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Trigonometry – Deriving Identities, Laws, & Area of a Triangle
Once students are familiar with the quotient, reciprocal, co-function, and negative angle identities, we
can use those to derive many of the other trigonometric identities. We often use these identities to
find the exact values of sin, cos, and tan of 15°, 75°, etc., angles and to do proofs. In deriving the
identities, we can essentially prove them. Included here are scaffolded tasks that allow you to derive
the angle sum & difference identities, double and half-angle identities, and the Laws of Sine and
Cosine. We also show that the area of a triangle can be found using a trigonometric formula.
Teacher notes follow. If you find any errors, please email me ([email protected]) so that I
may make the appropriate changes!! If you have any questions, I can try to answer those via email
also!
1
Derivation of the Pythagorean Identities
Write down the six trigonometric ratios in terms of x, y, & r and sin and cos. Also, write the
Pythagorean Theorem in terms of x, y, and r with r being the hypotenuse.
Divide the Pythagorean Thm through by r2. Substitute in the trig ratios. What is the result?
Divide the Pythagorean Thm through by x2. Substitute in the trig ratios. What is the result?
Divide the Pythagorean Thm through by y2. Substitute in the trig ratios. What is the result?
We could also derive the other two Pythagorean Identities by having one of them and
dividing through by the trig term. If we know 𝑠𝑖𝑛2 𝜃 + 𝑐𝑜𝑠 2 𝜃 = 1, then we can divide
through by 𝑠𝑖𝑛2 𝜃 or 𝑐𝑜𝑠 2 𝜃 to derive the other two. Show that here.
2
Derivation of the Sum and Difference Identities
In the picture at the right, point A lies on the unit
circle and on the terminal side of an angle
measuring u units. Point B lies on the unit circle
and on the terminal side of an angle measuring v
units. Also, note that the angle made between
points A and B and the origin measures s units.
This means that angle s = angle u – angle v.
Find the coordinates of A and B in terms of sin
and cos of u and v.
Find the distance between A and B.
The picture above has been rotated so that
point B now lies on the x-axis.
Find the coordinates of B.
The angle between points A and B still
measures s units.
Find the coordinates of A in terms of sin and
cos of s.
Find the distance between A and B.
The distance between A and B has been maintained so they are equal. Set the distance you found
between A and B from the first figure equal to the distance you found between A and B from the
second figure. What you end up with after simplifying is an important identity.
3

We can now use this identity to derive the other sum and difference identities.
Find cos (u + v) by using cos (u – (- v)). You will also need to use your negative angle identities to
simplify.
We can now find sin (u + v) by using cos (u – v) (the first identity we found here) and our cofunction
identities. We need to remember that sin  cos(90   ) . We then have:
sin (u + v) = cos (90 – (u + v)) = cos (90 – u) – v)) =

sin (u – v) = sin (u + (- v))
Using these identities with our quotient identities, we can derive the identities for tangent.
tan(u  v) 
sin (u  v)
cos(u  v)
tan(u  v) 
sin (u  v)
cos(u  v)
Substitute the identities in for the numerator and denominator and then “divide through” by
cos u cos v. You are really multiplying the fraction by 1 (multiplicative identity) in the form of

