Transformation of Trigonometric Identities

A trigonometric expression can have one of two shapes. Some are products, such as \( 2\sin 35^\circ \cos 25^\circ \), where two ratios are multiplied together. Others are sums or differences, such as \( \sin 55^\circ - \sin 25^\circ \), where two ratios are added or subtracted. The same value can be written in either shape, and moving from one shape to the other is often the key step in a calculation or a proof.

Why move at all? A product of two awkward angles such as 35° and 25° can become a sum involving 60° and 10°, and 60° has an exact value. In the other direction, a sum in the top of a fraction can become a product, and a product can be cancelled against a matching product in the bottom. Sums are hard to cancel. Products are easy.

There are two directions of travel, and each has four formulas. All eight grow out of the compound angle identities, so every one of them can be rebuilt in a minute if it is ever forgotten.

Four compound angle identities are the starting point. They are numbered here so that each step later can say which one it uses.

Identity (1), the sine of a sum:

\[ \sin(A+B) = \sin A\cos B + \cos A\sin B \]

Identity (2), the sine of a difference:

\[ \sin(A-B) = \sin A\cos B - \cos A\sin B \]

Identity (3), the cosine of a sum:

\[ \cos(A+B) = \cos A\cos B - \sin A\sin B \]

Identity (4), the cosine of a difference:

\[ \cos(A-B) = \cos A\cos B + \sin A\sin B \]

Look at them in pairs. Identities (1) and (2) have the same two terms on the right, and only the sign in the middle changes. Identities (3) and (4) are the same. So when a pair is added or subtracted, one kind of term cancels and the other kind doubles. That doubling is where the 2 in every formula of this topic comes from.

Each product formula comes from adding or subtracting one pair of compound angle identities. Here they are worked out in full.

Step 1: Adding identity (1) and identity (2). The left sides give \( \sin(A+B) + \sin(A-B) \), and the right sides together are:

\[ \begin{aligned} &\sin A\cos B + \cos A\sin B \\ &+ \sin A\cos B - \cos A\sin B \end{aligned} \]

Step 2: Cancelling the two \( \cos A\sin B \) terms, since one is added and the other subtracted. The two \( \sin A\cos B \) terms add to twice one of them, so:

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

Step 3: Subtracting identity (2) from identity (1). This time the \( \sin A\cos B \) terms cancel and the \( \cos A\sin B \) terms double:

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

Step 4: Adding identity (3) and identity (4). The \( \sin A\sin B \) terms cancel and the \( \cos A\cos B \) terms double:

\[ 2\cos A\cos B = \cos(A+B) + \cos(A-B) \]

Step 5: Subtracting identity (4) from identity (3). The \( \cos A\cos B \) terms cancel, and the two \( \sin A\sin B \) terms both carry a minus sign:

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

Step 6: Multiplying both sides by \( -1 \) to clear the minus sign:

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

A product in a question usually has no 2 in front of it. The fix is to multiply and divide by 2, which leaves the value unchanged, and then use the matching formula.

(a) Express \( \sin 35^\circ \cos 25^\circ \) as a sum.

Step 1: Multiplying and dividing by 2 to make the form \( 2\sin A\cos B \):

\[ \sin 35^\circ \cos 25^\circ = \frac{1}{2}(2\sin 35^\circ \cos 25^\circ) \]

Step 2: Using the formula for \( 2\sin A\cos B \) with \( A = 35^\circ \) and \( B = 25^\circ \):

\[ = \frac{1}{2}[\sin(35^\circ + 25^\circ) + \sin(35^\circ - 25^\circ)] \]

Step 3: Adding and subtracting the angles:

\[ = \frac{1}{2}[\sin 60^\circ + \sin 10^\circ] \]

(b) Express \( \cos 25^\circ \cos 10^\circ \) as a sum.

Step 1: Multiplying and dividing by 2 to make the form \( 2\cos A\cos B \):

\[ \cos 25^\circ \cos 10^\circ = \frac{1}{2}(2\cos 25^\circ \cos 10^\circ) \]

Step 2: Using the formula for \( 2\cos A\cos B \) with \( A = 25^\circ \) and \( B = 10^\circ \):

\[ = \frac{1}{2}[\cos(25^\circ + 10^\circ) + \cos(25^\circ - 10^\circ)] \]

Step 3: Adding and subtracting the angles:

\[ = \frac{1}{2}[\cos 35^\circ + \cos 15^\circ] \]

(c) Express \( \sin 10A \sin 6A \) as a difference.

