Chapter 3
Exercise 3A
Chapter 3
1. The calculator display given shows the 1st quadrant solution. There will also be a 2nd quadrant solution at x = 180 - 14.47751219 giving two solutions: x 14.5 and x 165.5
2.
sin
x
=
?
1 2
which
has
solutions
in
all
four
quad-
rants:
x=
6
,
x
=
-
+
6
=
-
5 6
,
x=
-
6
,
x
=
-
6
=
5 6
3. 2x =
or
6
x=
12
or 2x = 2 +
6 13 =
6 13 x= 12
2x = -
6 5 = 6 5 x= 12
or 2x = 3 -
6 17 =
6 17 x= 12
4. sin2 x + cos2 x = 1 so this question simplifies to
sin x
=
1
with
solution
x
=
2
.
5. Use the null factor law:
2 sin x - 1 = 0
or
2 sin x = 1
1 sin x =
2 x= 6
or x = -
6 5 = 6
cos x = 0
x=
2 3 or x = 2
6. First factorise:
7. Use the null factor law:
2 cos x + 1 = 0 2 cos x = -1 1 cos x = - 2
5 sin x - 1 = 0 5 sin x = 1 1 sin x = 5
The first factor will have solutions in the second and third quadrant; the second will have solutions in the first and second quadrants.
x = 180 - 60 = 120
or x = 180 + 60 = 240
x = 11.5
or x = 180 - 11.5 = 168.5
8. First use the pythagorean identity to replace cos2 x with sin2 x:
sin x + 2 cos2 x = 2
sin x + 2 (1 - sin2 x) = 2
sin x + 2 - 2 sin2 x = 2
sin x - 2 sin2 x = 0
sin x 1 - 2 sin x = 0
sin x + 2 sin2 x = 0 sin x(1 + 2 sin x) = 0
Now use the null factor law:
Now use the null factor law:
sin x = 0
or
x=0
or x = 180
or x = 360
1 + 2 sin x = 0
2 sin x = -1 1
sin x = - 2
x = 180 + 30 = 210
or x = 360 - 30 = 330
1
sin x = 0 or x=0
or x = - or x =
1 - 2 sin x = 0
2 sin x = 1
1 sin x =
2 x= 4
or x = -
4 3 = 4
Exercise 3A
9.
8 sin2 x + 4 cos2 x = 7
4 sin2 x + (4 sin2 x + 4 cos2 x) = 7
4 sin2 x + 4(sin2 x + cos2 x) = 7
4 sin2 x + 4 = 7
4 sin2 x = 3
sin2 x = 3 4 3
sin x = ? 2
x=
3
or x = - 3
2 =
3
or x = + 3
4 =
3
or x = 2 - 3
5 =
3
10. Rearrange and factorise:
tan2 x + tan x = 2 tan2 x + tan x - 2 = 0 (tan x + 2)(tan x - 1) = 0
Null factor law:
Solutions to A.J. Sadler's
Null factor law:
sin x = 0 x=0
or x = or x = 2
3 + 2 sin x = 0 2 sin x = - 3 3 sin x = - 2 x=+ 3 4 = 3 or x = 2 - 3 5 = 3
12.
5 - 4 cos x = 4 sin2 x
5 - 4 cos x = 4(1 - cos2 x)
5 - 4 cos x = 4 - 4 cos2 x
1 - 4 cos x = -4 cos2 x
4 cos2 x - 4 cos x + 1 = 0
(2 cos x - 1)2 = 0
2 cos x - 1 = 0
2 cos x = 1
1 cos x =
2 x = ?60
13.
3 = 2 cos2 x + 3 sin x
3 = 2(1 - sin2 x) + 3 sin x
3 = 2 - 2 sin2 x + 3 sin x
2 sin2 x - 3 sin x + 1 = 0
(sin x - 1)(2 sin x - 1) = 0
tan x + 2 = 0 tan x = -2
tan x - 1 = 0 tan x = 1
The first factor has solutions in the 2nd and 4th quadrants. The second factor has solutions in the 1st and 3rd quadrants.
x = 180 - 63.4 = 116.6
or x = -63.4
x = 45
x = -180 + 45 = -135
11.
