Sum and Difference Formulas

Sum and difference formulas are trigonometric identities used to find the values of angles by expressing them as the sum or difference of known standard angles like 30°, 45°, and 60° and are useful for solving problems, simplifying expressions, and proving identities.

The six main trigonometric formulas for sum and difference are

Proof of Sum and Difference Identities

To demonstrate, the trigonometric sum and difference formulas let us consider a unit circle with coordinates given as (cos θ, sin θ).

• Consider points A and B, which form angles of α and β with the positive X-axis, respectively.
• The coordinates of A and B are (cos α, sin α) and (cos β, sin β), respectively.

The angle AOB is equal to (α - β). Now, consider another two points, P and Q, on the unit circle such that Q is a point on the X-axis with coordinates (1, 0) and angle POQ is equal to (α - β), and thus the coordinates of the point P are (cos (α - β), sin (α - β)). 

Now, OA = OP, and OB = OQ as they are the radii of the same unit circle, and also the measure of one of the included angles of both triangles is (α - β).

Hence, by the side-angle-side congruence, AOB and POQ are congruent triangles.

We know that the corresponding parts of congruent triangles are congruent, hence AB = PQ.

So, AB = PQ.

Using the distance formula between two points, we get,

d_AB =\sqrt{(\cos\alpha - \cos\beta)^2 + (\sin\alpha - \sin\beta)^2}

= \sqrt{\cos^2\alpha - 2\cos\alpha\cos\beta + \cos^2\beta + \sin^2\alpha - 2\sin\alpha\sin\beta + \sin^2\beta}\quad {Since (a - b)² = a² - 2ab + b}

=\sqrt{(\cos^2\alpha + \sin^2\alpha) + (\cos^2\beta + \sin^2\beta) - 2(\cos\alpha \cos\beta + \sin\alpha \sin\beta)}

= \sqrt{1 + 1 - 2(\cos\alpha \cos\beta + \sin\alpha \sin\beta)}        {Since, sin² x + cos² x = 1}

= \sqrt{2 - 2(\cos\alpha \cos\beta + \sin\alpha \sin\beta)}.......(1)

d_PQ = \sqrt{(\cos(\alpha - \beta) - 1)^2 + (\sin(\alpha - \beta) - 0)^2}

= \sqrt{\cos^2(\alpha - \beta) - 2\cos(\alpha - \beta) + 1 + \sin^2(\alpha - \beta)}      {(a - b)² = a² - 2ab + b}

= \sqrt{(\cos^2(\alpha - \beta) + \sin^2(\alpha - \beta)) + 1 - 2\cos(\alpha - \beta)}

= \sqrt{1 + 1 - 2\cos(\alpha - \beta)}          {Since, sin² x + cos² x = 1}

= \sqrt{2 - 2\cos(\alpha - \beta)}......(2)

Since AB = PQ, equate both equations (1) and (2).

√[2 - 2(cos α cos β+ sin α sin β)] = \sqrt{2 - 2\cos(\alpha - \beta)} 

By squaring on both sides, we get,

2 - 2(cos α cos β+ sin α sin β) = 2 - 2 cos (α - β)......(3)

Sum and Difference Formulas for Cosine

Cos (α - β) formula

from eq (3)

2 (1 - cos α cos β - sin α sin β) = 2 (1 - cos (α - β))
1 - cos α cos β - sin α sin β = 1 - cos (α - β)

cos (α - β) = cos α cos β + sin α sin β

Cos (α + β) formula

To derive the sum formula of the cosine function substitute (-β) in the place of β in the difference of the cosine function.

Hence, cos (α + β) = cos (α - (β))
                 = cos α cos (-β) + sin α sin (-β)    {Since, cos (α - β) = cos α cos β + sin α sin β}
                  = cos α cos β - sin α sin β            {Since, cos (-θ) = cos θ, sin (-θ) = - sin θ}

cos (α + β) = cos α cos β - sin α sin β

Sum and Difference Formulas for Sine

Sin (α - β) formula

We know that, sin (90° - θ) = cos θ and cos (90° - θ) = sin θ. 

