , 750°
 2 x  45  150 ,390 ,510 ,750
2x = 105°,345°,465°,705°
(subtract 45°)
x = 52.5°,172.5°,232.5°,352.5°
(divide by 2)
Harder Trig Equations
Harder Trig Equations
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information on earlier slides, shown without
colour, so that they can be printed and
photocopied.
For most purposes the slides can be printed
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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