3 sin x - 2 cos2 x + 2 = 0
3 sin x - 2(1 - sin2 x) + 2 = 0
2
3 sin x - 2 + 2 sin2 x + 2 = 0
3 sin x + 2 sin2 x = 0
sin x - 1 = 0
sin x = 1
x= 2
or x = 2 + 2
5 =
2
2 sin x - 1 = 0
2 sin x = 1
1 sin x =
2 x= 6
or x = -
6 5 = 6
or x = 2 +
6 13 =
6
or x = 3 - 6
17 =
6
Unit 3B Specialist Mathematics
14. (sin x)(2 + sin x) + cos2 x = 0 2 sin x + sin2 x + cos2 x = 0 2 sin x + 1 = 0 2 sin x = -1 1 sin x = - 2 x=+ 6 7 = 6 or x = 2 - 6 11 = 6
15. 2 cos x - 3 cos x sin x = 0
cos x(2 - 3 sin x = 0
cos x = 0
x=-
2 or x = 2
2 - 3 sin x = 0
3 sin x = 2
2 sin x =
3 (no real solution)
16.
(sin x)(1 - sin x) = - cos2 x
(sin x)(1 - sin x) = -(1 - sin2 x)
(sin x)(1 - sin x) = -1 + sin2 x
sin x - sin2 x = -1 + sin2 x
2 sin2 x - sin x - 1 = 0
(2 sin x + 1)(sin x - 1) = 0
Exercise 3A
17. sin x tan x = 2 - cos x
sin x sin x = 2 - cos x
cos x sin2 x
= 2 - cos x cos x sin2 x = cos x(2 - cos x)
sin2 x = 2 cos x - cos2 x
1 - cos2 x = 2 cos x - cos2 x
1 = 2 cos x
1 cos x =
2 x= 3
or x = -
3
or x = 2 - 3
5 = or x
3 -5 =
3
= -2 +
3
18. L.H.S.:
2 cos2 + 3 = 2(1 - sin2 ) + 3 = 2 - 2 sin2 + 3 = 5 - 2 sin2 = R.H.S.
19. L.H.S.:
sin - cos2 = sin - (1 - sin2 ) = sin - 1 + sin2 = sin + sin2 - 1 = (sin )(1 + sin ) - 1 = R.H.S.
20. L.H.S.:
(sin + cos )2 = sin2 + 2 sin cos + cos2 = 2 sin cos + sin2 + cos2 = 2 sin cos + 1 = R.H.S.
2 sin x + 1 = 0
2 sin x = -1 1
sin x = - 2
x=+ 6
7 =
6
sin x - 1 = 0 sin x = 1 x= 2
21. R.H.S.: (sin - cos )2 = sin2 - 2 sin cos + cos2 = sin2 + cos2 - 2 sin cos = 1 - 2 sin cos = L.H.S.
3
Exercise 3A
Solutions to A.J. Sadler's
22. L.H.S.: sin4 - cos4 = (sin2 + cos2 )(sin2 - cos2 )
(difference of perfect squares)
= 1(sin2 - cos2 ) = (1 - cos2 ) - cos2 = 1 - 2 cos2 = R.H.S.
23. L.H.S.:
sin4 - sin2 = sin2 (sin2 - 1) = (1 - cos2 )(- cos2 ) = - cos2 + cos4 = cos4 - cos2 = R.H.S.
24. L.H.S.:
sin2 tan2 = (1 - cos2 ) tan2
= tan2 - cos2 tan2
=
tan2
-
cos2
sin2 cos2
= tan2 - sin2
= R.H.S.
25. L.H.S.:
(1 + sin )(1 - sin ) = 1 - sin2 = cos2 = 1 + cos2 - 1 = 1 + (cos + 1)(cos - 1) = R.H.S.
26. L.H.S.:
sin sin tan + cos = sin + cos
cos
sin2
=
+ cos
cos
1 - cos2
=
+ cos
cos
1 cos2
=
-
+ cos
cos cos
1
=
- cos + cos
cos
1 =
cos
= R.H.S.
