Compound and multiple angles

Ask for sin⁡75°\sin75° and a calculator gives you a decimal instantly — but an exam wants the exact value, built from angles you already know like 45°45° and 30°30°. The compound-angle formulae are the machine that does this: they expand sin⁡(x±y)\sin(x\pm y), cos⁡(x±y)\cos(x\pm y) and tan⁡(x±y)\tan(x\pm y) into combinations of the separate ratios of xx and yy. Once you own these four formulae, the rest of the chapter falls out almost for free. Set y=xy=x and you get the double-angle formulae; write 3x=2x+x3x=2x+x and you get the triple-angle formulae; add or subtract pairs of the compound formulae and you get the sum-to-product identities. This block is the single most heavily examined chunk of ISC Class 11 trigonometry, and everything in it traces back to just four lines you can reconstruct on the spot if your memory slips.

The core insight

Every compound-, double-, triple-angle and sum-to-product identity in this chapter is a rearrangement of the four formulae sin⁡(x±y)=sin⁡xcos⁡y±cos⁡xsin⁡y\sin(x\pm y)=\sin x\cos y\pm\cos x\sin y and cos⁡(x±y)=cos⁡xcos⁡y∓sin⁡xsin⁡y\cos(x\pm y)=\cos x\cos y\mp\sin x\sin y.

The core idea

The four compound-angle formulae. These express the sine, cosine and tangent of a sum or difference of two angles in terms of the separate ratios of each angle:

sin⁡(x±y)=sin⁡xcos⁡y±cos⁡xsin⁡y\sin(x\pm y) = \sin x\cos y \pm \cos x\sin y
cos⁡(x±y)=cos⁡xcos⁡y∓sin⁡xsin⁡y\cos(x\pm y) = \cos x\cos y \mp \sin x\sin y
tan⁡(x±y)=tan⁡x±tan⁡y1∓tan⁡xtan⁡y\tan(x\pm y) = \frac{\tan x\pm\tan y}{1\mp\tan x\tan y}

where: xx and yy are the two angles being combined, and the upper sign on the left always matches the upper sign on the right — read sin⁡(x+y)\sin(x+y) with all the top signs, and sin⁡(x−y)\sin(x-y) with all the bottom signs.

Double-angle formulae — set y=xy=x. Substituting y=xy=x directly into the compound-angle formulae collapses two angles into one:

sin⁡2x=2sin⁡xcos⁡x,cos⁡2x=cos⁡2x−sin⁡2x=2cos⁡2x−1=1−2sin⁡2x,tan⁡2x=2tan⁡x1−tan⁡2x\sin 2x = 2\sin x\cos x, \qquad \cos 2x = \cos^2x-\sin^2x = 2\cos^2x-1 = 1-2\sin^2x, \qquad \tan 2x = \frac{2\tan x}{1-\tan^2x}

where: the three forms of cos⁡2x\cos2x are all equivalent (using sin⁡2x+cos⁡2x=1\sin^2x+\cos^2x=1 to swap between them) — pick whichever is expressed in the ratio you already know, to save a step.

Triple-angle formulae — write 3x=2x+x3x=2x+x. Expanding sin⁡(2x+x)\sin(2x+x) and cos⁡(2x+x)\cos(2x+x) with the compound-angle formula, then substituting the double-angle results, gives:

sin⁡3x=3sin⁡x−4sin⁡3x,cos⁡3x=4cos⁡3x−3cos⁡x\sin 3x = 3\sin x-4\sin^3x, \qquad \cos 3x = 4\cos^3x-3\cos x

Sum-to-product identities. Adding or subtracting pairs of compound-angle formulae converts a sum or difference of two ratios into a product — useful whenever a factor needs to cancel against a denominator elsewhere:

sin⁡α±sin⁡β=2sin⁡(α±β2)cos⁡(α∓β2),cos⁡α+cos⁡β=2cos⁡(α+β2)cos⁡(α−β2)\sin\alpha\pm\sin\beta = 2\sin\left(\frac{\alpha\pm\beta}{2}\right)\cos\left(\frac{\alpha\mp\beta}{2}\right), \qquad \cos\alpha+\cos\beta = 2\cos\left(\frac{\alpha+\beta}{2}\right)\cos\left(\frac{\alpha-\beta}{2}\right)
cos⁡α−cos⁡β=−2sin⁡(α+β2)sin⁡(α−β2)\cos\alpha-\cos\beta = -2\sin\left(\frac{\alpha+\beta}{2}\right)\sin\left(\frac{\alpha-\beta}{2}\right)

where: α\alpha and β\beta are any two angles; the reverse move — turning a product like 2sin⁡Acos⁡B2\sin A\cos B back into a sum sin⁡(A+B)+sin⁡(A−B)\sin(A+B)+\sin(A-B) — is called product-to-sum, and follows from the same four compound-angle formulae read the other way. Notice the minus sign carried by cos⁡α−cos⁡β\cos\alpha-\cos\beta: it is the one identity in this family whose right-hand side is a product of two sines rather than two cosines, and it is the form most often needed when an identity to be proved contains a difference of cosines in a denominator that must be cancelled against a numerator built from sines. There is a compact way to see why all of this — compound, double, triple and sum-to-product — is really "one formula wearing different hats." The compound-angle formulae are proved directly from the coordinates of points on the unit circle together with the Pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1; that derivation is beyond what you are asked to reproduce, so treat the four boxed results as a trusted starting point. From there, every other result in this chapter is pure algebra: substitution (y=xy=x for double angle), decomposition (3x=2x+x3x=2x+x for triple angle), or addition/subtraction of two instances of the same formula (for sum-to-product). None of it requires new trigonometric insight beyond the original four formulae — only careful bookkeeping of signs and squares.

