Download 0C2/1C2 Example Sheet 4 - Trigonometric and Inverse

0C2/1C2 Example Sheet 4 - Trigonometric and Inverse
Trigonometric Functions. SOLUTIONS
1. (a)
sin(A + B)
sin A cos B + cos A sin B
=
cos(A + B)
cos A cos B − sin A sin B
Now, the trick is to divide top and bottom by cos A cos B:
tan(A + B) =
tan(A + B) =
sin A
cos A
cos A
cos A
cos B
cos B
cos B
cos B
A sin B
+ cos
tan A + tan B
cos A cos B
=
sin A sin B
1 − tan A tan B
− cos A cos B
+ A−B
and B = A+B
− A−B
and using the sum formula:
2
2
2
A+B A−B
A+B A−B
cos A − cos B = cos
+
− cos
−
2
2
2
2
A+B
A−B
A+B
A−B
= cos
cos
− sin
sin
2
2
2
2
A+B
A−B
A+B
A−B
− cos
cos
− sin
sin
2
2
2
2
A−B
A+B
sin
= −2 sin
2
2
(b) We can write A =
A+B
2
(c) We use the sum formulas repeatedly: sin 3x = sin(2x+x) = sin 2x cos x+cos 2x sin x =
(2 sin x cos x) cos x+(cos2 x−sin2 x) sin x = 2 sin x cos2 x+sin x cos2 x−sin3 x =
3 sin x(1 − sin2 x) − sin3 x = 3 sin x − 3 sin3 x − sin3 x = 3 sin x − 4 sin3 x
2. To find the expression cos x+sin x = A sin(x+B), apply the addition formula: sin(x+B) =
sin x cos B + cos x sin B, so A sin(x + B) = (A cos B) sin x + (A sin B) cos x. We need to
have A cos B = A sin B = 1; from here cos B = sin B, hence B = π/4 or B = 5π/4. We
√
√
choose B = π/4. Then A = 1/ sin π4 = 2 . Answer: cos x + sin x = 2 sin(x + π4 ).
3. (i) cos 3x = cos(2x+x) = cos 2x cos x−sin 2x sin x = (cos2 x−sin2 x) cos x−2 sin x cos x sin x =
(2 cos2 x − 1) cos x − 2 cos x(1 − cos2 x) = 4 cos3 x − 3 cos x .
(ii) cos 4x = cos(3x + x) = (4 cos3 x − 3 cos x) cos x − (3 sin x − 4 sin3 x) sin x = 4 cos4 x −
3 cos2 x − 3 sin2 x + 4 sin4 x = 4 cos4 x + 4 sin4 x − 3 = 4 cos4 x + 4(1 − cos2 x)2 − 3 =
4 cos4 x + 4(1 − 2 cos2 x + cos4 x) − 3 = 8 cos4 x − 8 cos2 x + 1 .
4. We have cos 3x = 4 cos3 x − 3 cos x, hence 4 cos3 x = cos 3x + 3 cos x and finally cos3 x =
1
(cos 3x
4
+ 3 cos x) .
5. (i) By using the double angle formula and dividing and multiplying by sec2 x2 , we obtain
sin x = sin
x
2
+
2 sin x2 cos x2 sec2
x
x
x
= 2 sin cos =
2
2
2
sec2 x2
x
2
=
2 tan x2
2 tan x2
2t
=
=
x
2 x
2
sec 2
1 + tan 2
1 + t2
(ii) Similarly we have
sin2
x
2 x
2 x
cos x = cos
+
= sin
− cos
=
2 2
2
2
x
− cos2 x2 sec2 x2
=
sec2 x2
1 − tan2 x2
1 − tan2
=
sec2 x2
1 + tan2
x
2
x
2
x
2
1 − t2
=
1 + t2
(iii) Now we can use the previous two parts to get the formula for tan x:
tan x =
sin x
2t
=
cos x
1 − t2
6. First write sin θ = x3 . Now
2
4 x3
4 sin2 θ
y =
=
2
cos2 θ
1 − x3
x2
4
2
y 1−
= x2
9
9
2
9y 2 − x2 y 2 = 4x2
is probably the simplest form.
7. (i) To solve sin y = 12 , the principal value is y =
other solution in the range 0 to 2π is π −
y=
π
6
π
+ 2nπ
6
=
5π
.
6
or
π
6
and we see from the graph that the
Thus, the complete solution is
5π
+ 2nπ
6
where n is any integer (positive or negative).
(ii) To solve cos 2y = 1, the principal value is 2y = 2π, and there are no other solutions
until 2π. Hence, we have
2y = 2nπ ⇔ y = nπ
where n is any integer.
1
, from
10
1
− arccos 10
and
(iii) We can rearrange sec y = 10 to cos y =
1
which we obtain y = arccos 10
as
the principal value. The other value is
so altogether
y = ± arccos
1
+ 2nπ
10
1
with n an integer. (Note: arccos 10
≈ 1.47.)
8. (i) Consider sin x = cos x, divide to get tan x = 1 The principal value is x = π4 , and the
other solution is x =
circle.)
5π
.
4
(Alternatively, we may see both solutions directly from the unit
(ii) From sin 5x = 0 we obtain 5x = nπ, for n any integer. Thus x =
nπ
,
5
and the solutions
between 0 and 2π are
6π 7π 8π 9π
π 2π 3π 4π
x = 0, , , , , π, , , ,
5 5 5 5
5 5 5 5
(iii) Now cos2 x =
1
4
implies cos x = ± 12 The principal value of cos x =
the other solution is x = 2π −
and x =
4π
.
3
π
3
=
5π
.
3
is x =
π
3
and
Similarly, the solutions of cos x = − 12 are x =
Altogether:
x=
1
2
π 2π 4π 5π
, , ,
3 3 3 3
2π
3