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Sum & Difference Identities

You already know the exact values of sine and cosine at 30∘30^\circ, 45∘45^\circ, and 60∘60^\circ. The sum and difference identities let you reach far past that short list. Since 15∘=45∘−30∘15^\circ = 45^\circ - 30^\circ and 105∘=60∘+45∘105^\circ = 60^\circ + 45^\circ, any angle you can build by adding or subtracting two special angles has an exact value you can write down without a calculator.

The identities also work the other way. When an expression already looks like sin⁡Acos⁡B+cos⁡Asin⁡B\sin A \cos B + \cos A \sin B, you can collapse it into a single sine of a single angle. That collapse is what makes these formulas worth memorizing rather than looking up.

The three identities

The sine identity is the friendly one: sin⁡(A±B)=sin⁡Acos⁡B±cos⁡Asin⁡B\sin(A \pm B) = \sin A \cos B \pm \cos A \sin B. The sign between the two products matches the sign inside the parentheses, and the two products are mixed, a sine with a cosine each time.

Cosine is the contrarian: cos⁡(A±B)=cos⁡Acos⁡B∓sin⁡Asin⁡B\cos(A \pm B) = \cos A \cos B \mp \sin A \sin B. The products stay matched, cosines with cosines and sines with sines, but the middle sign flips. A plus inside gives a minus in the middle. Tangent has its own form, tan⁡(A±B)=tan⁡A±tan⁡B1∓tan⁡Atan⁡B\tan(A \pm B) = \dfrac{\tan A \pm \tan B}{1 \mp \tan A \tan B}, where the denominator sign flips the same way cosine's does.

Splitting an angle before you touch a formula

The first move on an exact-value problem is arithmetic, not trigonometry: write the angle as a sum or a difference of two angles you know. Useful splits are 15∘=45∘−30∘15^\circ = 45^\circ - 30^\circ, 75∘=45∘+30∘75^\circ = 45^\circ + 30^\circ, 105∘=60∘+45∘105^\circ = 60^\circ + 45^\circ, and 165∘=120∘+45∘165^\circ = 120^\circ + 45^\circ. More than one split usually works and they all give the same value.

Then build a small table of the four values you need, sin⁡A\sin A, cos⁡A\cos A, sin⁡B\sin B, and cos⁡B\cos B, before you write the identity. Almost every lost point in this topic is a sign error, and sign errors die in tables. Angles past 90∘90^\circ are handled the same way, with one extra care: cos⁡120∘=−12\cos 120^\circ = -\dfrac{1}{2} and cos⁡150∘=−32\cos 150^\circ = -\dfrac{\sqrt{3}}{2} are negative, so the finished value can land below zero.

Reading the identities in reverse

Half of the questions on this topic never mention 15∘15^\circ at all. They hand you an expression like sin⁡75∘cos⁡15∘+cos⁡75∘sin⁡15∘\sin 75^\circ \cos 15^\circ + \cos 75^\circ \sin 15^\circ and ask for its value. Match it to the right side of the sine sum identity with A=75∘A = 75^\circ and B=15∘B = 15^\circ, and it is simply sin⁡90∘=1\sin 90^\circ = 1.

The other common setup gives you two ratios instead of two angles: sin⁡A=45\sin A = \dfrac{4}{5} and sin⁡B=513\sin B = \dfrac{5}{13}, both acute. Here the work is filling the gaps. Use sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1 to get the missing cosines, keep them positive because both angles are acute, and only then substitute into the identity.

Worked examples

Example 1: an exact value from a difference

Find the exact value of cos⁡15∘\cos 15^\circ.

Split into special angles15∘=45∘−30∘15^\circ = 45^\circ - 30^\circ
Cosine difference identity, middle sign flips to ++cos⁡45∘cos⁡30∘+sin⁡45∘sin⁡30∘\cos 45^\circ \cos 30^\circ + \sin 45^\circ \sin 30^\circ
Substitute the known values22⋅32+22⋅12\dfrac{\sqrt{2}}{2} \cdot \dfrac{\sqrt{3}}{2} + \dfrac{\sqrt{2}}{2} \cdot \dfrac{1}{2}
Put it over 4464+24=6+24\dfrac{\sqrt{6}}{4} + \dfrac{\sqrt{2}}{4} = \dfrac{\sqrt{6} + \sqrt{2}}{4}

Answer: cos⁡15∘=6+24\cos 15^\circ = \dfrac{\sqrt{6} + \sqrt{2}}{4}, about 0.9660.966.

Example 2: an exact value from a sum

Find the exact value of sin⁡105∘\sin 105^\circ.

