# Trigonometric Formulas

Sub Topics
Formulas are nothing but established mathematical relations mainly to find an unknown quantity with the given information. The formulas become prominent and important when the unknown quantity is very important and very useful. In trigonometry also, formulas play important roles. In this section let us try to study some important formulas that are related to trigonometric ratios of composites and multiples of angles.

## Sum and Difference Formulas

Let us start with finding a formula for an angle which can be expressed as a sum of two angles. To derive a formula, consider a scalene triangle in general as shown below.

ABC is a scalene triangle. Let BP is the altitude on side AC and CQ be the altitude on side AB produced. As per triangle sum formula,
angle C = Π - angle (A + B)
So, sin C = sin [Π - angle (A + B)] = sin (A + B)

In the triangle CBP, BP = BC $\times$ sin C = a $\times$ sin C = a $\times$ sin (A + B)
In the triangle ABP, BP = AB $\times$ sin A = (AQ - BQ) $\times$ sin A = AQ $\times$ sin A - BQsin A

Therefore, a $\times$ sin (A + B) = AQ $\times$ sin A - BQsin A

Now considering triangle BQC, BQ = BC  $\times$ cos CBQ = a $\times$ cos (180 - B) = - a $\times$ cos B
Therefore, a $\times$ sin (A + B) = AQ $\times$ sin A - BQsin A =  AQ $\times$ sin A - (- a $\times$ cos B)sin A = AQ $\times$ sin A + a $\times$ sin Acos B

Again from triangle BQC, QC = a $\times$ sin (180 - B) = a $\times$ sin B
In triangle AQC, AQ = AC $\times$ cos A and AC = QC/sin A = (a $\times$ sin B)/ sin A

Hence, AQ = [(a $\times$ sin B)/ sin A] $\times$ cos A = (a $\times$ cos A sin B )/(sin A)

Therefore, AQ $\times$ sin A = a $\times$ cos A sin B

Thus,
a $\times$ sin (A + B) = AQ $\times$ sin A + a $\times$ sin Acos B = a $\times$ cos A sin B + + a $\times$ sin Acos B

Cancelling 'a' and rearranging in sequence,
sin (A + B) = sin A $\times$ cos B + cos A $\times$ sin B

Thus, we we have derived a formula for sine of a composite angle.

Now, to derive the formula for sin (A - B), replace '-B' for 'B' in the above formula. That is,
sin (A - B) = sin A $\times$ cos(- B) + cos A $\times$ sin (-B)

Since cos (-B) = cos B and sin (-B) = -sin B,
sin (A - B) = sin A $\times$ cos B - cos A $\times$ sin B

As per the trigonometric ratio of complimentary angles,
cos (A + B) = sin [Π/2 - (A + B)] = sin [(Π/2 - A) - B]

Applying the difference formula of sine on the right side,
cos (A + B) = sin (Π/2 - A) $\times$ cos B â€“ cos (Π/2 - A) $\times$ sin B, which means,
cos (A + B) = cos A $\times$ cos B - sin A $\times$ sin B

Now, to derive the formula for cos (A - B), replace '-B' for 'B' in the above formula. That is,
cos (A - B) = cos A $\times$ cos (-B) - sin A $\times$ sin (-B)

Since cos (-B) = cos B and sin (-B) = -sin B,
cos (A - B) = cos A $\times$ cos B + sin A $\times$ sin B

Having known the sum and difference formulas for sine and cosine ratios, let us derive similar formulas for the tangent ratio from fundamental definition.

tan (A + B) = [sin (A + B)]/[cos (A + B)] = [sin A $\times$ cos B + cos A $\times$ sin B]/[cos A $\times$ cos B - sin A $\times$ sin B],

Now dividing by cos A $\times$ cos B both sides,
tan (A + B) = (tan A + tan B)/(1 - tan A $\times$ tan B)

Now replacing B by '-B' and since tan (-B) = - tan B,
tan (A - B) = (tan A - tan B)/(1 + tan A $\times$ tan B)

Thus, we have derived the sum and difference formulas for all the basic trigonometric ratios.