Step 1: Multiplying and dividing by 2 to make the form \( 2\sin A\sin B \):

\[ \sin 10A \sin 6A = \frac{1}{2}(2\sin 10A \sin 6A) \]

Step 2: Using \( 2\sin A\sin B = \cos(A-B) - \cos(A+B) \), with the difference first:

\[ = \frac{1}{2}[\cos(10A - 6A) - \cos(10A + 6A)] \]

Step 3: Simplifying the angles:

\[ = \frac{1}{2}[\cos 4A - \cos 16A] \]

Now read the product formulas from right to left. Each one then starts with a sum or difference and ends with a product, which is exactly the other direction of travel. There is one problem. The angles on the sum side are written as \( A+B \) and \( A-B \), but a real question gives plain angles such as 55° and 25°. So the sum and the difference are given names of their own.

Step 1: Writing the four product formulas the other way round:

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

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

\[ \cos(A+B) + \cos(A-B) = 2\cos A\cos B \]

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

Step 2: Naming the sum of the two angles C and their difference D:

\[ A + B = C \]

\[ A - B = D \]

Step 3: Adding these two equations. The B cancels, giving \( 2A = C + D \), so:

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

Step 4: Subtracting the second equation from the first. The A cancels, giving \( 2B = C - D \), so:

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

Step 5: Putting C, D and these values of A and B into the four formulas of Step 1. This gives the four sum formulas:

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

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

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

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

The last formula is often written with the minus sign in front instead. Since \( \sin(-x) = -\sin x \), turning \( D - C \) round into \( C - D \) brings a minus sign out to the front:

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

(a) Express \( \sin 55^\circ - \sin 25^\circ \) as a product.

Step 1: Using the formula for \( \sin C - \sin D \) with \( C = 55^\circ \) and \( D = 25^\circ \):

\[ \sin 55^\circ - \sin 25^\circ = 2\cos\frac{55^\circ + 25^\circ}{2}\sin\frac{55^\circ - 25^\circ}{2} \]

Step 2: Halving the sum, 80°, and the difference, 30°:

\[ = 2\cos 40^\circ \sin 15^\circ \]

(b) Express \( \cos 25^\circ + \cos 15^\circ \) as a product.

Step 1: Using the formula for \( \cos C + \cos D \) with \( C = 25^\circ \) and \( D = 15^\circ \):

\[ \cos 25^\circ + \cos 15^\circ = 2\cos\frac{25^\circ + 15^\circ}{2}\cos\frac{25^\circ - 15^\circ}{2} \]

Step 2: Halving the sum, 40°, and the difference, 10°:

\[ = 2\cos 20^\circ \cos 5^\circ \]

(c) Express \( \sin 8A - \sin 6A \) as a product.

Step 1: Using the formula for \( \sin C - \sin D \) with \( C = 8A \) and \( D = 6A \):

\[ \sin 8A - \sin 6A = 2\cos\frac{8A + 6A}{2}\sin\frac{8A - 6A}{2} \]

Step 2: Halving \( 14A \) and \( 2A \):

\[ = 2\cos 7A \sin A \]

To prove an identity, start from one side, usually the left-hand side (L.H.S.), and change it one step at a time until it becomes exactly the right-hand side (R.H.S.). Every step must be backed by a formula or a known exact value. The eight formulas give two kinds of move, and the expression in front of you tells you which one to make.

• Start from the side that looks more complicated.
• A sum or difference of two sines, or of two cosines, is turned into a product with a sum formula.
• A product of two ratios is multiplied and divided by 2, then turned into a sum with a product formula.
• If a sine and a cosine are mixed, one is changed into the other first, using \( \cos\theta = \sin(90^\circ - \theta) \).
• Special angles are replaced by their exact values, for example \( \sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}} \), and the result is simplified until it matches the R.H.S.

Prove that \( \frac{\cos 10^\circ + \sin 10^\circ}{\cos 10^\circ - \sin 10^\circ} = \cot 35^\circ \).