Figure
A tree diagram showing the compound angle formulae at the root branching down into double angle then triple angle formulae and sideways into sum to product identities
One family, four formulae at the root sin(x±y), cos(x±y) tan(x±y) set y = x: double angle 3x = 2x + x: triple angle add/subtract pairs: sum-to-product substitute double-angle results into y=x, x pattern Memorise the root, rebuild the branches on demand. All results in this article are algebraic consequences of the two boxed formulae at the top.
Derivation ADeriving sin 3x from the compound-angle and double-angle formulae

The goal is to derive the identity sin⁡3x=3sin⁡x−4sin⁡3x\sin3x = 3\sin x-4\sin^3x starting from nothing but the compound-angle formula for sin⁡(x+y)\sin(x+y) and the double-angle results sin⁡2x=2sin⁡xcos⁡x\sin2x=2\sin x\cos x and cos⁡2x=1−2sin⁡2x\cos2x=1-2\sin^2x, showing that the triple-angle formula is not a fact to memorise blind but a two-line consequence of results you already have.

Step 1. Write 3x3x as a sum of a doubled angle and a single angle, 3x=2x+x3x=2x+x, and apply the compound-angle formula for sine:

sin⁡3x=sin⁡(2x+x)=sin⁡2xcos⁡x+cos⁡2xsin⁡x\sin3x = \sin(2x+x) = \sin2x\cos x+\cos2x\sin x

Step 2. Substitute the double-angle results for sin⁡2x\sin2x and cos⁡2x\cos2x into this expression:

sin⁡3x=(2sin⁡xcos⁡x)cos⁡x+(1−2sin⁡2x)sin⁡x=2sin⁡xcos⁡2x+sin⁡x−2sin⁡3x\sin3x = (2\sin x\cos x)\cos x + (1-2\sin^2x)\sin x = 2\sin x\cos^2x + \sin x - 2\sin^3x

Step 3. Replace cos⁡2x\cos^2x using the Pythagorean identity cos⁡2x=1−sin⁡2x\cos^2x=1-\sin^2x, so that only sin⁡x\sin x remains:

sin⁡3x=2sin⁡x(1−sin⁡2x)+sin⁡x−2sin⁡3x=2sin⁡x−2sin⁡3x+sin⁡x−2sin⁡3x\sin3x = 2\sin x(1-\sin^2x) + \sin x - 2\sin^3x = 2\sin x-2\sin^3x+\sin x-2\sin^3x

Step 4. Collect like terms to reach the final identity:

sin⁡3x=3sin⁡x−4sin⁡3x■\sin3x = 3\sin x - 4\sin^3x \qquad \blacksquare

Worked example

Problem. Given sin⁡x=35\sin x=\dfrac{3}{5} and cos⁡y=513\cos y=\dfrac{5}{13}, with xx and yy both acute, find sin⁡(x+y)\sin(x+y) and cos⁡2x\cos2x.

Step 1. Find the missing ratios using sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1; since both angles are acute, all ratios are positive:

cos⁡x=1−sin⁡2x=1−925=45,sin⁡y=1−cos⁡2y=1−25169=1213\cos x = \sqrt{1-\sin^2x} = \sqrt{1-\tfrac{9}{25}} = \frac{4}{5}, \qquad \sin y = \sqrt{1-\cos^2y} = \sqrt{1-\tfrac{25}{169}} = \frac{12}{13}

Step 2. Apply the compound-angle formula for sin⁡(x+y)\sin(x+y):

sin⁡(x+y)=sin⁡xcos⁡y+cos⁡xsin⁡y=(35)(513)+(45)(1213)=1565+4865=6365\sin(x+y) = \sin x\cos y + \cos x\sin y = \left(\frac{3}{5}\right)\left(\frac{5}{13}\right) + \left(\frac{4}{5}\right)\left(\frac{12}{13}\right) = \frac{15}{65}+\frac{48}{65} = \frac{63}{65}

Step 3. Apply the double-angle formula for cos⁡2x\cos2x, choosing the form written in sin⁡x\sin x since that ratio is already known:

cos⁡2x=1−2sin⁡2x=1−2(35)2=1−1825=725\cos2x = 1-2\sin^2x = 1-2\left(\frac{3}{5}\right)^2 = 1-\frac{18}{25} = \frac{7}{25}

So sin⁡(x+y)=6365\sin(x+y)=\dfrac{63}{65} and cos⁡2x=725\cos2x=\dfrac{7}{25}.