(1/ cos u cos v) / (1/ cos u cos v).
4
Derivation of Double-Angle Identities
Find sin 2x, cos 2x, and tan 2x by looking at 2x as (x + x). You should derive 3 new identities.
For the double-angle identity for cosine (cos 2x), you can find alternative forms by using the
Pythagorean identities. For example, where you see cos2 x, you can substitute 1 – sin2 x and simplify.
Find two alternative identities for cos 2x.
Derivation of Half-Angle Identities
From the double-angle identities above, we have: cos 2x = 1 – 2sin2 x .
Solve this identity for sin x and substitute in x/2 for x to represent half of the angle.
The half-angle identities for cosine and tangent can be derived similarly using
cos 2x = 2cos2 x – 1
tan2 𝑥 =
sin2 𝑥
cos2 𝑥
=
1−cos 2𝑥
1+ cos 2𝑥
5
Derivation of the Law of Sines
When “solving a triangle,” you are expected to find the lengths of all its sides and the measures of all its
angles. Previously, we have been able to solve only RIGHT triangles. Not all triangles are right triangles.
What about oblique triangles?
We can solve ANY triangle, right or oblique, if we know the length of at least one side and any other two
measures due to triangle similarities (AAS, ASA, SSA, SAS, SSS). The AA similarity postulate is not
included here. Why not?
On triangle ABC, draw altitude, h, from angle C to side c. Find sin A and sin B in terms of h. Solve each
for h and set them equal.
C
A
B
This leads to the Law of Sines:
Why do we not need to worry about division by zero?
On triangle ABC, draw altitude, h, from angle C to side c. Find sin A and sin B in terms of h. Solve each
for h and set them equal.
C
A
B
This also leads to the Law of Sines. If we interchanged the angles and sides to C and c, we would have
two other variations of the Law of Sines. We can simply put them together into one law.
6
The Trigonometric Triangle Area Formula
Show that the area of right triangle ABC can be given as ½ ab sin C and ½ bc sin A.
A
B
C
Show that the areas of these oblique triangles can be given as ½ bc sin A. It may be helpful to
construct altitude h from C to c.
C
A
B
C
A
B
7
Derivation of the Law of Cosines
On a coordinate plane, draw obtuse triangle ABC so that angle C is the obtuse angle. The coordinates of
point C are (0,0), of point B are (a, 0), and of point A are (x,y). Find the values of x and y in terms of b
and C. Remember that x = r cos  and y = r sin .
Use the distance formula to find c2.
8
Trigonometry – Deriving Identities, Laws, & Area of a Triangle – TEACHER NOTES
