MathematicsClass 11ICSE

How to Turn a Sum of Sines Into a Product in One Step

Convert sums and differences of sines and cosines into products, turn products such as 2 sin A cos B into sums, and use both sets of transformation formulae to simplify expressions and solve trigonometric equations.

Why convert between sums and products in trigonometry?

A sum such as

sin 5 x + sin x

is hard to set equal to zero, but the equal product

2 sin 3 x cos 2 x

splits at once into two simpler equations. Going the other way, a product like

2 sin 5 x cos 3 x

is easier to evaluate or integrate as a sum.

This part covers sum-to-product formulae, product-to-sum formulae, and using both to simplify expressions and solve equations.

How do you convert sums and differences of sines and cosines into products?

**The sum-to-product formulae are

sin C + sin D = 2 sin \frac{C + D}{2} cos \frac{C - D}{2}

sin C - sin D = 2 cos \frac{C + D}{2} sin \frac{C - D}{2}

cos C + cos D = 2 cos \frac{C + D}{2} cos \frac{C - D}{2}

and

cos C - cos D = - 2 sin \frac{C + D}{2} sin \frac{C - D}{2}

.

Where they come from.** Adding

sin (A + B)

and

sin (A - B)

gives

2 sin A cos B

. Putting

A + B = C

and

A - B = D

, so that

A = \frac{C + D}{2}

and

B = \frac{C - D}{2}

, gives the first formula; the other three follow in the same way.

Worked example. Simplify

sin 75° + sin 15°

2 sin \frac{75° + 15°}{2} cos \frac{75° - 15°}{2} = 2 sin 45° cos 30° = 2 \times \frac{1}{2} \times \frac{3}{2} = \frac{3}{2} \approx 1.2247

Checking directly,

0.9659 + 0.2588 = 1.2247

.

Worked example 2.

cos 80° - cos 20° = - 2 sin 50° sin 30° = - sin 50° \approx - 0.766

.

An everyday example. Ripples from two pebbles dropped into a village pond overlap; writing the sum of the two waves as a product shows where the water stays calm.

The substance. **

cos C - cos D

carries a minus sign** — forgetting it is the commonest error in this topic.

How do you convert products such as 2 sin A cos B into sums or differences?

**The product-to-sum formulae are

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

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

2 cos A cos B = cos (A + B) + cos (A - B)

and

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

.

Where they come from.** Each is a pair of compound angle formulae added or subtracted; for example,

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

.

Worked example. Write

2 sin 5 x cos 3 x

as a sum:

2 sin 5 x cos 3 x = sin 8 x + sin 2 x

Worked example 2. Evaluate

2 sin 75° sin 15°

2 sin 75° sin 15° = cos 60° - cos 90° = \frac{1}{2} - 0 = \frac{1}{2}

Worked example 3. Show that

cos 20° cos 40° cos 80° = \frac{1}{8}

. First,

cos 40° cos 80° = \frac{1}{2} (cos 120° + cos 40°) = \frac{1}{2} (cos 40° - \frac{1}{2})

. Multiplying by

cos 20°

and using

cos 20° cos 40° = \frac{1}{2} (cos 60° + cos 20°)

\frac{1}{2} (\frac{1}{4} + \frac{1}{2} cos 20°) - \frac{1}{4} cos 20° = \frac{1}{8}

An everyday example. A radio receiver multiplies the incoming signal by a wave of its own; the product contains sum and difference frequencies, exactly as these formulae predict.

The substance. **

2 sin A sin B

begins with

cos (A - B)

, not

cos (A + B)

** — reversing the order flips the sign of the answer.

How are sum-to-product and product-to-sum formulae used to solve trigonometric equations?

An equation containing a sum of sines or cosines is solved by converting the sum into a product and setting each factor equal to zero, while an expression containing products is simplified by converting them into sums.

General solutions needed, for any integer n:

-

sin θ = 0

when

θ = nπ

cos θ = 0

when

θ = (2 n + 1) \frac{π}{2}

Worked example. Solve

sin 5 x + sin x = 0

for

0 \leq x \leq π

sin 5 x + sin x = 2 sin 3 x cos 2 x = 0

sin 3 x = 0

gives

x = 0, \frac{π}{3}, \frac{2 π}{3}, π

cos 2 x = 0

gives

x = \frac{π}{4}, \frac{3 π}{4}

So there are six solutions. Check

x = \frac{π}{4}

sin \frac{5 π}{4} + sin \frac{π}{4} = - 0.7071 + 0.7071 = 0

.

Worked example 2. Simplify

\frac{sin 7 x + sin 3 x}{cos 7 x + cos 3 x} = \frac{2 sin 5 x cos 2 x}{2 cos 5 x cos 2 x} = tan 5 x

An everyday example. Noise-cancelling headphones add a wave that is the opposite of the incoming noise; setting a sum of sines equal to zero, as here, shows exactly when the combined signal is silent.

The substance. Never divide both sides by a trigonometric factor — dividing by

cos 2 x

would lose the solutions

\frac{π}{4}

and

\frac{3 π}{4}

Exam tip

What earns full marks on transformation formulae?

Write the formula in C and D, or A and B, before substituting, and show the half-sum and half-difference as a separate step.

-

sin C + sin D = 2 sin \frac{C + D}{2} cos \frac{C - D}{2}

cos C - cos D = - 2 sin \frac{C + D}{2} sin \frac{C - D}{2}

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

- In equations, factorise instead of dividing

The trap. Writing

sin C - sin D = 2 sin \frac{C + D}{2} cos \frac{C - D}{2}

. **For a difference of sines, cosine takes the half-sum:

2 cos \frac{C + D}{2} sin \frac{C - D}{2}

.**

Did you know

Why do two nearly identical notes make a wobbling sound?

When two musical notes of slightly different pitch sound together, the loudness swells and fades in a steady rhythm called beats.

The sum-to-product formula explains it. Adding waves of 440 and 444 vibrations per second gives a tone at the average, 442, whose loudness is controlled by a slow cosine at half the difference, 2 per second. The loudness peaks twice in each cycle of that cosine, producing 4 beats every second.

Musicians tuning a sitar or a harmonium listen for these beats and adjust until they slow down and disappear.

Exam relevance

How are transformation formulae tested in JEE Main and JEE Advanced?

Sum-to-product and product-to-sum formulae are workhorses of Trigonometric Functions in JEE Main, and JEE Advanced uses them in trigonometric equations, series and integration.

What gets asked. **Products such as

cos 20° cos 40° cos 80°

, simplifying ratios of sums, counting solutions** of equations like

sin 5 x + sin x = 0

in an interval, and sums of sines of angles in arithmetic progression.

Question types. Multiple-choice and numerical-value questions.

The trap that costs marks. Losing solutions by cancelling a common factor instead of setting it equal to zero.

Key takeaways

What must you be able to do from this part?

- Sum to product:

sin C \pm sin D

and

cos C \pm cos D

become products of half-sum and half-difference terms
- Product to sum:

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

and its three companions
- Equations: convert a sum to a product, set each factor to zero, and never divide out a factor

How many solutions does

cos 3 x + cos x = 0

have for

0 \leq x \leq π

Why convert between sums and products in trigonometry?

How do you convert sums and differences of sines and cosines into products?

How do you convert products such as 2 sin A cos B into sums or differences?

How are sum-to-product and product-to-sum formulae used to solve trigonometric equations?

What earns full marks on transformation formulae?

Why do two nearly identical notes make a wobbling sound?

How are transformation formulae tested in JEE Main and JEE Advanced?

What must you be able to do from this part?

Related study notes

Ready to put this into practice?