Trigonometry - Weebly
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Core Mathematics 3 Trigonometry
Core Maths 3
Trigonometry
Page 1
C3 Trigonometry
In C2 you were introduced to radian measure and had to find areas of sectors and segments. In addition to this you solved trigonometric equations using the identities below.
Sin2 Cos2 1
Tan Sin Cos
By the end of this unit you should: Have a knowledge of secant, cosecant and cotangent and of arcsin, arcos and arctan. Their relationship to sine, cosine and tangent and their respective graphs including appropriate restrictions of the domain. Have a knowledge of 1 Cot2 Co sec2 and Tan2 1 Sec2. Have a knowledge of double angle formulae and "r" formulae.
Core Maths 3
Trigonometry
Page 2
New trigonometric functions
The following three trigonometric functions are the reciprocals of sine, cosine and tan. The way to remember them is by looking at the third letter.
Secx
1 Cosx
Co sec x 1 Sinx
Cotx 1 Tanx
Core Maths 3
Trigonometry
Page 3
These three trigonometric functions are use to derive two more identities.
New Identities Starting with Sin2 Cos2 1 and by dividing by Sin2 gives:
Sin2 Sin2
Cos 2 Sin2
1 Sin2
Using
the
new
functions
outlined
above
and
the
fact
that
Cot
Cos Sin
this becomes:
1 Cot2 Co sec2 .
Returning to Sin2 Cos2 1 and by dividing by Cos2 gives:
Therefore:
Sin2 Cos 2
Cos 2 Cos 2
1 Cos 2
Tan2 1 Sec2.
The three identities will be used time and again. Try to remember them but you should also be able to derive them as outlined above.
Sin2 Cos2 1
1 Cot2 Co sec2
Tan2 1 Sec2
Example 1 Solve for 0 360 the equation
5tan2 sec 1 ,
giving your answers to 1 decimal place.
You should have come across questions of this type in C2 using the identity cos2 sin2 1 . The given equation has a single power of sec therefore we must use an identity to get rid of the tan2.
tan2 1 sec2
So the equation becomes:
Core Maths 3
Trigonometry
Page 4
5 sec2 5 sec 1
5 sec2 sec 6 0 We now have a quadratic in sec so by factorising:
5 sec2 sec 6 0
(5 sec 6)(sec 1) 0
sec = -6 cos= -5 146.4,213.6
5
6
sec 1 cos=1 0,360
Inverse Trigonometric Functions.
Functions are introduced in C3 and we use the concept of inverses to find
the following functions (remember that the inverse of a function in
graphical terms is its reflection in the line y = x). The domain of the
original trigonometric function has to be restricted to ensure that it is
still one to one. It is also worth remembering that the domain and range
swap over as you go from the function to the inverse. ie in the first case
the domain of sinx is restricted to
2
sin x
and this becomes the 2
range of the inverse function.
y=arcsinx
Domain
1 x1
Range
2
arcsin x
2
Core Maths 3
Trigonometry
Page 5
Example Find
arcsin0.5 = y
Simply swap around
Siny = 0.5
y = /6
y=arccosx
Domain
1 x1
Range
0 arccos x
y=arctanx
Domain
x
Range
arctan x
2
2
Core Maths 3
Trigonometry
Page 6
Addition Formulae A majority of the formulae in C3 need to be learnt. One's in red are in the formula book.
Sin(A B ) SinACosB CosASinB
Cos (A B ) CosACosB SinASinB
Tan
(A
B
)
TanA TanB TanATanB
The examples below use addition formulae.
Example
Given that Sin A =
12 13
and that Cos B =
4 5
where A
is obtuse
and B is
reflex find:
a) Sin (A + B)
b) Cos (A ? B)
c) Cot (A ? B)
Before we start the question it is advisable to draw the graphs of Sin x and Cos x.
Core Maths 3
Trigonometry
Page 7
Since A is obtuse the cosine of A must be negative and by using the
Pythagorean triple Cos A = 5 . The angle B is slightly more tricky. We 13
are told that B is reflex but we know that Cos B is positive. Therefore B
must be between 270? and 360? and so Sin B is negative.
Hence Sin B = 3 . 5
We are now ready to attempt part (a)
Sin A = 12 13
Cos A = 5 13
Sin B = 3 5
Cos B = 4 5
a) Using the formula above to find Sin (A + B)
Sin (A + B) = Sin A Cos B + Sin B Cos A
Sin
(A
+
B)
=
12 13
4 5
3 5
5 13
Sin
(A
+
B)
=
33 65
b)
Cos (A ? B) = Cos A Cos B + Sin A Sin B
Cos (A ? B) = 5 4 12 3 13 5 13 5
Cos (A ? B) = 56 65
c)
Cot
(A
?
B)
=
1 Tan (A
B)
1 TanATanB TanA TanB
Core Maths 3
Trigonometry
Page 8
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