Derivation of the Pythagorean Identities
Write down the six trigonometric ratios in terms of x, y, & r and sin and cos. Also, write the
Pythagorean Theorem in terms of x, y, and r with r being the hypotenuse.
x2 + y2 = r2
sin 𝜃 =
1
csc 𝜃 = sin 𝜃 =
𝑦
cos 𝜃 =
𝑟
𝑟
sec 𝜃 =
𝑦
𝑥
tan 𝜃 =
𝑟
1
cos 𝜃
=
𝑟
𝑥
cot 𝜃 =
sin 𝜃
cos 𝜃
cos 𝜃
sin 𝜃
=
=
𝑦
𝑥
𝑥
𝑦
Divide the Pythagorean Thm through by r2. Substitute in the trig ratios. What is the result?
𝑥2
𝑟2
+
𝑦2
𝑟2
=
𝑟2
𝑟2
𝑥 2
𝑦 2
𝑟 2
 (𝑟 ) + ( 𝑟 ) = (𝑟)
 𝑐𝑜𝑠 2 𝜃 + 𝑠𝑖𝑛2 𝜃 = 1
Divide the Pythagorean Thm through by x2. Substitute in the trig ratios. What is the result?
𝑥2
𝑥2
+
𝑦2
𝑥2
=
𝑟2
𝑥2
𝑥 2
𝑦 2
𝑟 2
 (𝑥) + (𝑥 ) = (𝑥)
 1 + 𝑡𝑎𝑛 2 𝜃 = 𝑠𝑒𝑐 2 𝜃
Divide the Pythagorean Thm through by y2. Substitute in the trig ratios. What is the result?
𝑥2
𝑦2
+
𝑦2
𝑦2
=
𝑟2
𝑦2
𝑥 2
𝑦 2
𝑟 2
 (𝑦) + (𝑦) = (𝑦)
 𝑐𝑜𝑡 2 𝜃 + 1 = 𝑐𝑠𝑐 2 𝜃
We could also derive the other two Pythagorean Identities by having one of them and
dividing through by the trig term. If we know 𝑠𝑖𝑛2 𝜃 + 𝑐𝑜𝑠 2 𝜃 = 1, then we can divide
through by 𝑠𝑖𝑛2 𝜃 or 𝑐𝑜𝑠 2 𝜃 to derive the other two. Show that here.
𝑠𝑖𝑛2 𝜃
𝑠𝑖𝑛2
𝜃
𝑠𝑖𝑛2 𝜃
+
+
𝑐𝑜𝑠2 𝜃
𝑐𝑜𝑠2 𝜃
𝑠𝑖𝑛2
𝜃
𝑐𝑜𝑠2 𝜃
𝑐𝑜𝑠2 𝜃
=
cos 𝜃 2
1
𝑠𝑖𝑛2
1
𝜃
 1+ (
sin 𝜃
sin 𝜃 2
) = (
1
sin 𝜃
1
)
= 𝑐𝑜𝑠2 𝜃  (cos 𝜃 ) + 1 = (cos 𝜃)
2
2
 1 + 𝑐𝑜𝑡 2 𝜃 = 𝑐𝑠𝑐 2 𝜃
 𝑡𝑎𝑛2 𝜃 + 1 = 𝑠𝑒𝑐 2 𝜃
9
Derivation of the Sum and Difference Identities
In the picture at the right, point A lies on the unit
circle and on the terminal side of an angle
measuring u units. Point B lies on the unit circle
and on the terminal side of an angle measuring v
units. Also, note that the angle made between
points A and B and the origin measures s units.
This means that angle s = angle u – angle v.
Find the coordinates of A and B in terms of sin
and cos of u and v.
A (cos u, sin u) & B (cos v, sin v)
Find the distance between A and B.
𝑑 = √(cos 𝑢 − cos 𝑣)2 + (sin 𝑢 − sin 𝑣)2
The picture above has been rotated so that
point B now lies on the x-axis.
Find the coordinates of B. (1, 0)
The angle between points A and B still
measures s units.
Find the coordinates of A in terms of sin and
cos of s. (cos s, sin s)
Find the distance between A and B.
𝑑 = √(cos 𝑠 − 1)2 + (sin 𝑢 − 0)2
10
The distance between A and B has been maintained so they are equal. Set the distance you found
between A and B from the first figure equal to the distance you found between A and B from the
second figure. What you end up with after simplifying is an important identity.
√(cos 𝑢 − cos 𝑣)2 + (sin 𝑢 − sin 𝑣)2 = √(cos 𝑠 − 1)2 + (sin 𝑠 − 0)2