Split into special angles105∘=60∘+45∘105^\circ = 60^\circ + 45^\circ
Sine sum identity, sign matches the insidesin⁡60∘cos⁡45∘+cos⁡60∘sin⁡45∘\sin 60^\circ \cos 45^\circ + \cos 60^\circ \sin 45^\circ
Substitute the known values32⋅22+12⋅22\dfrac{\sqrt{3}}{2} \cdot \dfrac{\sqrt{2}}{2} + \dfrac{1}{2} \cdot \dfrac{\sqrt{2}}{2}
Put it over 4464+24=6+24\dfrac{\sqrt{6}}{4} + \dfrac{\sqrt{2}}{4} = \dfrac{\sqrt{6} + \sqrt{2}}{4}

Answer: sin⁡105∘=6+24\sin 105^\circ = \dfrac{\sqrt{6} + \sqrt{2}}{4}, about 0.9660.966.

Example 3: given two ratios instead of two angles

sin⁡A=45\sin A = \dfrac{4}{5} and sin⁡B=513\sin B = \dfrac{5}{13}, with AA and BB both between 0∘0^\circ and 90∘90^\circ. Find sin⁡(A+B)\sin(A + B).

Find cos⁡A\cos A with the Pythagorean identity, positive because AA is acutecos⁡A=1−1625=35\cos A = \sqrt{1 - \tfrac{16}{25}} = \dfrac{3}{5}
Find cos⁡B\cos B the same waycos⁡B=1−25169=1213\cos B = \sqrt{1 - \tfrac{25}{169}} = \dfrac{12}{13}
Sine sum identitysin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\sin(A + B) = \sin A \cos B + \cos A \sin B
Substitute and add45⋅1213+35⋅513=4865+1565\dfrac{4}{5} \cdot \dfrac{12}{13} + \dfrac{3}{5} \cdot \dfrac{5}{13} = \dfrac{48}{65} + \dfrac{15}{65}

Answer: sin⁡(A+B)=6365\sin(A + B) = \dfrac{63}{65}

Example 4: an angle past 90∘90^\circ

Find the exact value of cos⁡165∘\cos 165^\circ.

Split into angles with known values165∘=120∘+45∘165^\circ = 120^\circ + 45^\circ
List the four values, and note that cos⁡120∘\cos 120^\circ is negativecos⁡120∘=−12, sin⁡120∘=32\cos 120^\circ = -\dfrac{1}{2}, \ \sin 120^\circ = \dfrac{\sqrt{3}}{2}
Cosine sum identity, middle sign flips to −-cos⁡120∘cos⁡45∘−sin⁡120∘sin⁡45∘\cos 120^\circ \cos 45^\circ - \sin 120^\circ \sin 45^\circ
Substitute and combine−24−64=−6+24-\dfrac{\sqrt{2}}{4} - \dfrac{\sqrt{6}}{4} = -\dfrac{\sqrt{6} + \sqrt{2}}{4}

Answer: cos⁡165∘=−6+24\cos 165^\circ = -\dfrac{\sqrt{6} + \sqrt{2}}{4}, about −0.966-0.966.

Try one yourself

Anglesin⁡\sincos⁡\cos
30∘30^\circ12\dfrac{1}{2}32\dfrac{\sqrt{3}}{2}
45∘45^\circ22\dfrac{\sqrt{2}}{2}22\dfrac{\sqrt{2}}{2}

Common questions

Why does the sign flip in the cosine identity?

It comes out of the derivation, and it is the single fact most worth memorizing here. In cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B\cos(A + B) = \cos A \cos B - \sin A \sin B the plus inside becomes a minus in the middle, and in cos⁡(A−B)=cos⁡Acos⁡B+sin⁡Asin⁡B\cos(A - B) = \cos A \cos B + \sin A \sin B the minus inside becomes a plus. Sine keeps the sign it was given; cosine reverses it.

Does it matter which split I choose?

No. You can write 75∘75^\circ as 45∘+30∘45^\circ + 30^\circ or as 135∘−60∘135^\circ - 60^\circ, and both give 6−24\dfrac{\sqrt{6} - \sqrt{2}}{4}. Choose whichever pair you know the exact values for, which usually means 30∘30^\circ, 45∘45^\circ, 60∘60^\circ, 90∘90^\circ, or their reflections such as 120∘120^\circ and 150∘150^\circ.

Is sin⁡(A+B)\sin(A + B) ever just sin⁡A+sin⁡B\sin A + \sin B?

Almost never, and treating it that way is the biggest error in this unit. Test it: sin⁡(30∘+60∘)=sin⁡90∘=1\sin(30^\circ + 60^\circ) = \sin 90^\circ = 1, but sin⁡30∘+sin⁡60∘=12+32≈1.37\sin 30^\circ + \sin 60^\circ = \dfrac{1}{2} + \dfrac{\sqrt{3}}{2} \approx 1.37. Sine does not distribute over addition, which is exactly why these identities exist.

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