Ken Ward's Mathematics Pages

Trigonometry: Sum and Product of Sine and Cosine

On this page, we look at examples of adding two ratios, but we could go on and derive relationships for more than two.
Trigonometry Contents

Page Contents

Sum of Sine and Cosine

Sines:
Cosines (Sum)
Cosines Difference:
Sum of Cosine and Sine:

Products

Sum of Sine and Cosine

Sines Sum

We wish to show that

[1.1]

In particular, we wish to show that

[1.2]

Let
A=(x+y) [1.3]
B=(x-y) [1.4]

So, by adding and subtracting Equations 1.3, 1.4 we find:
A+B=2x, so:
x=(A+B)/2 [1.5], and
A-B=2y, so:
y=(A-B)/2 [1.6]

Substituting these, we find:
sin x+sin y=sin (A+B)+sin(A-B) [1.7]

=sinA·cosB+cosA·sinB+sinA·cosB−cosA·sinB (using the compound angle formulae)

=sinA·cosB+sinA·cosB (because two terms cancel)
=2·sinA·cosB
And reconverting the A's and B's to x and y from Equations 1.3 and 1.4 gives us:

[1.2]
■

Sines Difference

In a similar fashion:

Substituting from Equations 1.3 and 1.4:
sin x-sin y=sin (A+B)-sin(A-B) [1.7]

=sinA·cosB+cosA·sinB−(sinA·cosB−cosA·sinB) (using the compound angle formulae)

=sinB·cosA+sinB·cosA (because two terms cancel)
=2·sinB·cosA
Giving us:

[1.8]
■

Cosines Sum

The formulae for the sum of two cosines and for the difference are a little different (The addition is in terms of cosines: the substraction in terms of sines).

We wish to show that:

[2.1]

Let A=(x+y)/2 [2.2], and
B=(x-y)/2 [2.3]
So, x=A+B, and y=A-B [2.4]

And
cos x+cos y=cos(A+B)+cos(A-B)

Expanding the right-hand side using the compound angle formula:
cos(A+B)+cos(A-B)=cosA·cosB-sinA·sinB+cosA·cosB+sinA·sinB
=2·cosA·cosB

Using Equations 2.2 and 2.3 to convert the A and B back to x and y:

which is Equation 2.1, the result we sought.
■

Cosines Difference

The formula for the difference between two cosines is:

[3.1]
To prove this, Let A=(x+y)/2, and B=(x−y)/2 [2.3, repeated]
So, x=A+B, and y=A−B [2.4, repeated]
By substituting [2.4] in the left-hand side of Equation 3.1, we get:
cos x−cos y=cos(A+B)−cos(A−B)

Expanding the compound angles, using the compound angle formulae:
cos(A+B)−cos(A−B)=cosA·cosB−sinA·sinB−(cosA·cosB+sinA·sinB)
=−2·sin [(x+y)/2]·sin[(x−y)/2]
Noting that −sin (θ)=sin (-θ), we can write −sin[(x−y)/2]=sin[(y-x)/2] to remove the minus sign. And substituting for A and B [using 2.3 above], we get our equation:

which is Equation 3.1.■

Sum of Cosine and Sine

The sum of the cosine and sine of the same angle, x, is given by:

[4.1]
We show this by using the principle cos θ=sin (π/2−θ), and convert the problem into the sum (or difference) between two sines.

We note that sin π/4=cos π/4=1/√2, and re-use cos θ=sin (π/2−θ) to obtain the required formula.

Sum

The plus option gives:

[4.2]

We can write cos x as sin (π/2−x), so the left-hand side of Equation 4.2 becomes:
=sin (π/2−x)+sin x [4.3]

Which is the sum of two sines. Using the formula for the sum of two sines (above):

[1.2, repeated]

We get, by substituting in Equation 4.3:
cos x+ sin x=2·sin(π/4)·cos(π/4-x)

Noting sin π/4=cos π/4=1/√2:
cos x+ sin x=2/√2·cos(π/4-x)

Noting that: cos(π/4-x)=sin(π/2-(π/4-x))=sin (π/4+x) [cos θ=sin (π/2−θ)]

We have:

That is, Equation 4.2, which we wished to prove. ■

Difference

[5.1]
Using the same form as before...

We can write cos x as sin (π/2−x), so the left-hand side of Equation 5.1 becomes:

=sin (π/2−x)−sin x [5.2]

Which is the difference of two sines. Using the formula for the sum of two sines (above):

[repeated]

We get, by substituting in Equation 5.2:
cos x- sin x=2·sin(π/4-x)·cos(π/4)

Noting sin π/4=cos π/4=1/√2:
cos x- sin x=2/√2·sin(π/4-x)

That is:

That is, Equation 5.1, which we wished to prove. ■

Products

[6.2]

[6.3]

[6.4]

We can show these relationships are true by expanding the right-hand sides using the compound angle formulas, the result occurs immediately.

Trigonometry Contents

Ken Ward's Mathematics Pages