Multiple Angles

Suppose the value of sin 15° is known and you are asked for sin 30° or sin 45°. You do not need a new table. Look at how the angles are related. The angle 30° is exactly twice 15°, and 45° is exactly three times 15°. So each of these values can be written using the angle 15° itself:

\[ \sin 30^\circ = \sin (2 \times 15^\circ) \]

\[ \sin 45^\circ = \sin (3 \times 15^\circ) \]

If we had a rule that gives sin 2A or sin 3A from the ratios of A, then one known angle would unlock the ratios of its double and its triple. Building those rules is the whole aim of this topic.

The ratios of 2A and 3A are not new facts to memorise blindly. They come straight out of the compound angle formulas, which give the ratios of a sum of two angles:

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

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

\[ \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \]

The trick is simple. Put B equal to A. Then A + B becomes 2A, and every formula above turns into a formula for the double angle. For 3A, we write 3A as A + 2A and use the formulas once more.

We begin with sin 2A. Each step below says what is done, and the line under it shows the result.

Step 1: Writing 2A as the sum of two equal angles:

\[ \sin 2A = \sin(A + A) \]

Step 2: Using the formula for sin(A + B) with B replaced by A:

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

Step 3: Adding the two terms, which are exactly the same:

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

Now cos 2A. This one is special because it can be written in three different ways, and each way is useful in a different problem.

Step 1: Writing 2A as A + A and using the formula for cos(A + B) with B replaced by A:

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

Step 2: Writing each product as a square. This is the first form:

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

Step 3: Replacing \( \cos^2 A \) by \( 1 - \sin^2 A \), which is allowed because \( \sin^2 A + \cos^2 A = 1 \):

\[ \cos 2A = 1 - \sin^2 A - \sin^2 A \]

Step 4: Collecting the two sine terms. This is the second form, written in sin A only:

\[ \cos 2A = 1 - 2\sin^2 A \]

Step 5: Going back to the first form and this time replacing \( \sin^2 A \) by \( 1 - \cos^2 A \):

\[ \cos 2A = \cos^2 A - (1 - \cos^2 A) \]

Step 6: Removing the bracket and collecting the cosine terms. This is the third form, written in cos A only:

\[ \cos 2A = 2\cos^2 A - 1 \]

Finally tan 2A, from the formula for tan(A + B).

Step 1: Writing 2A as A + A and putting B = A in the tangent formula:

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

Step 2: Adding the top and writing the product at the bottom as a square:

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

Sometimes a question gives you only tan A. Then it is handy to have sin 2A and cos 2A written using tan A alone. The idea behind both results is the same: write the expression as a fraction, then divide the top and the bottom by \( \cos^2 A \), because \( \frac{\sin A}{\cos A} = \tan A \).

Step 1: Writing sin 2A as a fraction over 1:

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

Step 2: Replacing the 1 by \( \cos^2 A + \sin^2 A \), since that sum is always 1:

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

Step 3: Dividing the top and the bottom by \( \cos^2 A \). On top, \( \frac{\sin A \cos A}{\cos^2 A} = \tan A \); at the bottom, \( \frac{\sin^2 A}{\cos^2 A} = \tan^2 A \):

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

There is another way to reach the same result, using secant.

Step 1: Multiplying and dividing \( 2\sin A \cos A \) by cos A:

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

Step 2: Writing \( \frac{\sin A}{\cos A} \) as tan A and \( \cos^2 A \) as \( \frac{1}{\sec^2 A} \):

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

Step 3: Using \( \sec^2 A = 1 + \tan^2 A \):

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

Now cos 2A in terms of tan A.

Step 1: Writing \( \cos^2 A - \sin^2 A \) over \( \cos^2 A + \sin^2 A \), which is the same as dividing by 1:

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

Step 2: Dividing the top and the bottom by \( \cos^2 A \):

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

For the triple angle we split 3A as A + 2A. Then the compound angle formula brings in the ratios of 2A, and we replace them with the results already found. We choose each replacement so that the final answer uses only one ratio of A.

Step 1: Splitting the angle and using the formula for sin(A + B) with B = 2A:

\[ \sin 3A = \sin A \cos 2A + \cos A \sin 2A \]

Step 2: Putting \( \cos 2A = 1 - 2\sin^2 A \) (the sine form, because we want sin A only) and \( \sin 2A = 2\sin A \cos A \):

\[ \sin 3A = \sin A(1 - 2\sin^2 A) + 2\sin A \cos^2 A \]

Step 3: Replacing \( \cos^2 A \) by \( 1 - \sin^2 A \) and opening both brackets:

\[ \sin 3A = \sin A - 2\sin^3 A + 2\sin A - 2\sin^3 A \]

Step 4: Collecting like terms:

\[ \sin 3A = 3\sin A - 4\sin^3 A \]

The cosine of 3A is found in the same way, this time keeping everything in cos A.

