Download Unit 8.2 Relationships Between Sine, Cosine and Tangent

Unit 8.2
Relationships Between Sine, Cosine and Tangent
Sine and Cosine
Consider triangle ABC:

Since the sum of the interior angles of a triangle must add up to 180°, the two acute
angles of a right triangle must add to 90°.
Angle A + Angle B = 90°

Angle A = 90° - Angle B
Angle B = 90° - Angle A
Using the trigonometric ratios for right triangles, the following ratios from ▲ABC can be
obtained:
𝑎
𝑏
 Sin A = = Cos B
Cos A = = Sin B
𝑐
𝑐
So,
Sin θ = cos(90° - θ)
and
Cos θ = sin(90° - θ)
sin 30° = cos 60°
and
cos 30° = sin 60°
sin 55° = cos 35°
and
cos 55° = sin 35°
For Example:
Tangent
Using the trigonometric ratios for right triangles, the following ratios from ▲ABC can be
obtained:
Tan B =
𝑏
𝑎
sin B =
𝑏
𝑐
cos B =
𝑎
𝑐
So,
sin 𝐵
cos 𝐵
=
𝑏
𝑐
𝑎
𝑐
=
𝑏
𝑎
= tan B
Therefore,
tan θ =
sin 𝜃
cos 𝜃
Sine2 and Cosine2
Consider Pythagoras’ Theorem:


a2 + b2 = c2
Again we can use our trigonometric ratios for right triangles
The following ratios can be obtained:
sin B =
cos B =
𝑏
𝑐
𝑎
𝑐
b = c(sin B)
b2 = c2(sin2B)
a = c(cos B)
a2 = c2(cos2B)
* Note: (sin B)2 is usually written as sin2B, (cos B)2 is usually written as cos2 B
By substitution:
a2 + b2 = c2
c2sin2B + c2cos2B = c2
sin2B + cos2B = 1
So….
Sin2θ + Cos2θ = 1
Sine, Cosine and Tangent
If sin θ + cos θ = 1,
2
So…
2
tan2θ + 1 =
then
1
cos2 θ
sin2 θ
cos2 θ
+
cos2 θ
cos2 θ
=
1
cos2 θ
Summary
1. Sinθ = cos(90° - θ)
2. Sinθ = cos(90° - θ)
3. Tanθ =
𝑠𝑖𝑛𝜃
𝑐𝑜𝑠𝜃
4. Sin2θ + cos2θ = 1
5. Tan2θ + 1 =
1
cos2 θ
*
3
Example 1:
Given sinθ =
Solution 1:
Draw a right triangle with acute angle θ opposite side of length 3, and
hypotenuse of length 5. SOH CAH TOA
5
is an acute angle, find the exact value of cosθ and tanθ.
5
3
θ
a
Method 1:
Method 2:
By PYTHAGORUS’ Theorem
sin2θ + cos2 θ = 1
a2 + b2 = c2
3
( )2 + cos2θ = 1
5
a2 + 32 = 52
3
cos2θ = 1 – ( )2
5
a = 25 – 9
2
a2 = 16
cos2θ = 1 -
9
25
=
25
-
9
25 25
a=4
cosθ = √
So,
𝒂𝒅𝒋𝒂𝒄𝒆𝒏𝒕
cosθ = 𝒉𝒚𝒑𝒐𝒕𝒆𝒏𝒖𝒔𝒆
=
𝒐𝒑𝒑𝒐𝒔𝒊𝒕𝒆
tanθ = 𝒂𝒅𝒋𝒂𝒄𝒆𝒏𝒕
=
𝟑
𝟒
16
25
=
𝟒
𝟓
𝟒
𝟓
3
𝑠𝑖𝑛θ 5 𝟑
tanθ =
=
=
𝑐𝑜𝑠θ 4 𝟒
5
=
16
25
2
, find the exact value of sinθ and tanθ.
Example 2:
Given cosθ =
Solution 2:
Draw a right triangle with acute angle θ adjacent side of length 2, and
hypotenuse of length 3. SOH CAH TOA
3
3
b
θ
2
Method 1:
Method 2:
By PYTHAGORUS’ Theorem
sin2θ + cos2 θ = 1
a2 + b2 = c2
2
sin2θ + ( )2 = 1
3
22 + b2 = 32
2
sin2θ = 1 – ( )2
3
a =9–4
2
a2 = 5
sin2θ = 1 -
4
9
=
9 4 5
- =
9 9 9
=
√𝟓
𝟑
a = √5
sinθ = √
So,
𝒐𝒑𝒑𝒐𝒔𝒊𝒕𝒆
sinθ = 𝒉𝒚𝒑𝒐𝒕𝒆𝒏𝒖𝒔𝒆
=
𝒐𝒑𝒑𝒐𝒔𝒊𝒕𝒆
tanθ = 𝒂𝒅𝒋𝒂𝒄𝒆𝒏𝒕
=
√𝟓
𝟑
√𝟓
𝟐
5
9
√5
𝑠𝑖𝑛θ
3 √𝟓
tanθ =
= 2 =
𝑐𝑜𝑠θ
𝟐
3