Submultiple Angles

You know that \( \sin 60^\circ = \frac{\sqrt{3}}{2} \). Suppose that value is all you have, with no table and no calculator. Can you still find \( \sin 30^\circ \)? The key is to notice that 30° is exactly half of 60°. If we have a formula that connects an angle with its half, one value leads straight to the other.

The same idea reaches angles that are not in our standard table at all. The angle 15° is half of 30°, and the angle 22½° is half of 45°, so we may write \( \sin 15^\circ = \sin\frac{30^\circ}{2} \) and \( \sin 22\frac{1}{2}^\circ = \sin\frac{45^\circ}{2} \). Since we already know the ratios of 30° and 45°, a formula linking an angle with its half will give us the ratios of 15° and 22½° as well. Such formulas are what this topic builds.

Take any angle A and cut it into equal parts. Two equal parts give two angles of A/2 each. Three equal parts give three angles of A/3 each. Four equal parts give A/4 each, and so on. Each of these parts is called a submultiple angle of A.

For example, 30° is a submultiple angle of 60° because 60° divided by 2 is 30°. The angle 20° is also a submultiple angle of 60°, since 60° divided by 3 is 20°, and 15° is one too, since 60° divided by 4 is 15°. But 25° is not a submultiple angle of 60°, because 60° cannot be split into equal parts of 25°.

Compare this with multiple angles, which you met through the formulas for 2A and 3A. There we started from A and went UP to 2A or 3A. Here we start from A and go DOWN to A/2 or A/3. In fact every submultiple angle identity is a multiple angle identity in disguise: A is twice A/2 and three times A/3.

We start with sin A. The compound angle formula \( \sin(A + B) = \sin A\cos B + \cos A\sin B \) is true for any two angles, so we may use it with both angles equal to A/2.

Step 1: Writing A as the sum of its two halves:

\[ \sin A = \sin\left(\frac{A}{2} + \frac{A}{2}\right) \]

Step 2: Applying the formula for sin(A + B) with both angles equal to A/2:

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

Step 3: The two terms are equal, so adding them gives twice one term:

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

There is a quicker way to reach the same result. We know \( \sin 2A = 2\sin A\cos A \) for every angle A. Put A/2 in place of A. Then 2A becomes \( 2 \times \frac{A}{2} \), which is just A:

\[ \sin A = \sin\left(2\times\frac{A}{2}\right) = 2\sin\frac{A}{2}\cos\frac{A}{2} \]

This replacement trick works for every double angle formula. From \( \cos 2A = \cos^2 A - \sin^2 A \), putting A/2 in place of A gives the first form of cos A:

\[ \cos A = \cos^2\frac{A}{2} - \sin^2\frac{A}{2} \]

Replacing \( \cos^2\frac{A}{2} \) by \( 1 - \sin^2\frac{A}{2} \) leaves only sine, and replacing \( \sin^2\frac{A}{2} \) by \( 1 - \cos^2\frac{A}{2} \) leaves only cosine. This gives two more forms:

\[ \cos A = 1 - 2\sin^2\frac{A}{2} \]

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

From \( \tan 2A = \frac{2\tan A}{1 - \tan^2 A} \) we get tan A in the same way:

\[ \tan A = \frac{2\tan\frac{A}{2}}{1 - \tan^2\frac{A}{2}} \]

Sometimes only \( \tan\frac{A}{2} \) is known, so it helps to write sin A and cos A using tan alone. Here is how it is done for sin A.

Step 1: Dividing by 1, written as \( \cos^2\frac{A}{2} + \sin^2\frac{A}{2} \), which changes nothing:

\[ \sin A = \frac{2\sin\frac{A}{2}\cos\frac{A}{2}}{\cos^2\frac{A}{2} + \sin^2\frac{A}{2}} \]

Step 2: Dividing the numerator and the denominator by \( \cos^2\frac{A}{2} \), so that each sine over cosine becomes a tangent:

\[ \sin A = \frac{2\tan\frac{A}{2}}{1 + \tan^2\frac{A}{2}} \]

Doing the same to \( \cos A = \cos^2\frac{A}{2} - \sin^2\frac{A}{2} \) gives cos A in terms of tan:

\[ \cos A = \frac{1 - \tan^2\frac{A}{2}}{1 + \tan^2\frac{A}{2}} \]

Collected together, these are the seven results for A in terms of A/2:

