Sum, Difference, and Multiple-Angle Formulas

Study Sheet

Sum, Difference, and Multiple-Angle Formulas

Exact values, double & half-angles, power-reducing, product-to-sum

Sum and Difference Formulas

Concept
The six core formulas
sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡Bsin⁡(A−B)=sin⁡Acos⁡B−cos⁡Asin⁡Bcos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡Bcos⁡(A−B)=cos⁡Acos⁡B+sin⁡Asin⁡Btan⁡(A+B)=tan⁡A+tan⁡B1−tan⁡Atan⁡Btan⁡(A−B)=tan⁡A−tan⁡B1+tan⁡Atan⁡B\begin{aligned} \sin(A+B) &= \sin A\cos B + \cos A\sin B &\qquad \sin(A-B) &= \sin A\cos B - \cos A\sin B\\[2pt] \cos(A+B) &= \cos A\cos B - \sin A\sin B &\qquad \cos(A-B) &= \cos A\cos B + \sin A\sin B\\[2pt] \tan(A+B) &= \dfrac{\tan A + \tan B}{1-\tan A\tan B} &\qquad \tan(A-B) &= \dfrac{\tan A - \tan B}{1+\tan A\tan B} \end{aligned}

Watch the signs: cosine flips the sign (++ becomes −-); sine and tangent keep it.

Example
Exact value: cos⁡15∘\cos 15^\circ via a difference

Write 15∘=45∘−30∘15^\circ = 45^\circ - 30^\circ and use cos⁡(A−B)\cos(A-B):

cos⁡15∘=cos⁡(45∘−30∘)=cos⁡45∘cos⁡30∘+sin⁡45∘sin⁡30∘=22⋅32+22⋅12=64+24=6+24\begin{aligned} \cos 15^\circ &= \cos(45^\circ-30^\circ) = \cos45^\circ\cos30^\circ + \sin45^\circ\sin30^\circ\\ &= \frac{\sqrt2}{2}\cdot\frac{\sqrt3}{2} + \frac{\sqrt2}{2}\cdot\frac{1}{2} = \frac{\sqrt6}{4} + \frac{\sqrt2}{4} = \boxed{\dfrac{\sqrt6+\sqrt2}{4}} \end{aligned}
Example
Exact value: tan⁡75∘\tan 75^\circ via a sum

Write 75∘=45∘+30∘75^\circ = 45^\circ + 30^\circ:

tan⁡75∘=tan⁡45∘+tan⁡30∘1−tan⁡45∘tan⁡30∘=1+131−13=3+13−1=(3+1)2(3−1)(3+1)=4+232=2+3\begin{aligned} \tan 75^\circ &= \frac{\tan45^\circ+\tan30^\circ}{1-\tan45^\circ\tan30^\circ} = \frac{1+\tfrac{1}{\sqrt3}}{1-\tfrac{1}{\sqrt3}} = \frac{\sqrt3+1}{\sqrt3-1}\\ &= \frac{(\sqrt3+1)^2}{(\sqrt3-1)(\sqrt3+1)} = \frac{4+2\sqrt3}{2} = \boxed{2+\sqrt3} \end{aligned}
Tip

Tip: A “non-special” angle is usually a sum or difference of two special angles (30∘,45∘,60∘,90∘30^\circ,45^\circ,60^\circ,90^\circ). Pick the pair whose sum or difference lands on your target.

Finding Exact Values of Non-Special Angles

Concept
Common decompositions
15∘=45∘−30∘,75∘=45∘+30∘,105∘=60∘+45∘,7π12=105∘=π3+π415^\circ = 45^\circ-30^\circ,\quad 75^\circ = 45^\circ+30^\circ,\quad 105^\circ = 60^\circ+45^\circ,\quad \tfrac{7\pi}{12}=105^\circ=\tfrac{\pi}{3}+\tfrac{\pi}{4}

Then apply the matching sum or difference formula and simplify the radicals.