√(𝑐𝑜𝑠 2 𝑢 − 2 cos 𝑢 cos 𝑣 + 𝑐𝑜𝑠 2 𝑣) + (𝑠𝑖𝑛2 𝑢 − 2 sin 𝑢 sin 𝑣 + 𝑠𝑖𝑛2 𝑣) = √(𝑐𝑜𝑠 2 𝑠 − 2 cos 𝑠 + 1) + 𝑠𝑖𝑛2 𝑠 
Square both sides.
(𝑐𝑜𝑠 2 𝑢 − 2 cos 𝑢 cos 𝑣 + 𝑐𝑜𝑠 2 𝑣) + (𝑠𝑖𝑛2 𝑢 − 2 sin 𝑢 sin 𝑣 + 𝑠𝑖𝑛2 𝑣) = (𝑐𝑜𝑠 2 𝑠 − 2 cos 𝑠 + 1) + 𝑠𝑖𝑛2 𝑠 
Note the three pairs of the Pythagorean identity that each sum to 1.
1 − 2 cos 𝑢 cos 𝑣 + 1 − 2 sin 𝑢 sin 𝑣 = 1 − 2 cos 𝑠 + 1 
Simplify both sides.
2 − 2 cos 𝑢 cos 𝑣 − 2 sin 𝑢 sin 𝑣 = 2 − 2 cos 𝑠 
Subtract 2 from both sides.
−2 cos 𝑢 cos 𝑣 − 2 sin 𝑢 sin 𝑣 = −2 cos 𝑠 
Divide both sides by -2.
cos 𝑢 cos 𝑣 + sin 𝑢 sin 𝑣 = cos 𝑠 
Apply reflexive property.
cos 𝑠 = cos 𝑢 cos 𝑣 + sin 𝑢 sin 𝑣 
Remember: s = u – v. Substitute.
𝐜𝐨𝐬(𝒖 − 𝒗) = 𝐜𝐨𝐬 𝒖 𝐜𝐨𝐬 𝒗 + 𝐬𝐢𝐧 𝒖 𝐬𝐢𝐧 𝒗
We can now use this identity to derive the other sum and difference identities.
Find cos (u + v) by using cos (u – (- v)). You will also need to use your negative angle identities to
simplify.
Remember the negative angle identities!
sin (-v) = – sin v
cos (-v) = cos v
tan (-v) = – tan v
cos(𝑢 + 𝑣) = cos(𝑢 − (−𝑣)) 
Minus a negative is same as plus.
cos(𝑢 + 𝑣) = cos 𝑢 cos(−𝑣) + sin 𝑢 sin(−𝑣) 
Substitute into previous identity.
cos(𝑢 + 𝑣) = cos 𝑢 cos 𝑣 + sin 𝑢 ∙ − sin 𝑣 
Substitute negative angle identities.
𝐜𝐨𝐬(𝒖 + 𝒗) = 𝐜𝐨𝐬 𝒖 𝐜𝐨𝐬 𝒗 − 𝐬𝐢𝐧 𝒖 𝐬𝐢𝐧 𝒗
Simplify.
11
We can now find sin (u + v) by using cos (u – v) (the first identity we found here) and our cofunction
identities. We need to remember that 𝑠𝑖𝑛𝜃 = cos(90° − 𝜃). We then have:
We will also need: cos 𝜃 = sin(90° − 𝜃)
sin (u + v) = cos (90 – (u + v)) = cos (90 – u) – v)) 
1st angle: 90 – u
2nd angle: v
sin (u + v) = cos(90 − 𝑢) cos 𝑣 + sin(90 − 𝑢) sin 𝑣 
Substitute into cosine of diff identity.
sin (u + v) = 𝐬𝐢𝐧 𝒖 𝐜𝐨𝐬 𝒗 + 𝐜𝐨𝐬 𝒖 𝐬𝐢𝐧 𝒗
Substitute cofunction identities.
sin (u – v) = sin (u + (- v)) 
sin (u – v) = sin u cos (-v) + cos u sin (-v) 
Substitute in previous identity.
sin (u – v) = sin u cos v + cos u ∙ −sin v 
Substitute in neg angle identities.
sin (u – v) = sin u cos v − cos u sin v
Simplify.
Using these identities with our quotient identities, we can derive the identities for tangent.
tan(𝑢 + 𝑣) =
sin(𝑢+𝑣)
tan(𝑢 − 𝑣) =
cos(𝑢+𝑣)
sin(𝑢−𝑣)
cos(𝑢−𝑣)
Substitute the identities in for the numerator and denominator and then “divide through” by
cos u cos v. You are really multiplying the fraction by 1 (multiplicative identity) in the form of
(1/ cos u cos v) / (1/ cos u cos v).
tan(𝑢 + 𝑣) =
sin(𝑢+𝑣)
cos(𝑢+𝑣)
=
sin 𝑢 cos 𝑣 + cos 𝑢 sin 𝑣
cos 𝑢 cos 𝑣 − sin 𝑢 sin 𝑣
tan(𝑢 + 𝑣) =
sin 𝑢 cos 𝑣
cos 𝑢 sin 𝑣
+
cos 𝑢 cos 𝑣 cos 𝑢 cos 𝑣
cos 𝑢 cos 𝑣 sin 𝑢 sin 𝑣
–
cos 𝑢 cos 𝑣 cos 𝑢 cos 𝑣
tan(𝑢 + 𝑣) =
sin 𝑢
sin 𝑣
+
cos 𝑢 cos 𝑣
sin 𝑢 sin 𝑣
1–
cos 𝑢 cos 𝑣
𝐭𝐚𝐧(𝒖 + 𝒗) =