27. L.H.S.:
1
1
1 + tan2
=
1+
sin2 cos2
1
=
cos2 cos2
+
sin2 cos2
1 = cos2 +sin2
cos2
cos2 = cos2 + sin2
cos2 =
1
= cos2
= R.H.S.
28. R.H.S.:
1 + cos 1 + cos 1 + cos
=
?
1 - cos 1 - cos 1 + cos
1 + 2 cos + cos2
=
1 - cos2
cos2 + 2 cos + 1
=
sin2
= L.H.S.
29. L.H.S.:
sin cos sin 1 + cos cos
-
=
?
-
1 - cos sin 1 - cos 1 + cos sin
sin (1 + cos ) cos
=
1 - cos2
- sin
sin (1 + cos ) cos
=
sin2
- sin
1 + cos cos
=
-
sin sin
1 + cos - cos =
sin
1 =
sin
= R.H.S.
30. L.H.S.:
1 - sin cos - cos2 1 - sin cos - (1 - sin2 ) sin2 + sin cos - 1 = (1 - cos2 ) + sin cos - 1
1 - sin cos - 1 + sin2 = 1 - cos2 + sin cos - 1
- sin cos + sin2 =
- cos2 + sin cos = sin (- cos + sin )
cos (- cos + sin ) sin = cos = tan
= R.H.S.
4
Unit 3B Specialist Mathematics
Exercise 3B
Exercise 3B
1. Compare this with identity { on page 57 of Sadler. Substitute A = 2x and B = x:
sin 2x cos x + cos 2x sin x = sin(2x + x) = sin 3x
2. Compare this with identity x on page 56 of Sadler. Substitute A = 3x and B = x:
7. sin 75 = sin(45 + 30)
= sin 45 cos 30 + cos 45 sin 30
1
311
= ? + ?
2 2
22
= 3 + 1
2 2 2 2
= 3+ 1
22
Compare this with the answer for question 5.
We could have arrived at this more simply using the identity sin(90 - A) = cos(A) substituting A = 15.
cos 3x cos x + sin 3x sin x = cos(3x - x) = cos 2x
3. Compare this with identity z on page 57 of Sadler. Substitute A = 5x and B = x:
8. cos 75 = cos(45 + 30)
= cos 45 cos 30 - sin 45 sin 30
1
311
= ? - ?
2 2
22
31
=-
2 2 2 2
3-1 =
22
sin 5x cos x - cos 5x sin x = sin(5x - x) = sin 4x
4. Compare this with identity y on page 56 of Sadler. Substitute A = 7x and B = x:
cos 7x cos x - sin 7x sin x = cos(7x + x) = cos 8x
5. cos 15 = cos(45 - 30)
= cos 45 cos 30 + sin 45 sin 30
1
311
= ? + ?
2 2
22
31 =+
2 2 2 2
3+1 =
22
6. tan 15 = tan(45 - 30)
tan 45 - tan 30
= 1 + tan 45 tan 30
5
1 - 1
=
3 1
9. tan 75 = tan(45 + 30)
tan 45 + tan 30 = 1 - tan 45 tan 30
1 + 1
=
1-1
3
? 1
3
3 + 1
= 3
3
3 - 1
3
3
=
3+1
3
? 3
3-1
3
3
= 3 + 1
3 - 1
3+1 3+1 = ?
3 -1 3 + 1
3+2 3+1
=
3 - 1
4+2 3
=
2
=2+ 3
10. 2 sin( + 45) = 2(sin cos 45 + cos sin 45)
a=b= 2
= 2 1 sin + 1 cos
2
2
2
2
= sin + cos
2
2
= 2 sin + 2 cos
Exercise 3B
Solutions to A.J. Sadler's
11. 8 cos -
3
= 8 cos cos + sin sin
3
3
1
3
= 8 cos + sin
2
2
= 4 cos + 4 3 sin
= 4 3 sin + 4 cos
c = 4 3, d = 4
12. 4 cos( + 30) = 4(cos cos 30 - sin sin 30)
3
1
=4
cos - sin
2
2
= 2 3 cos - 2 sin
e = 2 3, f = -2
tan A + tan B
13. tan(A + B) =
1 - tan A tan B
=
5 3-
3 4
1-5
3?-
3 4
=
3(5
-
1 4
)
1
+
5
?