Step 1: Using the formula for cos(A + B) with B = 2A:

\[ \cos 3A = \cos A \cos 2A - \sin A \sin 2A \]

Step 2: Putting \( \cos 2A = 2\cos^2 A - 1 \) (the cosine form) and \( \sin 2A = 2\sin A \cos A \):

\[ \cos 3A = 2\cos^3 A - \cos A - 2\sin^2 A \cos A \]

Step 3: Replacing \( \sin^2 A \) by \( 1 - \cos^2 A \) and opening the bracket:

\[ \cos 3A = 2\cos^3 A - \cos A - 2\cos A + 2\cos^3 A \]

Step 4: Collecting like terms:

\[ \cos 3A = 4\cos^3 A - 3\cos A \]

Last comes tan 3A. To keep each line short, write t for tan A while working and put tan A back at the end.

Step 1: Using the formula for tan(A + B) with B = 2A:

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

Step 2: Putting \( \tan 2A = \frac{2t}{1 - t^2} \):

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

Step 3: Multiplying the top and the bottom by \( 1 - t^2 \) to clear the small fractions:

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

Step 4: Collecting like terms and writing tan A in place of t:

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

Here are all the results of this topic together. Every question that follows uses one or more of them, so learn them well enough to write them out without looking.

Example 1. If \( \sin\theta = \frac{8}{17} \) and θ is acute, find sin 2θ, cos 2θ and tan 2θ.

Every double angle formula for sine needs both sin θ and cos θ, so the first job is to find cos θ. A right triangle with the side opposite θ equal to 8 and the hypotenuse equal to 17 shows it at once.

Step 1: Finding cos θ from \( \sin^2\theta + \cos^2\theta = 1 \):

\[ \cos\theta = \sqrt{1 - \frac{64}{289}} = \sqrt{\frac{225}{289}} = \frac{15}{17} \]

Step 2: Using \( \sin 2\theta = 2\sin\theta\cos\theta \):

\[ \sin 2\theta = 2 \times \frac{8}{17} \times \frac{15}{17} = \frac{240}{289} \]

Step 3: Using \( \cos 2\theta = \cos^2\theta - \sin^2\theta \):

\[ \cos 2\theta = \frac{225}{289} - \frac{64}{289} = \frac{161}{289} \]

Step 4: Dividing sin 2θ by cos 2θ to get tan 2θ:

\[ \tan 2\theta = \frac{240}{289} \times \frac{289}{161} = \frac{240}{161} \]

So \( \sin 2\theta = \frac{240}{289} \), \( \cos 2\theta = \frac{161}{289} \) and \( \tan 2\theta = \frac{240}{161} \). You can check the last answer another way. Since \( \tan\theta = \frac{8}{15} \), the formula \( \frac{2\tan\theta}{1 - \tan^2\theta} \) gives \( \frac{16}{15} \div \frac{161}{225} = \frac{240}{161} \), the same value.

Example 2. If \( \cos\theta = \frac{1}{2} \), find sin 3θ and cos 3θ.

Step 1: Finding sin θ from \( \sin^2\theta + \cos^2\theta = 1 \):

\[ \sin\theta = \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \]

Step 2: Putting this into \( \sin 3\theta = 3\sin\theta - 4\sin^3\theta \). Since \( \left(\frac{\sqrt{3}}{2}\right)^3 = \frac{3\sqrt{3}}{8} \), the second term is \( 4 \times \frac{3\sqrt{3}}{8} = \frac{3\sqrt{3}}{2} \):

\[ \sin 3\theta = \frac{3\sqrt{3}}{2} - \frac{3\sqrt{3}}{2} = 0 \]

Step 3: Putting \( \cos\theta = \frac{1}{2} \) into \( \cos 3\theta = 4\cos^3\theta - 3\cos\theta \):

\[ \cos 3\theta = 4 \times \frac{1}{8} - \frac{3}{2} = \frac{1}{2} - \frac{3}{2} = -1 \]

So sin 3θ = 0 and cos 3θ = -1. This makes sense: cos θ = 1/2 means θ = 60°, so 3θ = 180°, and sin 180° = 0 while cos 180° = -1. Checking an answer against a known angle like this is a good habit.

Useful habits when a multiple angle appears in a proof:

• Change every 2A or 3A ratio into ratios of A, unless the other side is also written in 2A.
• Pick the form of cos 2A that cancels a 1 already present: next to 1 + cos 2A use \( 2\cos^2 A - 1 \), next to 1 - cos 2A use \( 1 - 2\sin^2 A \).
• Look for a common factor to cancel once everything is in single angles.
• If the expression contains only tan or cot, stay with tan or cot.

Example 3. Prove that \( \cot 3A = \frac{\cot^3 A - 3\cot A}{3\cot^2 A - 1} \).