𝐭𝐚𝐧 𝒖 + 𝐭𝐚𝐧 𝒗
𝟏 − 𝐭𝐚𝐧 𝒖 𝐭𝐚𝐧 𝒗


Substitute in identities.
Divide through by cos 𝑢 cos 𝑣.
Simplify each fraction.
Substitute in quotient identities.
12
tan(𝑢 − 𝑣) = tan(𝑢 + (−𝑣)) 
tan(𝑢 − 𝑣) =
tan(𝑢 − 𝑣) =
𝐭𝐚𝐧(𝒖 − 𝒗) =
tan 𝑢+tan(−𝑣)
1−tan 𝑢 tan(−𝑣)
tan 𝑢−tan 𝑣
1− tan 𝑢∙ − tan 𝑣
𝐭𝐚𝐧 𝒖−𝐭𝐚𝐧 𝒗
𝟏+ 𝐭𝐚𝐧 𝒖 𝐭𝐚𝐧 𝒗
Minus same as plus a negative.

Substitute into previous identity.

Simplify.
Simplify.
13
Derivation of Double-Angle Identities
Find sin 2x, cos 2x, and tan 2x by looking at 2x as (x + x). You should derive 3 new identities.
sin 2x = sin (x + x) 
sin 2x = sin x cos x + cos x sin x 
Substitute into identity.
sin 2x = 2 sin x cos x
Simplify.
cos 2x = cos (x + x) 
cos 2x = cos x cos x – sin x sin x 
Substitute into identity.
cos 2x = cos2 x – sin2 x
Simplify.
tan 2𝑥 = 𝑡𝑎𝑛(𝑥 + 𝑥) 
tan 2𝑥 =
𝐭𝐚𝐧 𝟐𝒙 =
tan 𝑥 + tan 𝑥
1 − tan 𝑥 tan 𝑥

𝟐 𝐭𝐚𝐧 𝒙
𝟏 − 𝐭𝐚𝐧𝟐 𝒙
Substitute into identity.
Simplify.
For the double-angle identity for cosine (cos 2x), you can find alternative forms by using the
Pythagorean identities. For example, where you see cos2 x, you can substitute 1 – sin2 x and simplify.
Find two alternative identities for cos 2x.
cos 2x = cos2 x – sin2 x 
cos 2x = (1 – sin2 x) – sin2 x 
Substitute Pythagorean Identity.
cos 2x = 1 – 2sin2 x
Simplify.
cos 2x = cos2 x – sin2 x 
sin2 x = 1 – cos2 x
cos 2x = cos2 x – (1 – cos2 x) 
Substitute Pythagorean Identity.
cos 2x = 2cos2 x – 1
Simplify.
14
Derivation of Half-Angle Identities
From the double-angle identities above, we have: cos 2x = 1 – 2sin2 x .
Solve this identity for sin x and substitute in x/2 for x to represent half of the angle.
cos 2x = 1 – 2sin2 x 
2sin2 x = 1 – cos 2x 
sin2 𝑥 =
1 − cos 2𝑥
1 − cos 2𝑥
𝒙
Apply division property of equality.

2
sin 𝑥 = ±√
Apply addition property of equality.
Take square root of both sides.

2
𝟏 − 𝐜𝐨𝐬 𝒙
𝐬𝐢𝐧 𝟐 = ±√
Substitute x/2 for x.
𝟐
The half-angle identities for cosine and tangent can be derived similarly using
cos 2x = 2cos2 x – 1
tan2 𝑥 =
sin2 𝑥
cos2 𝑥
=
1−cos 2𝑥
1+ cos 2𝑥
cos 2x = 2cos2 x – 1 
Double Angle Identity.
cos 2x + 1 = 2cos2 x 
Addition Property of Equality.
2cos2 x = cos 2x + 1 
Apply Reflexive Property.
cos 2 𝑥 =
cos 2𝑥+1
2
cos 2𝑥+1
cos 𝑥 = ±√
𝒙
2
tan2 𝑥 =
Substitute x/2 for x.
𝟐
sin2 𝑥
cos2 𝑥
Apply the Quotient Identity.