So, sin (α - β) = cos (90° - (α - β))
                 = cos (90° - α + β)
                 = cos [(90° - α) + β]
                 = cos (90° - α) cos β - sin (90° - α) sin β      {Since,  cos (α + β) = cos α cos β - sin α sin β}

sin (α - β) = sin α cos β - cos α sin β

Sin (α + β) formula

We know that, sin (α + β) = cos (90° - (α + β))
                 = cos (90° - α - β)
                 = cos [(90° - α) - β]
                 = cos (90° - α) cos β + sin (90° - α) sin β    {Since, cos (α - β) = cos α cos β + sin α sin β}

sin (α + β) = sin α cos β + cos α sin β

Sum and Difference Formulas for Tangent

Tan (α - β) formula

We know that, tan θ = sin θ/cos θ

So, tan (α - β) = sin (α - β)/cos (α - β)
                 = \frac{\sin\alpha \cos\beta - \cos\alpha \sin\beta}{\cos\alpha \cos\beta + \sin\alpha \sin\beta}

Now, divide the numerator and denominator with cos α cos β
               = \frac{(\sin\alpha \cos\beta - \cos\alpha \sin\beta)\cos\alpha \cos\beta}{\dfrac{\cos\alpha \cos\beta + \sin\alpha \sin\beta}{\cos\alpha \cos\beta}}
                 = \frac{\frac{\sin\alpha}{\cos\alpha} - \frac{\sin\beta}{\cos\beta}}{1 + \left(\frac{\sin\alpha}{\cos\alpha}\right)\left(\frac{\sin\beta}{\cos\beta}\right)}
                 = \frac{\tan\alpha - \tan\beta}{1 + \tan\alpha \tan\beta}
tan (α - β) =\frac{\tan\alpha - \tan\beta}{1 + \tan\alpha \tan\beta}

Tan (α + β) formula

To derive the tan (α + β) formula substitute (-β) in the place of β in the tan (α - β) formula.

Hence, we get, 

tan (α + β) = tan(α - (-β))

                  = \frac{\tan\alpha - \tan(-\beta)}{1 + \tan\alpha \tan(-\beta)}           {Since, tan (α - β) = (tan α - tan β)/(1 + tan α tan β)}
                  = \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha \tan\beta}                  {Since, tan (-θ) = - tan θ}
tan (α + β) = \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha \tan\beta}

Applying Sum and Difference Formulas

To Apply Sum and difference formulas, study the following example:

Example: Find the value of sin 15°

Solution:

Step 1: Write the given function in the sum and difference of the standard function,
sin 15° = sin (45 -30)°

Step 2: Use the required Sum and Difference Formulas, here we use, sin (α - β) = sin α cos β - cos α sin β
sin (45 -30)° = sin 45° cos 30° - cos 45° sin 30°

Step 3: Substitute the value of these standard trigonometric functions using the trigonometric table.
sin (45 -30)° = 1/√2 × √3/2 - 1/√2 × 1/2

Step 4: Simplify the value obtained in the above step.
sin (45 -30)° = 1/√2 × √3/2 - 1/√2 × 1/2
                    = (√3 -1)/ 2√2 

sin 15° = (√3 -1)√2 / 4

Related Articles

• Pythagoras Theorem
• Trigonometric Ratios
• Heights and Distances

Solved Examples

Example 1: Prove the triple angle formulae of sine and cosine functions using the sum and difference formulae.