Competitive stretch

Using the same xx, find sin⁡3x\sin3x without re-deriving the identity — just plug sin⁡x=35\sin x=\tfrac{3}{5} into the standard result sin⁡3x=3sin⁡x−4sin⁡3x\sin3x=3\sin x-4\sin^3x:

sin⁡3x=3(35)−4(35)3=95−4⋅27125=95−108125=225125−108125=117125\sin3x = 3\left(\frac{3}{5}\right)-4\left(\frac{3}{5}\right)^3 = \frac{9}{5}-4\cdot\frac{27}{125} = \frac{9}{5}-\frac{108}{125} = \frac{225}{125}-\frac{108}{125} = \frac{117}{125}

sin⁡3x=117125\sin3x=\dfrac{117}{125}.

Common traps

  • Mixing up the sign in cos⁡(x±y)=cos⁡xcos⁡y∓sin⁡xsin⁡y\cos(x\pm y)=\cos x\cos y\mp\sin x\sin y — the sign flips relative to the input, opposite to the sine formula, which keeps the same sign.
  • Forgetting the denominator sign flip in tan⁡(x±y)=tan⁡x±tan⁡y1∓tan⁡xtan⁡y\tan(x\pm y)=\dfrac{\tan x\pm\tan y}{1\mp\tan x\tan y} — the denominator sign is opposite to the numerator sign.
  • Picking a form of cos⁡2x\cos2x that is not expressed in the ratio already known, adding an unnecessary extra step (as avoided in Step 3 of the worked example above).
  • Trying to recall sin⁡3x\sin3x or cos⁡3x\cos3x from memory under pressure instead of rebuilding it via 3x=2x+x3x=2x+x when memory is shaky — the derivation above takes under a minute once the compound- and double-angle formulae are secure.
  • Forgetting to check whether an angle is acute, obtuse, or in a specified quadrant before taking a square root such as cos⁡x=1−sin⁡2x\cos x=\sqrt{1-\sin^2x} — the square root only gives the positive value automatically; if the angle's quadrant were not given, both signs would need to be considered.
  • Applying a sum-to-product identity to sin⁡α+cos⁡β\sin\alpha+\cos\beta (a mixed sine-and-cosine sum) — the identities in this chapter only combine two sines together or two cosines together; a mixed expression must first be rewritten, for instance using cos⁡β=sin⁡(90°−β)\cos\beta=\sin(90°-\beta), before a sum-to-product identity can apply.

Ask the tutor: Want to practise deriving cos⁡3x\cos3x from the compound-angle formula the same way this article derived sin⁡3x\sin3x?

Do I need to memorise all three forms of cos 2x separately?

No — memorise just one, say cos⁡2x=1−2sin⁡2x\cos2x=1-2\sin^2x, and derive the other two on the spot using sin⁡2x+cos⁡2x=1\sin^2x+\cos^2x=1. Swap sin⁡2x=1−cos⁡2x\sin^2x=1-\cos^2x into it to get cos⁡2x=2cos⁡2x−1\cos2x=2\cos^2x-1, or start from cos⁡2x−sin⁡2x\cos^2x-\sin^2x directly. In practice, look at which ratio the question already gives you and pick the matching form — that is exactly what saved a step in the worked example above, where sin⁡x\sin x was known so the 1−2sin⁡2x1-2\sin^2x form was fastest.

Why does the sign flip in the cosine formula but not the sine formula?

It comes straight from expanding cos⁡(x+y)\cos(x+y) and cos⁡(x−y)\cos(x-y) on the unit circle and comparing: the cross terms in cosine's expansion involve sin⁡xsin⁡y\sin x\sin y, and replacing yy with −y-y flips the sign of that term because sin⁡(−y)=−sin⁡y\sin(-y)=-\sin y. In the sine formula the cross term is cos⁡xsin⁡y\cos x\sin y, and cos⁡(−y)=cos⁡y\cos(-y)=\cos y stays unchanged, so the sign does not flip there. The safest habit is not to memorise "which formula flips" as a separate fact, but to always substitute y→−yy\to-y mentally and track how sin⁡y\sin y and cos⁡y\cos y each behave.

Is it worth memorising the sum-to-product identities separately, or can I derive them each time?

Either works, but for speed under exam pressure it helps to memorise at least sin⁡α+sin⁡β=2sin⁡(α+β2)cos⁡(α−β2)\sin\alpha+\sin\beta=2\sin\left(\frac{\alpha+\beta}{2}\right)\cos\left(\frac{\alpha-\beta}{2}\right) as an anchor, since the other three sum-to-product forms follow the same pattern with sign changes. If your memory of the exact pattern is uncertain, you can always fall back on adding the compound-angle formulae for sin⁡(A+B)\sin(A+B) and sin⁡(A−B)\sin(A-B) and solving for the sum — slower, but it never fails, and it is good insurance the night before an exam.

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