Download Section 9.3 Right Angle Trigonometry:

Section 9.3 Right Angle Trigonometry:
Consider a right triangle where one of the angles is labeled θ. The longest side is called
the Hypotenuse (hyp), the side opposite the angle θ is called the Opposite Side (opp)
and the side adjacent to the angle is called the Adjacent Side (adj). Using the lengths of
these sides you can form 6 ratios which are the fundamental trigonometric functions of
the angle θ.
hypotenuse
opposite
θ
adjacent
Sine of θ:
sin(θ ) =
opp
hyp
Cosecant of θ:
csc(θ ) =
hyp
opp
Cosine of θ:
cos(θ ) =
adj
hyp
Secant of θ:
sec(θ ) =
hyp
adj
opp
adj
Cotangent of θ: cot(θ ) =
adj
opp
Tangent of θ: tan(θ ) =
Ex 1:
13
5
α
θ
12
sin(θ ) =
opp 5
=
hyp 13
csc(θ ) =
hyp 13
=
opp 5
cos(θ ) =
adj 12
=
hyp 13
sec(θ ) =
hyp 13
=
adj 12
tan(θ ) =
opp 5
=
adj 12
cot(θ ) =
adj 12
=
opp 5
The same thing can be done for α but now the opposite side is different:
sin(α ) =
opp
hyp
csc(α ) =
hyp
=
opp
cos(α ) =
adj
=
hyp
sec(α ) =
hyp
=
adj
tan(α ) =
opp
=
adj
cot(α ) =
adj
=
opp
Trigonometric Identities: These are always true. You must have these memorized for the
test.
1
csc θ
1
cos(θ ) =
sec θ
sin(θ ) =
1
sin θ
1
sec(θ ) =
cos θ
csc(θ ) =
tan(θ ) =
1
cot θ
cot(θ ) =
1
tan θ
tan(θ ) =
sin θ
cos θ
cot(θ ) =
cos θ
sin θ
Pythagorean identities:
sin 2 (θ ) + cos 2 (θ ) = 1
1 + tan 2 θ = sec 2 θ
1 + cot 2 θ = csc 2 θ
Ex 2: Suppose tan θ = 5 and 0 ≤ θ ≤ π/2, solve for the other five trigonometric functions.
You know that tan(θ) is the ration opp so in our triangle we know that the side opposite
adj
θ is 5 and the side adjacent is 1. We can draw a triangle and solve for the hypotenuse.
Then we read the values of the trig functions from the triangle.
sin θ = 5 / 26
26
5
csc θ =
cos θ = 1 / 26
tan θ = 5 / 1
θ
sec θ =
26 / 5
26 / 5
cot θ = 5/1
1
Two Special Triangles:
For the angles 45º, 30º and 60º we have two special triangles which allow us to find the
their trigonometric functions. Memorize these for the test.
2
2
1
1
30
45
3
1
How do we use these two triangles?
Ex 3: Suppose we have one side of a right triangle and an angle:
We have two triangles for 60º.
sin(60) = 3 / 2 and sin(60) = y / 18
18
y
60
So
3
y
=
⇒ y=9 3
2 18
Ex 4:
We have two triangles for 45º.
r
sin(45) =
20
45
1
2
and sin( 45) =
20
r
So
1
2
=
20
⇒ r = 20 2
r
Proving Identities
If we want to prove an identity we want to show that it is true for all values. If we have
an equation and we want to know if it is an identity we work with one side and try to
make it look like the other.
Ex 5. Prove the following:
a) cos( x) sec( x) = 1
We will work with the left side. Convert everything to cos(x).
cos( x) sec( x) = cos( x)
1
=1
cos( x)
So the identity is true.
b) sin 2 ( x) − cos 2 ( x) = 2 sin 2 ( x) − 1
For this problem we will work with the left hand side again but now we need to use one
of our Pythagorean identities:
sin 2 ( x) + cos 2 ( x) = 1
⇒
cos 2 ( x) = 1 − sin 2 ( x)
Now we take this expression for cos2(x) and substitute into the original equation:
sin 2 ( x) − cos 2 ( x) = sin 2 ( x) − (1 − sin 2 ( x))
= sin 2 ( x) − 1 − sin 2 ( x)
= 2 sin 2 ( x) − 1
So the statement is true.
Ex 6. A 6 foot person standing 20 feet from the base of a street light casts a 10 foot
shadow. What is the height of the street light?