1 − cos 2𝑥
1 + cos 2𝑥
1− cos 2𝑥
tan 𝑥 = ± √1 + cos 2𝑥
𝒙
Take square root of both sides.

𝐜𝐨𝐬 𝒙 + 𝟏
𝐜𝐨𝐬 𝟐 = ±√
tan2 𝑥 =
Apply division property of equality.

𝟏 − 𝐜𝐨𝐬 𝒙
𝐭𝐚𝐧 𝟐 = ± √𝟏 + 𝐜𝐨𝐬 𝒙
Apply the Double Angle Identities.


Take square root of both sides.
Substitute x/2 for x.
15
Derivation of the Law of Sines
When “solving a triangle,” you are expected to find the lengths of all its sides and the measures of all its
angles. Previously, we have been able to solve only RIGHT triangles. Not all triangles are right triangles.
What about oblique triangles?
We can solve ANY triangle, right or oblique, if we know the length of at least one side and any other two
measures due to triangle similarities (AAS, ASA, SSA, SAS, SSS). The AA similarity postulate is not
included here. Why not? It is implied in AAS & ASA.
On triangle ABC, draw altitude, h, from angle C to side c. Find sin A and sin B in terms of h. Solve each
for h and set them equal.
C
b
a
h
A
sin 𝐴 =
c
ℎ
𝑏
B
 ℎ = 𝑏 ∙ sin 𝐴
and
sin 𝐵 =
ℎ
𝑎
 ℎ = 𝑎 ∙ sin 𝐵
𝑏 ∙ sin 𝐴 = 𝑎 ∙ sin 𝐵 

divide both sides by sin A and sin B
𝑏
𝑎
=
sin 𝐵
sin 𝐴
This leads to the Law of Sines:
𝑎
𝑏
=
𝑜𝑟
sin 𝐴
sin 𝐵
sin 𝐴
sin 𝐵
=
𝑎
𝑏
Why do we not need to worry about division by zero?
In a triangle each angle must be between 0° and 180°. The only angle measures in this interval that you
can take the sine of and get zero are 0° and 180° where sin 0° = sin 180° = 0. In other words, the outputs
on the input interval (0°, 180°) are (0, 1]. Therefore, sin A ≠ 0, sin B ≠ 0, and sin C ≠ 0. Also, none of
the side lengths will measure zero units so a ≠ 0, b ≠ 0, and c ≠ 0.
16
On triangle ABC, draw altitude, h, from angle C to side c. Find sin A and sin B in terms of h. Solve each
for h and set them equal.
C
a
h
b
180 - A A
c
B
We can’t find sin A in the given triangle but we can find sin (180 – A). We would want to do this because
sin A = sin (180 – A).
sin(180 − 𝐴) =
and
sin 𝐵 =
ℎ
𝑎
ℎ
𝑏