3 4
19 3
=
4
1
+
15 4
19 3
=
4 19
4
=3
tan(A + B) is positive, so A + B is in quadrant
3:
A+
B
=
+
3
=
4 3
.
14. To proceed, we need to know cos A and sin B.
Given that we are working with acute angles, cos A = 1 - sin2 A. (If the angle was not known to be acute we'd have to also consider cos A = - 1 - sin2 A.) This gives us
4 cos A = 1 -
5 3 = 5
5 sin B = 1 -
13 12 = 13
Another way of looking at this is to think about
the given values in the context of a right angle
triangle.
sin A =
4 5
so think of a triangle with
hypotenuse of 5 units and opposite of 4 units.
Pythagoras' theorem gives us 3 for the other side
resulting in
cos A
=
3 5
and
tan A
=
4 3
.
Simi-
larly,
given
cos B
=
5 13
think
of
a
triangle
with
hypotenuse of 13 and adjacent of 5; Pythago-
ras gives us 12 as the remaining side and hence
sin B
=
12 13
and
tan B
=
12 5
.
5 45B
13
A3
12
(a) sin(A + B) = sin A cos B + cos A sin B
4 5 3 12 =? +?
5 13 5 13 20 36 =+ 65 65 56 = 65
(b) cos(A - B) = cos A cos B + sin A sin B
3 5 4 12 =? +?
5 13 5 13 15 48 =+ 65 65 63 = 65
15. Pythagoras (see question 14 above) gives us
cos D
=
24 25
and
cos E
=
4 5
.
(a) sin(D - E) = sin D cos E - cos D sin E
7 4 24 3 = ?- ?
25 5 25 5 28 72 =- 125 125
44 =-
125
(b) cos(D + E) = cos D cos E - sin D sin E
24 4 7 3 = ?- ?
25 5 25 5 96 21 =- 125 125 75 = 125 3 = 5
16.
To prove:
sin
x
+
2
= cos x
Proof:
L.H.S.:
sin x +
2
= sin x cos + cos x sin
2
2
= sin x ? 0 + cos x ? 1
= cos x
= R.H.S
17. (a) To prove: sin(x + 2) = sin x Proof:
L.H.S. = sin(x + 2) = sin x cos 2 + cos x sin 2 = sin x ? 1 + cos x ? 0 = sin x = R.H.S
6
Unit 3B Specialist Mathematics
Exercise 3B
(b) To prove: sin(x - 2) = sin x Proof:
L.H.S. = sin(x - 2) = sin x cos 2 - cos x sin 2 = sin x ? 1 - cos x ? 0 = sin x = R.H.S
18. To prove: cos(x + 2) = cos x Proof:
L.H.S. = cos(x + 2) = cos x cos 2 - sin x sin 2 = cos x ? 1 + sin x ? 0 = cos x = R.H.S
19. To prove: tan(x + ) = tan x
Proof:
L.H.S.:
tan x + tan tan(x + ) =
1 - tan x tan tan x + 0
= 1 - tan x ? 0 tan x
= 1
= tan x
= R.H.S
20. To prove: tan(-x) = - tan x Proof:
L.H.S.:
tan(-x) = tan(0 - x)
tan 0 - tan x =
1 + tan 0 tan x - tan x
= 1 + 0 ? tan x - tan x
= 1
= - tan x
= R.H.S
sin cos tan
A
5 13
-
12 13
-
5 12
B
3 5
-
4 5
-
3 4
(a) sin(A + B) = sin A cos B + cos A sin B
5
4
12 3
= ?- +- ?