## Double-Angle formulas

The trigonometric ratios of sum of angles are known to us. From the same concept we can derive the formulas for double angles. In all the formulas we derived, we considered the sum of the angles as (A + B). Suppose, B = A, then the sum of the angles (A + B) become as double angle 2A. Hence, to derive the formulas for double angle, we need to plug in B = A in all the sum formulas. Thus,

sin 2A = sin (A + A) = sin A $\times$ cos A+ cos A $\times$ sin A, or,
sin 2A = 2 sin A $\times$ cos A

cos 2A = cos (A + A) = cos A $\times$ cos A – sin A $\times$ sin A, or,
cos 2A = $cos^2$ A – $sin^2$ A

By using the fundamental identity $sin^2$ A + $cos^2$ A = 1, the above formula can be rewritten as,
cos 2A = 2$cos^2$ A – 1 = 1 – 2$sin^2$ A

Alternately, cos 2A = $cos^2$ A – $sin^2$ A =  $cos^2$ A(1 – $tan^2$ A) = (1 – $tan^2$ A)/$sec^2$ A or,
cos 2A = (1 – $tan^2$ A)/(1 + $tan^2$ A)

tan 2A = tan (A + A) = (tan A + tan A)/(1 – tan A $\times$ tan A), or,
tan 2A =  (2tan A)/(1 – $tan^2$ A)

## Half Angle Formulas

The formulas of trigonometric ratios of half of known angles can be derived by first figuring out the cosine ratio of the full angle and considering that full angle as double of half angle. Thus, the formulas for half angles are usually expressed in terms of cosine ratio of the full angle.
Now let us derive the individual formulas.

2 $cos^2$ A – 1 = cos 2A or, $cos^2$ A = [(1 + cos 2A)/2] or cos A = √[(1 + cos 2A)/2]
Similarly it can be shown that, sin A = √[(1 – cos 2A)/2]

Now plugging in A = A/2, we get the half angle formulas as,
cos (A/2) = √[(1 + cos A)/2]
sin (A/2) = √[(1 – cos A)/2]

Since tan (A/2) = [sin (A/2)]/[cos (A/2)],
tan (A/2) = √[(1 – cos A)/(1 + cos A)]

## Triple Angle Formulas

The triple angle formulas are derived by just extending the concept we did for double angles. That is, a triple angle 3A is written as a sum (2A + A) and the sum formulas are used. The values of double angles are plugged in and the resulting expressions are simplified. That is,

sin 3A = sin (2A + A) = sin 2A $\times$ cos A+ cos 2A $\times$ sin A = 2sin A $\times$ $cos^2$ A + (1 – 2$sin^2$ A) $\times$ sin A
= 2sin A $\times$ (1 – $sin^2$ A) + (1 – $sin^2$ A) $\times$ sin A = 3 sin A – 4 $sin^3$ A. Thus,
sin 3A = 3 sin A – 4 $sin^3$ A

Similarly,
cos 3A = cos (2A + A) = cos 2A $\times$ cos A – sin 2A $\times$ sin A = (2$cos^2$ A – 1) $\times$ cos A – (2$sin^2$ A $\times$ cos A)
= [(2$cos^2$ A – 1) $\times$ cos A] – (2 – 2$cos^2$ A) $\times$ (cos A) = 4$cos^3$ A – 3cos A. Thus,
cos 3A = 4$cos^3$ A – 3cos A
and same way,

tan 3A = tan (2A + A) = [{2 tan A/(1- $tan^2$ A)}+tan A]/ [1 – {2 tan A/(1- $tan^2$ A)}tan A]
= [(3 tan A – $tan^3$ A)/(1 – tan A)]/ [(1 – 3$tan^2$ A)/ (1 – $tan^2$ A)]
= (3tan A – $tan^3$ A)/ (1 – 3$tan^2$ A). Thus,

tan 3A = (3tan A – $tan^3$ A)/ (1 – 3$tan^2$ A)

## Product to Sum Formulas

Let us recall that, cos A $\times$ cos B – sin A $\times$ sin B = cos (A + B) and cos A $\times$ cos B + sin A $\times$ sin B = cos (A – B)
By adding both sides of the two equations,
2cos A $\times$ cos B = cos (A + B) + cos (A – B) or,
cos A $\times$ cos B = (1/2)[cos (A + B) + cos (A – B)]

On the other hand, by subtraction on both sides of the two equations,
-2sin A $\times$ sin B = cos (A + B) - cos (A – B) or,
sin A $\times$ sin B = -(1/2)[cos (A + B) - cos (A – B)]

Now, let us consider the formulas sin A* cos B + cos A*sin B =  sin (A + B) and sin A* cos B – cos A*sin B =  sin (A – B).