Step 1: The L.H.S. adds a cosine to a sine, and no sum formula does that. So the cosine is written with the complementary angle, using \( 10^\circ = 90^\circ - 80^\circ \):

\[ \text{L.H.S.} = \frac{\cos(90^\circ - 80^\circ) + \sin 10^\circ}{\cos(90^\circ - 80^\circ) - \sin 10^\circ} \]

Step 2: Since \( \cos(90^\circ - \theta) = \sin\theta \), both cosines become sines:

\[ = \frac{\sin 80^\circ + \sin 10^\circ}{\sin 80^\circ - \sin 10^\circ} \]

Step 3: Using \( \sin C + \sin D \) on the top and \( \sin C - \sin D \) on the bottom, with \( C = 80^\circ \) and \( D = 10^\circ \):

\[ = \frac{2\sin\frac{80^\circ + 10^\circ}{2}\cos\frac{80^\circ - 10^\circ}{2}}{2\cos\frac{80^\circ + 10^\circ}{2}\sin\frac{80^\circ - 10^\circ}{2}} \]

Step 4: Halving the sum, 90°, and the difference, 70°:

\[ = \frac{2\sin 45^\circ \cos 35^\circ}{2\cos 45^\circ \sin 35^\circ} \]

Step 5: Putting in \( \sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}} \). These are equal, so they cancel along with the 2s:

\[ = \frac{\cos 35^\circ}{\sin 35^\circ} \]

Step 6: Writing a cosine over a sine of the same angle as a cotangent:

\[ = \cot 35^\circ = \text{R.H.S.} \]

Prove that \( 2\cos(45^\circ + A)\sin(45^\circ - A) = (\sin A - \cos A)^2 \).

Step 1: The L.H.S. already has the form \( 2\cos A\sin B \), with first angle \( 45^\circ + A \) and second angle \( 45^\circ - A \):

\[ \text{L.H.S.} = 2\cos(45^\circ + A)\sin(45^\circ - A) \]

Step 2: Using \( 2\cos A\sin B = \sin(A+B) - \sin(A-B) \):

\[ \begin{aligned} = {} &\sin(45^\circ + A + 45^\circ - A) \\ &- \sin(45^\circ + A - 45^\circ + A) \end{aligned} \]

Step 3: Simplifying the angles. In the first the A terms cancel, and in the second the 45° terms cancel:

\[ = \sin 90^\circ - \sin 2A \]

Step 4: Putting in \( \sin 90^\circ = 1 \):

\[ = 1 - \sin 2A \]

Step 5: Writing 1 as \( \sin^2 A + \cos^2 A \) and \( \sin 2A \) as \( 2\sin A\cos A \):

\[ = \sin^2 A + \cos^2 A - 2\sin A\cos A \]

Step 6: Recognising the expansion of a perfect square, \( a^2 + b^2 - 2ab = (a - b)^2 \):

\[ = (\sin A - \cos A)^2 = \text{R.H.S.} \]

Prove that \( \sin\theta \sin(60^\circ + \theta)\sin(60^\circ - \theta) = \frac{1}{4}\sin 3\theta \).

A formula joins only two ratios at a time. With three, one is kept aside while the other two are joined.

Step 1: Keeping \( \sin\theta \) aside, and multiplying and dividing the other two by 2:

\[ \text{L.H.S.} = \frac{1}{2}\sin\theta\,[2\sin(60^\circ + \theta)\sin(60^\circ - \theta)] \]

Step 2: Using \( 2\sin A\sin B = \cos(A-B) - \cos(A+B) \). The difference of the angles is \( 2\theta \) and their sum is 120°, so:

\[ = \frac{1}{2}\sin\theta\,[\cos 2\theta - \cos 120^\circ] \]

Step 3: Putting in \( \cos 120^\circ = -\frac{1}{2} \):

\[ = \frac{1}{2}\sin\theta\,\left[\cos 2\theta + \frac{1}{2}\right] \]

Step 4: Multiplying out the bracket:

\[ = \frac{1}{2}\sin\theta\cos 2\theta + \frac{1}{4}\sin\theta \]

Step 5: Making the form \( 2\cos A\sin B \) in the first term by writing \( \frac{1}{2} \) as \( \frac{1}{4} \times 2 \):

\[ = \frac{1}{4}(2\cos 2\theta\sin\theta) + \frac{1}{4}\sin\theta \]

Step 6: Using \( 2\cos A\sin B = \sin(A+B) - \sin(A-B) \) with \( A = 2\theta \) and \( B = \theta \):

\[ = \frac{1}{4}[\sin 3\theta - \sin\theta] + \frac{1}{4}\sin\theta \]

Step 7: Opening the bracket:

\[ = \frac{1}{4}\sin 3\theta - \frac{1}{4}\sin\theta + \frac{1}{4}\sin\theta \]

Step 8: Cancelling the two \( \sin\theta \) terms:

\[ = \frac{1}{4}\sin 3\theta = \text{R.H.S.} \]

Prove that \( \frac{\cos\theta - \cos 2\theta + \cos 3\theta}{\sin\theta - \sin 2\theta + \sin 3\theta} = \cot 2\theta \).

