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Trigonometry Contents

[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 A and B for x and y, we find:

sin x+sin y=sin ( (A+B) )/2+sin( (A-B)/2 ) [1.7]

Using the compound angle formulae with each part:

sin ( (A+B) )/2=sin A/2 ·cos B/2 +cos A/2·sin B/2 (i)

And

sin( (A-B)/2 )=sin A/2 ·cos B/2 − cos A/2·sin B/2 (ii)

Adding (i) and (ii) we find:

sin ( (A+B) )/2 + sin( (A-B)/2 ) =2 · sin A/2 · cos B/2 [1.7b]

Because (A+B)/2=x [From 1.5] and (A-B)/2=y [From 1.6], substituting these in 1.7b, we find the relationship below:

[1.2]

■

Substituting from Equations 1.3 and 1.4, that is:

A=(x+y) [1.3 repeated]

B=(x-y) [1.4 repeated]

We find:

sin x−sin y=sin (A+B)−sin(A−B) [1.7]

Using the compound angle formulae with each part:

sin ( (A+B) )/2=sin A/2 ·cos B/2 +cos A/2·sin B/2 (i)

And

sin( (A-B)/2 )=sin A/2 ·cos B/2 − cos A/2·sin B/2 (ii)

Subtracting ii from i

sin ( (A+B) )/2 − sin( (A−B)/2 )=

2 ·cos A/2·sin B/2 (because two terms cancel) (iii)

Substituting in (iii), the relationships (A+B)/2=x [From 1.5], (A-B)/2=y,

A/2=(x+y)/2, and B/2=(x-y)/2 we find the relationship below:

[1.8]

■

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.

■

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

which is Equation 3.1.■

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.

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. ■

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. ■

[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

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