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How do we expand sin(A±B)\sin(A \pm B), cos(A±B)\cos(A \pm B) and tan(A±B)\tan(A \pm B), and what are these identities used for?

Use the sum and difference identities for sine, cosine and tangent to expand or simplify trigonometric expressions

A focused answer to the HSC Maths Extension 1 dot point on sum and difference identities. The expansions of sin(A±B)\sin(A \pm B), cos(A±B)\cos(A \pm B) and tan(A±B)\tan(A \pm B) seen on the unit circle, derivation of the double-angle and half-angle formulas, and exact values for non-standard angles, with worked examples.

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What this dot point is asking

NESA wants you to know the sum and difference identities for sine, cosine and tangent, use them to expand or simplify expressions, and use them to compute exact values for non-standard angles like 1515^\circ or 7575^\circ. These six formulas are the foundation of the whole trig-identities strand: the double-angle formulas, the half-angle formulas, the product-to-sum identities, the t-formula and the auxiliary-angle method are all consequences of them. The single feature that catches students is the sign flip in the cosine expansion (it goes the opposite way to the input sign), so that is the thing to lock down.

The answer

What "angle sum" means on the unit circle

The whole topic is about the angle A+BA + B. On the unit circle, an angle AA reaches a point with coordinates (cosA,sinA)(\cos A, \sin A).

Stage 1: start with angle AOn the unit circle, the angle A is measured from the positive x-axis to a radius, reaching a point whose coordinates are cosine A and sine A.A(cos A, sin A)

Adding a further angle BB swings the radius round to the point at angle A+BA + B, whose coordinates are cos(A+B)\cos(A + B) and sin(A+B)\sin(A + B).

Stage 2: add angle B to reach A plus BAdding the angle B on top of A swings the radius to the point at angle A plus B, whose coordinates cosine of A plus B and sine of A plus B are what the identities expand.AB(cos(A+B), sin(A+B))

The sum and difference identities are exactly the formulas that express those coordinates of the combined angle in terms of the sines and cosines of AA and BB separately. That is why they let you reach a "new" angle (7575^\circ) from angles you already know (4545^\circ and 3030^\circ).

The sum and difference identities

For sine,

sin(A+B)=sinAcosB+cosAsinB,sin(AB)=sinAcosBcosAsinB.\sin(A + B) = \sin A \cos B + \cos A \sin B, \qquad \sin(A - B) = \sin A \cos B - \cos A \sin B.

For cosine, note the sign flips relative to the input,

cos(A+B)=cosAcosBsinAsinB,cos(AB)=cosAcosB+sinAsinB.\cos(A + B) = \cos A \cos B - \sin A \sin B, \qquad \cos(A - B) = \cos A \cos B + \sin A \sin B.

For tangent,

tan(A+B)=tanA+tanB1tanAtanB,tan(AB)=tanAtanB1+tanAtanB.\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}, \qquad \tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}.

The tangent denominator must be non-zero; if 1tanAtanB=01 \mp \tan A\tan B = 0 the formula is undefined and you fall back on sin/cos\sin/\cos.

Double-angle formulas: just set A=B=θA = B = \theta

The double-angle formulas are not separate facts to memorise; they are the sum formulas with B=AB = A:

sin2θ=2sinθcosθ,\sin 2\theta = 2\sin\theta\cos\theta,

cos2θ=cos2θsin2θ=12sin2θ=2cos2θ1,\cos 2\theta = \cos^2\theta - \sin^2\theta = 1 - 2\sin^2\theta = 2\cos^2\theta - 1,

tan2θ=2tanθ1tan2θ.\tan 2\theta = \frac{2\tan\theta}{1 - \tan^2\theta}.

The three forms of cos2θ\cos 2\theta come from substituting cos2θ+sin2θ=1\cos^2\theta + \sin^2\theta = 1; pick whichever form suits the next step (the 12sin2θ1 - 2\sin^2\theta form is handy when you know sinθ\sin\theta). Setting both angles equal to θ\theta is exactly the unit-circle picture of doubling an angle.

Double angle: theta and 2 thetaSetting A and B both equal to theta turns the angle sum into the double angle 2 theta, which is how the double-angle identities arise.θθangle 2θ

Half-angle formulas

Rearranging the cos2θ\cos 2\theta forms gives the power-reduction identities, and hence the half-angle formulas:

sin2θ=1cos2θ2,cos2θ=1+cos2θ2,\sin^2\theta = \frac{1 - \cos 2\theta}{2}, \qquad \cos^2\theta = \frac{1 + \cos 2\theta}{2},

sinϕ2=±1cosϕ2,cosϕ2=±1+cosϕ2,\sin\frac{\phi}{2} = \pm\sqrt{\frac{1 - \cos\phi}{2}}, \qquad \cos\frac{\phi}{2} = \pm\sqrt{\frac{1 + \cos\phi}{2}},

with the sign chosen from the quadrant of ϕ2\frac{\phi}{2}. These give exact values such as cos15\cos 15^\circ and feed directly into the t-formula.