Example
Exact value: sin⁡7π12\sin\dfrac{7\pi}{12}

Since 7π12=π3+π4\dfrac{7\pi}{12} = \dfrac{\pi}{3}+\dfrac{\pi}{4} (that is 105∘=60∘+45∘105^\circ = 60^\circ+45^\circ):

sin⁡7π12=sin⁡π3cos⁡π4+cos⁡π3sin⁡π4=32⋅22+12⋅22=64+24=6+24\begin{aligned} \sin\frac{7\pi}{12} &= \sin\frac{\pi}{3}\cos\frac{\pi}{4} + \cos\frac{\pi}{3}\sin\frac{\pi}{4} = \frac{\sqrt3}{2}\cdot\frac{\sqrt2}{2} + \frac{1}{2}\cdot\frac{\sqrt2}{2}\\ &= \frac{\sqrt6}{4}+\frac{\sqrt2}{4} = \boxed{\dfrac{\sqrt6+\sqrt2}{4}} \end{aligned}

For cos⁡7π12\cos\dfrac{7\pi}{12} the same setup gives 2−64\dfrac{\sqrt2-\sqrt6}{4} (negative, as expected in Quadrant II).

Tip

Tip: Sanity-check the sign against the quadrant. 105∘105^\circ sits in Quadrant II, so its sine is positive and its cosine is negative --- exactly what the radicals show.

Double-Angle Formulas

Concept
Double-angle identities
sin⁡2A=2sin⁡Acos⁡A,tan⁡2A=2tan⁡A1−tan⁡2A\sin 2A = 2\sin A\cos A,\qquad \tan 2A = \dfrac{2\tan A}{1-\tan^2 A}
cos⁡2A=cos⁡2A−sin⁡2A=2cos⁡2A−1=1−2sin⁡2A\cos 2A = \cos^2 A - \sin^2 A = 2\cos^2 A - 1 = 1 - 2\sin^2 A

Choose the cosine form that matches what you already know: use 2cos⁡2A−12\cos^2A-1 when you know cos⁡A\cos A, and 1−2sin⁡2A1-2\sin^2A when you know sin⁡A\sin A.

Example
Given sin⁡θ=35\sin\theta=\tfrac{3}{5} with θ\theta in Quadrant II

In Quadrant II cosine is negative, so cos⁡θ=−45\cos\theta = -\tfrac45.

sin⁡2θ=2sin⁡θcos⁡θ=2⋅35⋅(−45)=−2425cos⁡2θ=1−2sin⁡2θ=1−2⋅925=1−1825=725tan⁡2θ=sin⁡2θcos⁡2θ=−24/257/25=−247\begin{aligned} \sin 2\theta &= 2\sin\theta\cos\theta = 2\cdot\tfrac35\cdot\left(-\tfrac45\right) = -\frac{24}{25}\\ \cos 2\theta &= 1-2\sin^2\theta = 1-2\cdot\tfrac{9}{25} = 1-\tfrac{18}{25} = \frac{7}{25}\\ \tan 2\theta &= \frac{\sin2\theta}{\cos2\theta} = \frac{-24/25}{7/25} = -\frac{24}{7} \end{aligned}
Tip

Tip: 2θ2\theta can land in a different quadrant than θ\theta. Trust the algebra: the sign comes out automatically once sin⁡θ\sin\theta and cos⁡θ\cos\theta carry their correct signs.

Half-Angle Formulas

Concept
Half-angle identities
sin⁡A2=±1−cos⁡A2,cos⁡A2=±1+cos⁡A2\sin\frac{A}{2} = \pm\sqrt{\dfrac{1-\cos A}{2}},\qquad \cos\frac{A}{2} = \pm\sqrt{\dfrac{1+\cos A}{2}}
tan⁡A2=1−cos⁡Asin⁡A=sin⁡A1+cos⁡A=±1−cos⁡A1+cos⁡A\tan\frac{A}{2} = \frac{1-\cos A}{\sin A} = \frac{\sin A}{1+\cos A} = \pm\sqrt{\dfrac{1-\cos A}{1+\cos A}}

The ±\pm is decided by the quadrant of A2\dfrac{A}{2}, not of AA. (The two tangent quotient forms need no sign choice.)