• \( \sin A = 2\sin\frac{A}{2}\cos\frac{A}{2} \)
• \( \cos A = \cos^2\frac{A}{2} - \sin^2\frac{A}{2} \)
• \( \cos A = 1 - 2\sin^2\frac{A}{2} \)
• \( \cos A = 2\cos^2\frac{A}{2} - 1 \)
• \( \tan A = \frac{2\tan\frac{A}{2}}{1 - \tan^2\frac{A}{2}} \)
• \( \sin A = \frac{2\tan\frac{A}{2}}{1 + \tan^2\frac{A}{2}} \)
• \( \cos A = \frac{1 - \tan^2\frac{A}{2}}{1 + \tan^2\frac{A}{2}} \)

The triple angle formulas can be turned round in exactly the same way. Because \( 3 \times \frac{A}{3} = A \), putting A/3 in place of A in each triple angle formula writes a ratio of A in terms of a ratio of A/3.

Step 1: Writing A as three times A/3 and using \( \sin 3A = 3\sin A - 4\sin^3 A \):

\[ \sin A = \sin\left(3\times\frac{A}{3}\right) \]

Step 2: Replacing A by A/3 on the right side of the triple angle formula:

\[ \sin A = 3\sin\frac{A}{3} - 4\sin^3\frac{A}{3} \]

In the same way, \( \cos 3A = 4\cos^3 A - 3\cos A \) gives cos A:

\[ \cos A = 4\cos^3\frac{A}{3} - 3\cos\frac{A}{3} \]

And \( \tan 3A = \frac{3\tan A - \tan^3 A}{1 - 3\tan^2 A} \) gives tan A:

\[ \tan A = \frac{3\tan\frac{A}{3} - \tan^3\frac{A}{3}}{1 - 3\tan^2\frac{A}{3}} \]

Every identity on the left below is a multiple angle identity. The identity beside it on the right is the same statement with A replaced by A/2 or A/3. Read across each row and you will see that nothing changes except the angle.

If \( \sin\frac{\theta}{2} = \frac{4}{5} \), find the values of sin θ, cos θ and tan θ.

Solution. The formula for sin θ needs both \( \sin\frac{\theta}{2} \) and \( \cos\frac{\theta}{2} \), so we find the cosine first.

Step 1: Finding \( \cos\frac{\theta}{2} \) from \( \sin^2 x + \cos^2 x = 1 \):

\[ \cos\frac{\theta}{2} = \sqrt{1 - \left(\frac{4}{5}\right)^2} = \sqrt{1 - \frac{16}{25}} \]

Step 2: Subtracting under the root and taking the square root:

\[ \cos\frac{\theta}{2} = \sqrt{\frac{9}{25}} = \frac{3}{5} \]

Step 3: Using \( \sin\theta = 2\sin\frac{\theta}{2}\cos\frac{\theta}{2} \):

\[ \sin\theta = 2\times\frac{4}{5}\times\frac{3}{5} = \frac{24}{25} \]

Step 4: Using \( \cos\theta = \cos^2\frac{\theta}{2} - \sin^2\frac{\theta}{2} \):

\[ \cos\theta = \frac{9}{25} - \frac{16}{25} = -\frac{7}{25} \]

Step 5: Dividing sin θ by cos θ to get tan θ:

\[ \tan\theta = \frac{24}{25}\div\left(-\frac{7}{25}\right) = -\frac{24}{7} \]

So \( \sin\theta = \frac{24}{25} \), \( \cos\theta = -\frac{7}{25} \) and \( \tan\theta = -\frac{24}{7} \). The negative cosine is not an error. Here θ/2 is a little more than 53°, so θ is a little more than 106°, an obtuse angle, and the cosine of an obtuse angle is negative.

If \( \sin\frac{A}{3} = \frac{1}{2} \), find the value of cos A.

Solution. The formula \( \cos A = 4\cos^3\frac{A}{3} - 3\cos\frac{A}{3} \) needs the cosine of A/3, so we find that first.

Step 1: Finding \( \cos\frac{A}{3} \) from the sine:

\[ \cos\frac{A}{3} = \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \]

Step 2: Substituting into the formula for cos A:

\[ \cos A = 4\left(\frac{\sqrt{3}}{2}\right)^3 - 3\times\frac{\sqrt{3}}{2} \]

Step 3: Cubing, since \( (\sqrt{3})^3 = 3\sqrt{3} \) and \( 2^3 = 8 \):

\[ \cos A = 4\times\frac{3\sqrt{3}}{8} - \frac{3\sqrt{3}}{2} \]

Step 4: Simplifying the first term and subtracting:

\[ \cos A = \frac{3\sqrt{3}}{2} - \frac{3\sqrt{3}}{2} = 0 \]

So cos A = 0. This agrees with what we know: sin(A/3) = 1/2 means A/3 = 30°, so A = 90°, and cos 90° = 0.