Step 1: Rearranging so that the \( 3\theta \) and \( \theta \) terms sit together. Their half sum is \( 2\theta \), the same angle as the term left over:

\[ \text{L.H.S.} = \frac{\cos 3\theta + \cos\theta - \cos 2\theta}{\sin 3\theta + \sin\theta - \sin 2\theta} \]

Step 2: Using \( \cos C + \cos D \) on the top and \( \sin C + \sin D \) on the bottom, with \( C = 3\theta \) and \( D = \theta \). Half the sum is \( 2\theta \) and half the difference is \( \theta \):

\[ = \frac{2\cos 2\theta\cos\theta - \cos 2\theta}{2\sin 2\theta\cos\theta - \sin 2\theta} \]

Step 3: Taking out the common factor from the top and from the bottom:

\[ = \frac{\cos 2\theta\,(2\cos\theta - 1)}{\sin 2\theta\,(2\cos\theta - 1)} \]

Step 4: Cancelling the bracket \( (2\cos\theta - 1) \):

\[ = \frac{\cos 2\theta}{\sin 2\theta} \]

Step 5: Writing this as a cotangent:

\[ = \cot 2\theta = \text{R.H.S.} \]

Prove that \( \sin 20^\circ \sin 40^\circ \sin 80^\circ = \frac{\sqrt{3}}{8} \).

Step 1: Keeping \( \sin 20^\circ \) aside and joining 80° with 40°, because their sum is 120°, an angle with a known cosine:

\[ \text{L.H.S.} = \frac{1}{2}\sin 20^\circ\,[2\sin 80^\circ \sin 40^\circ] \]

Step 2: Using \( 2\sin A\sin B = \cos(A-B) - \cos(A+B) \):

\[ = \frac{1}{2}\sin 20^\circ\,[\cos 40^\circ - \cos 120^\circ] \]

Step 3: Putting in \( \cos 120^\circ = -\frac{1}{2} \):

\[ = \frac{1}{2}\sin 20^\circ\,\left[\cos 40^\circ + \frac{1}{2}\right] \]

Step 4: Multiplying out the bracket:

\[ = \frac{1}{2}\sin 20^\circ\cos 40^\circ + \frac{1}{4}\sin 20^\circ \]

Step 5: Making the form \( 2\cos A\sin B \) in the first term:

\[ = \frac{1}{4}(2\cos 40^\circ \sin 20^\circ) + \frac{1}{4}\sin 20^\circ \]

Step 6: Using \( 2\cos A\sin B = \sin(A+B) - \sin(A-B) \) with \( A = 40^\circ \) and \( B = 20^\circ \):

\[ = \frac{1}{4}[\sin 60^\circ - \sin 20^\circ] + \frac{1}{4}\sin 20^\circ \]

Step 7: Putting in \( \sin 60^\circ = \frac{\sqrt{3}}{2} \) and opening the bracket:

\[ = \frac{\sqrt{3}}{8} - \frac{1}{4}\sin 20^\circ + \frac{1}{4}\sin 20^\circ \]

Step 8: Cancelling the two \( \sin 20^\circ \) terms:

\[ = \frac{\sqrt{3}}{8} = \text{R.H.S.} \]

Compare this with Example 5. The pattern is identical: keep one sine aside, join the other two to reach 120°, and the unwanted term cancels at the end. In fact Example 7 is Example 5 with \( \theta = 20^\circ \), since \( 60^\circ + 20^\circ = 80^\circ \), \( 60^\circ - 20^\circ = 40^\circ \) and \( \frac{1}{4}\sin 60^\circ = \frac{\sqrt{3}}{8} \).

In the longer proofs the formulas themselves are easy. The real skill is deciding which two terms to join. These habits make that choice for you.

• Join angles whose sum or difference is a special angle such as 60°, 90° or 120°. With 20°, 40° and 80°, the pair 80° and 40° reaches 120°.
• Join terms so that a common factor appears. With \( \theta \), \( 2\theta \) and \( 3\theta \), joining \( \theta \) with \( 3\theta \) gives half sum \( 2\theta \), which matches the term left alone.
• When a fraction has the same pattern on the top and the bottom, transform both in the same way, then look for a factor to cancel.
• When a step leaves an extra term, keep going. In products of three ratios the extra term usually cancels in the last line.