•  sin 3A = 3 sin A - 4 sin³ A
• cos 3A = 4 cos³³ A - 3 cos A

Solution:

To Prove: sin 3A = 3 sin A - 4 sin³ A

sin 3A = sin (2A + A)     [sin (A + B) = sin A cos B + cos A sin B]

sin (2A + A) = sin 2A cos A + cos 2A sin A

We know that,

sin 2A = 2 sin A cos A, and cos 2A = 1 - 2sin² A, and cos² A = 1 - sin² A

sin (2A + A) = (2 sin A cos A) cos A + (1 - 2sin² A)sin A
                    = 2 sin A cos² A + sin A - 2 sin³ A 
                    = 2 sin A (1 - sin² A) + sin A - 2 sin³ A
                    = 2 sin A - 2sin³ A + sin A - 2 sin³ A
                    = 3 sin A - 4 sin³ A

Thus, sin 3A = 3 sin A - 4 sin³ A  (proved)

To Prove: cos 3A = 4 cos³³ A - 3 cos A

cos 3A = cos (2A + A)   [cos (A + B) = cos A cos B - sin A sin B]

So, cos (2A + A) = cos 2A cos A - sin 2A sin A

We know that, 

sin 2A = 2sin A cos A and cos 2A = 2cos² A - 1, and sin² A = 1- cos² A

cos (2A + A) = (2 cos² A - 1) cos A - (2 sin A cos A) sin A
                    = 2 cos³ A - cos A - 2 sin² A cos A
                    = 2 cos³ A - cos A - 2 (1- cos² A) cos A
                    = 2 cos³ A - cos A - 2 cos A + 2 cos³ A 
                    = 4 cos³ A - 3 cos A

Thus, cos 3A = 4 cos³ A - 3 cos A  (proved)

Example 2: Find the value of cos 75° using the sum and difference formulas.

Solution:

We can write 75° as the sum of 45° and 30°

cos 75° = cos (45° + 30°)

            = cos 45° cos 30° - sin 45° sin 30°             {Since, cos (A + B) = cos A cos B - sin A sin B}
            = (1/√2) (√3/2) - (1/√2)(1/2)                    {Since, cos 45° = sin 45° = (1√2) , cos 30° = √3/2, sin 30° = 1/2}
            = (√3 -1)/2√2

Hence, cos 75° = (√3 - 1)/2√2

Example 3: Find the value of tan 105° using the sum and difference formulas.

Solution:

We can write 105° as the sum of 60° and 45°.

tan 105° = tan (60° + 45°)

              = (tan 60° + tan 45°)/(1 - tan 60° tan 45°)   {Since, tan (A + B) = (tan A + tan B)
              = (√3 + 1)/(1 - (√3 × 1))                              {Since, tan 60° = √3, tan 45° = 1}
              = (√3 + 1)/(1 - √3)

Rationalize the above expression with the conjugate of the denominator,

            = \left[\frac{\sqrt{3}+1}{1-\sqrt{3}}\right]\times\left[\frac{1+\sqrt{3}}{1+\sqrt{3}}\right]

            = (√3 + 1)²/(1 - (√3)²)
            = (3 + 2√3 + 1)/(1 - 3)
            = (4 + 2√3)/(-2)
            = -2 - √3

Hence, tan 105° = -2 - √3.

Example 4: Evaluate the value of sin 15° using the sum and difference formulas.

Solution:

We can write 15° as the difference between 45° and 30°

sin 15° = sin (45° - 30°)

            = sin 45° cos 30° - cos 45° sin 30°   {Since, sin (A - B) = sin A cos B - cos A sin B}
            = (1/√2) (√3/2) - (1/√2)(1/2)          {Since, cos 45° = sin 45° = (1√2) , cos 30° = √3/2, sin 30° = 1/2}
            = (√3 - 1)/2√2

Hence, sin 15° = (√3 - 1)/2√2

Example 5: Prove that sin (π/4 - a) cos (π/4 - b) + cos (π/4 - a) sin (π/4 - b) = cos (a + b).

Solution:

L.H.S = sin (π/4 - a) cos (π/4 - b) + cos (π/4 - a) sin (π/4 - b)   {sin (A + B) = sin A cos B + cos A sin B}

         = sin [(π/4 - a) + (π/4 - b)]
         = sin [(π/2) - (a + b)]
          = cos (a + b)             {Since, sin (90° - θ) = cos θ}
           = R. H. S    (proved)