Exact values for non-standard angles

The standard angles 0,30,45,60,900, 30, 45, 60, 90^\circ have well-known exact values:

sin30=12, cos30=32, tan30=13,sin45=cos45=22, tan45=1,\sin 30^\circ = \tfrac{1}{2}, \ \cos 30^\circ = \tfrac{\sqrt{3}}{2}, \ \tan 30^\circ = \tfrac{1}{\sqrt{3}}, \qquad \sin 45^\circ = \cos 45^\circ = \tfrac{\sqrt{2}}{2}, \ \tan 45^\circ = 1,

sin60=32, cos60=12, tan60=3.\sin 60^\circ = \tfrac{\sqrt{3}}{2}, \ \cos 60^\circ = \tfrac{1}{2}, \ \tan 60^\circ = \sqrt{3}.

Writing a non-standard angle as a sum or difference of two standard angles, 15=453015^\circ = 45^\circ - 30^\circ, 75=45+3075^\circ = 45^\circ + 30^\circ, 105=60+45105^\circ = 60^\circ + 45^\circ, then applying the matching identity, produces its exact value.

How exam questions ask about sum and difference identities

  • "Find the exact value of cos75\cos 75^\circ / sin15\sin 15^\circ / tan105\tan 105^\circ": split into two standard angles, apply the right identity, simplify and rationalise.
  • "Given sinα\sin\alpha and cosβ\cos\beta (with quadrants), find sin(α+β)\sin(\alpha+\beta)": get the missing ratios by Pythagoras with correct signs, then substitute into the sum identity.
  • "Expand and simplify sin(θ+π4)+sin(θπ4)\sin(\theta + \frac{\pi}{4}) + \sin(\theta - \frac{\pi}{4})": expand both, cancel the opposite terms.
  • "Prove the identity ...": expand each compound angle and collect; many proofs hinge on the cosine sign flip or on terms cancelling.
  • "Find sin3θ\sin 3\theta / cos3θ\cos 3\theta": write 3θ=2θ+θ3\theta = 2\theta + \theta and combine the sum identity with the double-angle formulas.
  • "Hence solve / find the maximum ...": the identity is a stepping stone to an equation or to the auxiliary-angle form.

Edge cases worth knowing

  • The tangent formula can be undefined. tan(45+45)\tan(45^\circ + 45^\circ) would need 1tan45tan45=11=01 - \tan 45^\circ\tan 45^\circ = 1 - 1 = 0 in the denominator. The formula fails (as it must, since tan90\tan 90^\circ is undefined); reason via sin90/cos90\sin 90^\circ/\cos 90^\circ instead.
  • Choose the efficient decomposition. sin105=sin(60+45)\sin 105^\circ = \sin(60^\circ + 45^\circ) is cleaner than sin(75+30)\sin(75^\circ + 30^\circ), because both 6060^\circ and 4545^\circ are standard while 7575^\circ is not.
  • The half-angle sign depends on the quadrant. cosϕ2=±1+cosϕ2\cos\frac{\phi}{2} = \pm\sqrt{\frac{1+\cos\phi}{2}} takes the sign of cosine in the quadrant of ϕ2\frac{\phi}{2}, not of ϕ\phi; decide it before writing the surd.

Exam-style practice questions

Practice questions written in the style of NESA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2022 HSC Q93 marksUse a sum or difference identity to find the exact value of cos75\cos 75^\circ.
Show worked answer →

Write 75=45+3075^\circ = 45^\circ + 30^\circ and use cos(A+B)=cosAcosBsinAsinB\cos(A + B) = \cos A \cos B - \sin A \sin B.

cos75=cos45cos30sin45sin30\cos 75^\circ = \cos 45^\circ \cos 30^\circ - \sin 45^\circ \sin 30^\circ.

=22322212= \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \cdot \frac{1}{2}

=6424=624= \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4} = \frac{\sqrt{6} - \sqrt{2}}{4}.

Markers reward identifying a useful decomposition into standard angles, the correct sum identity, and a final exact answer.

2020 HSC Q113 marksIf sinα=35\sin \alpha = \frac{3}{5} with α\alpha in the first quadrant and cosβ=1213\cos \beta = -\frac{12}{13} with β\beta in the second quadrant, find sin(α+β)\sin(\alpha + \beta).
Show worked answer →

In Q1: cosα=19/25=45\cos \alpha = \sqrt{1 - 9/25} = \frac{4}{5}.

In Q2: sinβ=1144/169=513\sin \beta = \sqrt{1 - 144/169} = \frac{5}{13} (positive in Q2).

Apply sin(α+β)=sinαcosβ+cosαsinβ\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta.

=35(1213)+45513= \frac{3}{5} \cdot \left( -\frac{12}{13} \right) + \frac{4}{5} \cdot \frac{5}{13}

=3665+2065=1665= -\frac{36}{65} + \frac{20}{65} = -\frac{16}{65}.

Markers reward correct quadrant signs for cosα\cos \alpha and sinβ\sin \beta, the sum identity, and clean arithmetic.

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