sin 𝐴 =
ℎ
𝑏
 ℎ = 𝑏 ∙ sin 𝐴
 ℎ = 𝑎 ∙ sin 𝐵
𝑏 ∙ sin 𝐴 = 𝑎 ∙ sin 𝐵 
divide both sides by sin A and sin B
𝑏
𝑎
=
sin 𝐵
sin 𝐴
This also leads to the Law of Sines. If we interchanged the angles and sides to C and c, we would have
the other variations of the Law of Sines. We can simply put them together into one law.
𝑎
𝑏
𝑐
=
=
𝑜𝑟
sin 𝐴
sin 𝐵
sin 𝐶
sin 𝐴
sin 𝐵
sin 𝐶
=
=
𝑎
𝑏
𝑐
17
Trigonometric Triangle Area Formula
Show that the area of right triangle ABC can be given as ½ ab sin C and ½ bc sin A.
A
c
b
C
B
a
The area of a triangle is typically found by taking half of the base times height, which in our case
1
1
is 𝐴∆ = 2 𝑎 𝑏. We want to show that 𝐴∆ = 2 𝑎 𝑏 ∙ sin 𝐶. We can do this simply by remembering
that sin 90° = 1.
𝐴∆ =
1
2
𝑎 𝑏 ∙ sin 𝐶  𝐴∆ =
We also want to show 𝐴∆ =
our possible formula.
𝐴∆ =
1
2
𝑏𝑐 ∙ sin 𝐴  𝐴∆ =
1
2
1
2
1
2
𝑎 𝑏 ∙ sin 90°  𝐴∆ =
1
2
𝑎 𝑏 ∙ 1  𝐴∆ =
1
2
𝑎𝑏
𝑎
𝑏𝑐 ∙ sin 𝐴. In triangle ABC, sin 𝐴 = 𝑐 , which we can substitute into
𝑏𝑐 ∙
𝑎
𝑐
 𝐴∆ =
1 𝑎𝑏𝑐
2 𝑐
 𝐴∆ =
1
2
𝑎𝑏
Show that the areas of these oblique triangles can be given as ½ bc sin A. It may be helpful to
construct altitude h from C to c.
C
b
A
a
h
B
c
The area of a triangle is typically found by taking half of the base x height, which is 𝐴∆ =
𝐴∆ =
1
2
𝑏𝑐 ∙ sin 𝐴  𝐴∆ =
1
2
ℎ
𝑏𝑐 ∙ 𝑏

𝐴∆ =
1 ℎ𝑏𝑐
2
𝑏
 𝐴∆ =
1
2
1
2
ℎ𝑐.
ℎ𝑐
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C
b
h
a
180 - A A
c
B
The area of a triangle is typically found by taking half of the base x height, which is 𝐴∆ =
𝐴∆ =
1
2
𝑏𝑐 ∙ sin 𝐴  𝐴∆ =
1
2
𝑏𝑐 ∙ sin(180 − 𝐴)  𝐴∆ =
1
2
ℎ
𝑏𝑐 ∙ 𝑏  𝐴∆ =
1 ℎ𝑏𝑐
2
𝑏
1
2
ℎ𝑐.
 𝐴∆ =
1
2
ℎ𝑐
19
Derivation of the Law of Cosines
On a coordinate plane, draw obtuse triangle ABC so that angle C is the obtuse angle. The coordinates of
point C are (0,0), of point B are (a, 0), and of point A are (x,y). Find the values of x and y in terms of b
and C. Remember that x = r cos  and y = r sin .
Use the distance formula to find c2.
The distance formula is 𝑑 = √(𝑥2 − 𝑥1 )2 + (𝑦2 − 𝑦1 )2 OR 𝑑2 = (𝑥2 − 𝑥1 )2 + (𝑦2 − 𝑦1 )2 .
In our case (x, y) can be written as (b cos C, b sin C). In finding the distance between points A & B, we
have:
𝑐 2 = (𝑏 cos 𝐶 − 𝑎)2 + (𝑏 sin 𝐶 − 0)2 
Substitution into distance formula.
𝑐 2 = (𝑏 cos 𝐶)2 − 2𝑎𝑏 cos 𝐶 + 𝑎2 + (𝑏 sin 𝐶)2 
Simplify.
𝑐 2 = 𝑏 2 cos2 𝐶 − 2𝑎𝑏 cos 𝐶 + 𝑎2 + 𝑏 2 sin2 𝐶 
Simplify.
𝑐 2 = 𝑏 2 cos2 𝐶 + 𝑏 2 sin2 𝐶 − 2𝑎𝑏 cos 𝐶 + 𝑎2 
Apply Commutative & Associative
Properties of Addition.
𝑐 2 = 𝑏 2 (cos2 𝐶 + sin2 𝐶) − 2𝑎𝑏 cos 𝐶 + 𝑎2 
Factor out b2.
𝑐 2 = 𝑏 2 (1) − 2𝑎𝑏 cos 𝐶 + 𝑎2 
Apply the Pythagorean Identity.
𝒄𝟐 = 𝒂𝟐 + 𝒃𝟐 − 𝟐𝒂𝒃 𝐜𝐨𝐬 𝑪
Apply Commutative & Associative
Properties of Addition.
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