13
5
13 5
20 36 =- -
65 65 56 =- 65
(b) cos(A - B) = cos A cos B + sin A sin B
12
4
=- ?-
13
5
48 15 =+
65 65 63 = 65
tan A + tan B (c) tan(A + B) =
1 - tan A tan B
=
1
-
5 12
-
-
-
5 12
3
4
-
3 4
-5-9
=
1
12
-
5 16
-14
=
12 11
16
7 16 =- ?
6 11
56 =-
33
53 +?
13 5
22. To prove: sin(A + B) - sin(A - B) = 2 cos A sin B
Proof:
L.H.S. = sin(A + B) - sin(A - B) = sin A cos B + cos A sin B -(sin A cos B - cos A sin B) = sin A cos B + cos A sin B - sin A cos B + cos A sin B = 2 cos A sin B = R.H.S
21. Given that they are obtuse angles, A and B fall into the 2nd quadrant so their sines are positive and cosines and tangents negative. Use Pythagoras to find the necessary ratios from those given using the triangle approach outlined in question 14 above:
13 A
5
53
12
B4
23. To prove: cos(A-B)+cos(A+B) = 2 cos A cos B Proof:
L.H.S. = cos(A - B) + sin(A + B) = cos A cos B + sin A sin B + cos A cos B - sin A sin B = 2 cos A cos B = R.H.S
7
Exercise 3B
24.
To prove:
2 cos
x
-
6
= sin x +
3 cos x
Proof:
L.H.S. = 2 cos x -
6
= 2 cos x cos + sin x sin
6
6
3
1
=2
cos x + sin x
2
2
= 3 cos x + sin x
= sin x + 3 cos x
= R.H.S
Solutions to A.J. Sadler's
Proof:
L.H.S. = 2(sin x - cos x) sin(x + 45)
= 2(sin x - cos x) ( sin x cos 45
+ cos x sin 45)
= 2(sin x - cos x)
1 sin x + 1 cos x
2
2
= (sin x - cos x)(sin x + cos x)
= sin2 x - cos2 x
= (1 - cos2 x) - cos2 x
= 1 - 2 cos2 x
= R.H.S
25.
To prove:
tan
+
4
=
1+tan 1-tan
Proof:
L.H.S. = tan +
4
=
tan
+
tan
4
1
-
tan
tan
4
tan + 1 =
1 - tan ? 1
1 + tan =
1 - tan
= R.H.S
26.
To prove:
cos(A+B) cos(A-B)
=
1-tan A tan B 1+tan A tan B
Proof:
cos(A + B) L.H.S. =
cos(A - B)
cos A cos B - sin A sin B =
cos A cos B + sin A sin B
cos A cos B-sin A sin B
=
cos A cos B cos A cos B+sin A sin B
cos A cos B
=
1 1
- +
sin A sin B cos A cos B sin A sin B cos A cos B
=
1 1
- +
sin A cos A sin A cos A
? ?
sin B cos B sin B cos B
1 - tan A tan B =
1 + tan A tan B
= R.H.S
27. To prove: 2(sin x - cos x) sin(x + 45) = 1 - 2 cos2 x
28.
To prove:
tan
+
4
=
1+2 sin cos 1-2 sin2
Proof:
L.H.S. = tan +
4
=
tan
+
tan
4
1 - tan tan 4
tan + 1 =
1 - tan ? 1
=
sin cos
+
1
1
-
sin cos
sin +cos
=
cos cos -sin
cos
sin + cos =
cos - sin
sin + cos sin + cos
=
?
cos - sin cos + sin
sin2 + 2 sin cos + cos2
=
cos2 - sin2
sin2 + cos2 + 2 sin cos
=
1 - sin2 - sin2
1 + 2 sin cos = 1 - 2 sin2
= R.H.S
29.
sin
x
cos
6
+
cos
x
sin
6
=
sin(x
+
6
)
so
the
equa-
tion to solve becomes
sin(x +
)
=
1
6
2
This
has
solutions
with
x+
6
in
the
1st
and
2nd
quadrant.
x+ =
64
x= - 46
x= 12
x+ =-
6
4
x=- -
46
7 x=
12
30. cos x cos 20 + sin x sin 20 = cos(x - 20) so the equation to solve becomes
cos(x - 20) = 1 2
8
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