If \( \sin 30^\circ = \frac{1}{2} \), prove that \( \sin 15^\circ = \frac{\sqrt{3} - 1}{2\sqrt{2}} \).

Solution. Since 15° is half of 30°, we use a cos 30° formula that contains \( \sin^2 15^\circ \), namely \( \cos 30^\circ = 1 - 2\sin^2 15^\circ \).

Step 1: Finding cos 30° from sin 30°:

\[ \cos 30^\circ = \sqrt{1 - \frac{1}{4}} = \frac{\sqrt{3}}{2} \]

Step 2: Writing 30° as 2 × 15° and using cos 2A = 1 − 2 sin²A:

\[ 1 - 2\sin^2 15^\circ = \frac{\sqrt{3}}{2} \]

Step 3: Moving \( 2\sin^2 15^\circ \) to one side and the numbers to the other:

\[ 2\sin^2 15^\circ = 1 - \frac{\sqrt{3}}{2} = \frac{2 - \sqrt{3}}{2} \]

Step 4: Dividing both sides by 2:

\[ \sin^2 15^\circ = \frac{2 - \sqrt{3}}{4} \]

Step 5: Multiplying the numerator and the denominator by 2, so the top can become a perfect square:

\[ \sin^2 15^\circ = \frac{4 - 2\sqrt{3}}{8} = \frac{3 + 1 - 2\sqrt{3}}{8} \]

Step 6: Recognising the numerator as \( (\sqrt{3})^2 - 2\times\sqrt{3}\times 1 + 1^2 = (\sqrt{3} - 1)^2 \), and 8 as \( (2\sqrt{2})^2 \):

\[ \sin^2 15^\circ = \frac{(\sqrt{3} - 1)^2}{(2\sqrt{2})^2} \]

Step 7: Taking the square root of both sides:

\[ \sin 15^\circ = \pm\frac{\sqrt{3} - 1}{2\sqrt{2}} \]

Since 15° is an acute angle, sin 15° is positive. Hence \( \sin 15^\circ = \frac{\sqrt{3} - 1}{2\sqrt{2}} \), which was to be proved.

Prove that \( \frac{1 + \cos\theta}{\sin\theta} = \cot\frac{\theta}{2} \).

Solution. The right side is in θ/2, so we change the left side into θ/2 as well. Of the three forms of cos θ, we choose \( 2\cos^2\frac{\theta}{2} - 1 \) because its −1 will cancel the +1 already in the numerator.

Step 1: Writing cos θ and sin θ in terms of θ/2:

\[ \text{LHS} = \frac{1 + 2\cos^2\frac{\theta}{2} - 1}{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}} \]

Step 2: Cancelling 1 and −1 in the numerator:

\[ \text{LHS} = \frac{2\cos^2\frac{\theta}{2}}{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}} \]

Step 3: Cancelling the common factor \( 2\cos\frac{\theta}{2} \):

\[ \text{LHS} = \frac{\cos\frac{\theta}{2}}{\sin\frac{\theta}{2}} = \cot\frac{\theta}{2} = \text{RHS} \]

Hence proved.

Prove that \( \frac{1 + \sin\theta - \cos\theta}{1 + \sin\theta + \cos\theta} = \tan\frac{\theta}{2} \).

Solution. The numerator has 1 − cos θ in it, and the denominator has 1 + cos θ. We use \( \cos\theta = 1 - 2\sin^2\frac{\theta}{2} \) on top and \( \cos\theta = 2\cos^2\frac{\theta}{2} - 1 \) underneath, so that the 1s disappear in both. The numerator and denominator are long, so we simplify each on its own first.

Step 1: Writing the numerator in terms of θ/2, since \( 1 - \cos\theta = 2\sin^2\frac{\theta}{2} \):

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

Step 2: Taking out the common factor \( 2\sin\frac{\theta}{2} \):

\[ = 2\sin\frac{\theta}{2}\left(\sin\frac{\theta}{2} + \cos\frac{\theta}{2}\right) \]

Step 3: Writing the denominator in terms of θ/2, since \( 1 + \cos\theta = 2\cos^2\frac{\theta}{2} \):

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

Step 4: Taking out the common factor \( 2\cos\frac{\theta}{2} \):

\[ = 2\cos\frac{\theta}{2}\left(\cos\frac{\theta}{2} + \sin\frac{\theta}{2}\right) \]

Step 5: Dividing the numerator by the denominator. The brackets are the same, so they cancel, and so does the 2:

\[ \text{LHS} = \frac{\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}} = \tan\frac{\theta}{2} = \text{RHS} \]

Hence proved.