By adding both sides of the two equations,
2sin A $\times$ cos B = sin (A + B) + sin (A – B) or,
sin A $\times$ cos B = (1/2)[sin (A + B) + sin (A – B)]

On the other hand, by subtraction on both sides of the two equations,
2cos A $\times$ sin B = sin (A + B) - sin (A – B) or,
cos A $\times$ sin B = (1/2)[sin (A + B) - sin (A – B)]

These formulas are called product to sum formulas.

## Sum to Product Formulas

The ‘product to sum formulas’ can be rewritten and simplified to give ‘sum to product formulas’. Let two angles be, ‘x’ and ‘y’ and let x + y = A and x – y = B. By solving for ‘x’ and ‘y’, we get x = (A + B)/2 and y = (A – B)/2. Now,

cos A + cos B = cos (x + y) + cos (x – y) = 2 cos x $\times$ cos y = 2 [cos (A + B)/2] $\times$ [cos (A – B)/2]. Or,
cos A + cos B = 2 [cos (A + B)/2] $\times$ [cos (A – B)/2]

cos A - cos B = cos (x + y) - cos (x – y) = -2 sin x $\times$ sin y = -2 [sin (A + B)/2] $\times$ [sin (A – B)/2]. Or,
cos A - cos B = -2 [sin (A + B)/2] $\times$ [sin (A – B)/2]

sin A + sin B = sin (x + y) + sin (x – y) = 2sin x $\times$ cos y = 2[sin (A + B)/2] $\times$ [cos (A – B)/2]. Or,
sin A + sin B = 2 [sin (A + B)/2] $\times$ [cos (A – B)/2]

sin A - sin B = sin (x + y) - sin (x – y) = 2cos x $\times$ sin y = 2[cos (A + B)/2] $\times$ [sin (A – B)/2]. Or,
sin A - sin B = 2 [cos (A + B)/2] $\times$ [sin (A – B)/2]

## Herion's FORMULA

The trigonometric formulas help us to express the area of a triangle in terms of all its sides. This formula is called as Heron's formula. It states, for any triangle with sides as 'a', 'b' and 'c', the area of the triangle T is given by, T = √[s(s - a)(s - b)(s - c)] where, s = (a + b + c)/2.

To derive this formula, we use an important trigonometric law related to any triangle which is known as 'Cosine law of triangle'. If A, B and C are the angles of any triangle and if 'a', 'b' and 'c' are their corresponding sides, the cosine law states as,
$c^2$ = $a^2$ + $b^2$ - 2abcos C.

Now let us try to derive Heron's formula with the help of the following diagram.

As per the area formula of triangles, the area T of the triangle is given by,
T = (1/2)b $\times$ h = (1/2)b $\times$ a $\times$ sin C , since h/a = sin C.
or, 4T^2 = b^2 a^2 $\times$ sin^2 C
= b^2 a^2(1 - cos^2 C)

As per cosine law, c^2 = a^2 + b^2 - 2abcos C or,
2abcos C = a^2 + b^2 - c^2 or, cos C = (a^2 + b^2 - c^2)/(2ab)
or, cos^2 C = (a^2 + b^2 - c^2)^2/(4a^2b^2)

Hence, 1 - cos^2 C = [(4a^2 b^2) - (a^2 + b^2 - c^2)^2] /(4a^2b^2) and so,
4T^2 = b^2a^2(1 - cos^2 C) = [(4a^2b^2) - (a^2 + b^2 - c^2)^2] /(4)

Therefore, 16T^2 = [(4a^2b^2) - (a^2 + b^2 - c^2)^2] = [(2ab)^2 - (a^2 + b^2 - c^2)^2]
Using the identity for difference of two squares,

16T^2 = (2ab + a^2 + b^2 - c^2)( 2ab - a^2 - b^2 + c^2)
= (2ab + a^2 + b^2 - c^2)[c^2 - (a^2 + b^2 - 2ab)]
= [(a + b)^2 - c^2)][c^2 - (a - b)^2]

Once again using the identity for difference of two squares,
16T^2 = (a + b + c)(a + b - c)(c + a - b)(c - a + b)
= (a + b + c)(a + b + c - 2c)(c + a + b - 2b)(c + a - 2a + b)
= (a + b + c)(a + b + c - 2c)(a + b + c - 2b)(a + b + c - 2a)
= (2s)(2s - 2c)(2s - 2b)(2s - 2a) = 16 s(s - a)(s - b)(s - c), where s = (a + b + c)/2

or, T^2 = s(s - a)(s - b)(s - c) and hence,
T = [s(s - a)(s - b)(s - c)]