Step 1: Writing cot 3A as the reciprocal of tan 3A and using the tan 3A formula:

\[ \text{L.H.S.} = \frac{1}{\tan 3A} = \frac{1 - 3\tan^2 A}{3\tan A - \tan^3 A} \]

Step 2: Replacing tan A by \( \frac{1}{\cot A} \) everywhere:

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

Step 3: Multiplying the top and the bottom by \( \cot^3 A \) to clear the fractions:

\[ = \frac{\cot^3 A - 3\cot A}{3\cot^2 A - 1} = \text{R.H.S.} \]

Example 4. Prove that \( \tan A = \pm\sqrt{\frac{1 - \cos 2A}{1 + \cos 2A}} \). Here we start from a known identity and rearrange it until tan A stands alone.

Step 1: Starting from cos 2A in terms of tan A:

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

Step 2: Cross multiplying and opening the bracket:

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

Step 3: Moving the tan terms to the left and cos 2A to the right:

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

Step 4: Taking \( \tan^2 A \) out as a common factor:

\[ \tan^2 A (1 + \cos 2A) = 1 - \cos 2A \]

Step 5: Dividing by \( 1 + \cos 2A \) and taking the square root. The ± sign is needed because a square root of \( \tan^2 A \) may be positive or negative:

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

Example 5. Prove that \( \frac{1 + \cos A + \cos 2A}{\sin A + \sin 2A} = \cot A \).

Step 1: Replacing cos 2A by \( 2\cos^2 A - 1 \), chosen because its -1 cancels the 1, and sin 2A by \( 2\sin A \cos A \):

\[ \text{L.H.S.} = \frac{1 + \cos A + 2\cos^2 A - 1}{\sin A + 2\sin A \cos A} \]

Step 2: Cancelling 1 and -1, then taking out cos A on top and sin A at the bottom:

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

Step 3: Cancelling the common factor \( 1 + 2\cos A \):

\[ = \frac{\cos A}{\sin A} = \cot A = \text{R.H.S.} \]

Example 6. Prove that \( \frac{1 + \tan^2(\frac{\pi}{4} - A)}{1 - \tan^2(\frac{\pi}{4} - A)} = \operatorname{cosec} 2A \). Notice that the left side is the fraction \( \frac{1 - \tan^2 x}{1 + \tan^2 x} \) turned upside down, with \( x = \frac{\pi}{4} - A \). That fraction is cos 2x.

Step 1: Writing the left side as 1 over cos 2x with \( x = \frac{\pi}{4} - A \):

\[ \text{L.H.S.} = \frac{1}{\cos 2(\frac{\pi}{4} - A)} \]

Step 2: Since \( \frac{\pi}{4} = 45^\circ \), the angle \( 2(45^\circ - A) \) is \( 90^\circ - 2A \):

\[ = \frac{1}{\cos(90^\circ - 2A)} \]

Step 3: Using \( \cos(90^\circ - \theta) = \sin\theta \):

\[ = \frac{1}{\sin 2A} = \operatorname{cosec} 2A = \text{R.H.S.} \]

High powers such as \( \cos^6\phi \) can be brought down to the cosine of a multiple angle. The tools are algebra you already know, \( a^3 + b^3 = (a + b)(a^2 - ab + b^2) \) and \( a^2 + b^2 = (a + b)^2 - 2ab \), together with \( \sin^2\phi + \cos^2\phi = 1 \).

Example 7. Prove that \( \cos^6\phi + \sin^6\phi = \frac{1}{8}(5 + 3\cos 4\phi) \).

Step 1: Writing the left side as a sum of two cubes, with \( a = \cos^2\phi \) and \( b = \sin^2\phi \):

\[ \text{L.H.S.} = (\cos^2\phi)^3 + (\sin^2\phi)^3 \]

Step 2: Using \( a^3 + b^3 = (a + b)(a^2 - ab + b^2) \). Here \( a + b = 1 \), so only the second bracket is left:

\[ = \cos^4\phi + \sin^4\phi - \cos^2\phi \sin^2\phi \]

Step 3: Writing \( a^2 + b^2 \) as \( (a + b)^2 - 2ab \), where \( a + b = 1 \):

\[ = 1 - 3\sin^2\phi \cos^2\phi \]

Step 4: Writing 3 as \( \frac{3}{4} \times 4 \) so that \( 2\sin\phi\cos\phi \) appears, which is sin 2φ:

\[ = 1 - \frac{3}{4}(2\sin\phi\cos\phi)^2 = 1 - \frac{3}{4}\sin^2 2\phi \]

Step 5: Using \( \cos 2x = 1 - 2\sin^2 x \) with \( x = 2\phi \), so that \( \sin^2 2\phi = \frac{1 - \cos 4\phi}{2} \):

\[ = 1 - \frac{3(1 - \cos 4\phi)}{8} \]

Step 6: Writing over the common denominator 8 and simplifying the top:

\[ = \frac{8 - 3 + 3\cos 4\phi}{8} = \frac{1}{8}(5 + 3\cos 4\phi) \]

This is the R.H.S., so the identity is proved. Notice how step 5 used a double angle formula with 2φ in place of A. The identities hold for any angle, so A may stand for 2φ, 4θ or \( 45^\circ - A \) just as well.