Prove that \( \sec A - \tan A = \tan\left(\frac{\pi}{4} - \frac{A}{2}\right) \).

Solution. This time the right side is the harder one, so we start there and work towards the left side. Remember that π/4 is 45° and \( \tan\frac{\pi}{4} = 1 \).

Step 1: Expanding with the formula for tan(A − B):

\[ \text{RHS} = \frac{\tan\frac{\pi}{4} - \tan\frac{A}{2}}{1 + \tan\frac{\pi}{4}\tan\frac{A}{2}} = \frac{1 - \tan\frac{A}{2}}{1 + \tan\frac{A}{2}} \]

Step 2: Writing tan as sine over cosine and multiplying the numerator and the denominator by \( \cos\frac{A}{2} \):

\[ \text{RHS} = \frac{\cos\frac{A}{2} - \sin\frac{A}{2}}{\cos\frac{A}{2} + \sin\frac{A}{2}} \]

Step 3: Multiplying the numerator and the denominator by \( \cos\frac{A}{2} - \sin\frac{A}{2} \), which makes the denominator a difference of two squares:

\[ \text{RHS} = \frac{\left(\cos\frac{A}{2} - \sin\frac{A}{2}\right)^2}{\cos^2\frac{A}{2} - \sin^2\frac{A}{2}} \]

Step 4: Expanding the square on top. The squares add up to 1, and \( 2\sin\frac{A}{2}\cos\frac{A}{2} = \sin A \), while the denominator is cos A:

\[ \text{RHS} = \frac{1 - 2\sin\frac{A}{2}\cos\frac{A}{2}}{\cos A} = \frac{1 - \sin A}{\cos A} \]

Step 5: Splitting the fraction into two:

\[ \text{RHS} = \frac{1}{\cos A} - \frac{\sin A}{\cos A} = \sec A - \tan A = \text{LHS} \]

Hence proved.

The same identities also run the other way. If a ratio of A is given and a ratio of A/2 is asked for, we rearrange the formula. From \( \cos A = 1 - 2\sin^2\frac{A}{2} \) and \( \cos A = 2\cos^2\frac{A}{2} - 1 \) we get two useful results:

\[ \sin^2\frac{A}{2} = \frac{1 - \cos A}{2} \]

\[ \cos^2\frac{A}{2} = \frac{1 + \cos A}{2} \]

For example, suppose \( \cos A = \frac{1}{9} \) and A is acute. We find sin A/2 and cos A/2.

Step 1: Substituting cos A into the formula for the square of the sine:

\[ \sin^2\frac{A}{2} = \frac{1 - \frac{1}{9}}{2} = \frac{8}{18} = \frac{4}{9} \]

Step 2: Taking the positive square root, because A/2 is acute:

\[ \sin\frac{A}{2} = \frac{2}{3} \]

Step 3: Substituting cos A into the formula for the square of the cosine:

\[ \cos^2\frac{A}{2} = \frac{1 + \frac{1}{9}}{2} = \frac{10}{18} = \frac{5}{9} \]

Step 4: Taking the positive square root:

\[ \cos\frac{A}{2} = \frac{\sqrt{5}}{3} \]

A quick check: \( \left(\frac{2}{3}\right)^2 + \left(\frac{\sqrt{5}}{3}\right)^2 = \frac{4}{9} + \frac{5}{9} = 1 \), as it must be. If tan A is given instead, put \( t = \tan\frac{A}{2} \) in \( \tan A = \frac{2t}{1 - t^2} \) and solve the quadratic equation in t.

Most mistakes in this topic come from picking a correct formula that does not help. These habits make the choice easier:

• When one ratio of A/2 is given, find the other ratio of A/2 first from \( \sin^2 x + \cos^2 x = 1 \), then use the formulas.
• When an expression contains 1 + cos A, write cos A as \( 2\cos^2\frac{A}{2} - 1 \) so that the 1 cancels.
• When an expression contains 1 − cos A, write cos A as \( 1 - 2\sin^2\frac{A}{2} \) so that the 1 cancels.
• When an expression contains 1 + sin A or 1 − sin A, write 1 as \( \cos^2\frac{A}{2} + \sin^2\frac{A}{2} \) and sin A as \( 2\sin\frac{A}{2}\cos\frac{A}{2} \); the result is a perfect square.
• When tan A/2 is given, use the three tan forms for sin A, cos A and tan A.
• When A/3 appears, use the triple angle formulas with A replaced by A/3.
• In a proof, start from the more complicated side and change it until it looks like the other side.