Example
Given cos⁡θ=35\cos\theta=\tfrac{3}{5} with θ\theta in Quadrant IV

Here 270∘<θ<360∘270^\circ<\theta<360^\circ, so 135∘<θ2<180∘135^\circ<\dfrac{\theta}{2}<180^\circ: the half-angle is in Quadrant II, where sine is positive and cosine is negative.

sin⁡θ2=+1−cos⁡θ2=1−352=15=55cos⁡θ2=−1+cos⁡θ2=−1+352=−45=−255\begin{aligned} \sin\frac{\theta}{2} &= +\sqrt{\frac{1-\cos\theta}{2}} = \sqrt{\frac{1-\tfrac35}{2}} = \sqrt{\frac{1}{5}} = \frac{\sqrt5}{5}\\ \cos\frac{\theta}{2} &= -\sqrt{\frac{1+\cos\theta}{2}} = -\sqrt{\frac{1+\tfrac35}{2}} = -\sqrt{\frac{4}{5}} = -\frac{2\sqrt5}{5} \end{aligned}
Tip

Tip: Always halve the interval first. If θ\theta is in Quadrant IV, θ2\dfrac{\theta}{2} is in Quadrant II --- a fresh sign decision every time.

Power-Reducing Formulas

Concept
Lowering even powers
sin⁡2A=1−cos⁡2A2,cos⁡2A=1+cos⁡2A2,tan⁡2A=1−cos⁡2A1+cos⁡2A\sin^2 A = \frac{1-\cos 2A}{2},\qquad \cos^2 A = \frac{1+\cos 2A}{2},\qquad \tan^2 A = \frac{1-\cos 2A}{1+\cos 2A}

These are just the double-angle cosine identity solved for sin⁡2A\sin^2A and cos⁡2A\cos^2A. Apply repeatedly to reduce higher powers.

Example
Reduce sin⁡4x\sin^4 x to first powers of cosine
sin⁡4x=(sin⁡2x)2=(1−cos⁡2x2)2=1−2cos⁡2x+cos⁡22x4=1−2cos⁡2x+1+cos⁡4x24=3−4cos⁡2x+cos⁡4x8\begin{aligned} \sin^4 x &= \left(\sin^2 x\right)^2 = \left(\frac{1-\cos 2x}{2}\right)^2 = \frac{1-2\cos 2x+\cos^2 2x}{4}\\ &= \frac{1-2\cos 2x+\tfrac{1+\cos 4x}{2}}{4} = \boxed{\dfrac{3-4\cos 2x+\cos 4x}{8}} \end{aligned}
Tip

Tip: Power-reducing is the key to integrating and simplifying sin⁡2\sin^2, cos⁡2\cos^2, sin⁡4,…\sin^4,\dots --- keep lowering the power until every term is a first-power cosine.

Product-to-Sum and Sum-to-Product

Concept
Two families of identities

Product-to-sum:

sin⁡Acos⁡B=12[sin⁡(A+B)+sin⁡(A−B)]cos⁡Asin⁡B=12[sin⁡(A+B)−sin⁡(A−B)]cos⁡Acos⁡B=12[cos⁡(A−B)+cos⁡(A+B)]sin⁡Asin⁡B=12[cos⁡(A−B)−cos⁡(A+B)]\begin{aligned} \sin A\cos B &= \tfrac12\big[\sin(A+B)+\sin(A-B)\big]\\ \cos A\sin B &= \tfrac12\big[\sin(A+B)-\sin(A-B)\big]\\ \cos A\cos B &= \tfrac12\big[\cos(A-B)+\cos(A+B)\big]\\ \sin A\sin B &= \tfrac12\big[\cos(A-B)-\cos(A+B)\big] \end{aligned}

Sum-to-product:

sin⁡A+sin⁡B=2sin⁡A+B2cos⁡A−B2sin⁡A−sin⁡B=2cos⁡A+B2sin⁡A−B2cos⁡A+cos⁡B=2cos⁡A+B2cos⁡A−B2cos⁡A−cos⁡B=−2sin⁡A+B2sin⁡A−B2\begin{aligned} \sin A+\sin B &= 2\sin\tfrac{A+B}{2}\cos\tfrac{A-B}{2}\\ \sin A-\sin B &= 2\cos\tfrac{A+B}{2}\sin\tfrac{A-B}{2}\\ \cos A+\cos B &= 2\cos\tfrac{A+B}{2}\cos\tfrac{A-B}{2}\\ \cos A-\cos B &= -2\sin\tfrac{A+B}{2}\sin\tfrac{A-B}{2} \end{aligned}
Example
Product-to-sum: exact value of cos⁡75∘cos⁡15∘\cos 75^\circ\cos 15^\circ
cos⁡75∘cos⁡15∘=12[cos⁡(75∘−15∘)+cos⁡(75∘+15∘)]=12[cos⁡60∘+cos⁡90∘]=12[12+0]=14\begin{aligned} \cos 75^\circ\cos 15^\circ &= \tfrac12\big[\cos(75^\circ-15^\circ)+\cos(75^\circ+15^\circ)\big]\\ &= \tfrac12\big[\cos 60^\circ+\cos 90^\circ\big] = \tfrac12\left[\tfrac12+0\right] = \boxed{\tfrac14} \end{aligned}
Example
Sum-to-product: exact value of sin⁡75∘+sin⁡15∘\sin 75^\circ+\sin 15^\circ
sin⁡75∘+sin⁡15∘=2sin⁡75∘+15∘2cos⁡75∘−15∘2=2sin⁡45∘cos⁡30∘=2⋅22⋅32=62\begin{aligned} \sin 75^\circ+\sin 15^\circ &= 2\sin\frac{75^\circ+15^\circ}{2}\cos\frac{75^\circ-15^\circ}{2} = 2\sin 45^\circ\cos 30^\circ\\ &= 2\cdot\frac{\sqrt2}{2}\cdot\frac{\sqrt3}{2} = \boxed{\dfrac{\sqrt6}{2}} \end{aligned}
Tip

Tip: Use product-to-sum when a product blocks you (integrals, exact values); use sum-to-product to factor a sum so it can cancel (proving identities, solving equations).

Simplifying, Verifying Identities, and Finding Values

Concept
Strategy
  • [leftmargin=*,itemsep=1pt]
  • To simplify, spot the pattern: 2sin⁡Acos⁡A=sin⁡2A2\sin A\cos A=\sin 2A, cos⁡2A−sin⁡2A=cos⁡2A\cos^2A-\sin^2A=\cos 2A, 1−cos⁡2A=2sin⁡2A1-\cos 2A = 2\sin^2A.
  • To verify, transform the busier side using double/half-angle or power-reducing until it matches the other.
  • To find values given quadrants, first recover sin⁡\sin and cos⁡\cos of each angle (with correct signs), then substitute into a sum/difference or double-angle formula.
Example
Verify sin⁡2x1+cos⁡2x=tan⁡x\dfrac{\sin 2x}{1+\cos 2x}=\tan x
sin⁡2x1+cos⁡2x=2sin⁡xcos⁡x1+(2cos⁡2x−1)=2sin⁡xcos⁡x2cos⁡2x=sin⁡xcos⁡x=tan⁡x ✓\begin{aligned} \frac{\sin 2x}{1+\cos 2x} &= \frac{2\sin x\cos x}{1+(2\cos^2 x-1)} = \frac{2\sin x\cos x}{2\cos^2 x} = \frac{\sin x}{\cos x} = \tan x\ \checkmark \end{aligned}
Example
Find sin⁡(α+β)\sin(\alpha+\beta) given quadrants

Let sin⁡α=35\sin\alpha=\tfrac35 with α\alpha in Quadrant II, and cos⁡β=513\cos\beta=\tfrac{5}{13} with β\beta in Quadrant IV.

α in QII:cos⁡α=−45;β in QIV:sin⁡β=−1213sin⁡(α+β)=sin⁡αcos⁡β+cos⁡αsin⁡β=35⋅513+(−45)(−1213)=1565+4865=6365\begin{aligned} &\alpha\ \text{in QII}: \cos\alpha=-\tfrac45; \qquad \beta\ \text{in QIV}: \sin\beta=-\tfrac{12}{13}\\ \sin(\alpha+\beta) &= \sin\alpha\cos\beta+\cos\alpha\sin\beta = \tfrac35\cdot\tfrac{5}{13} + \left(-\tfrac45\right)\left(-\tfrac{12}{13}\right)\\ &= \frac{15}{65}+\frac{48}{65} = \boxed{\dfrac{63}{65}} \end{aligned}
Tip

Tip: Before plugging in, draw each angle's reference triangle and label signs by quadrant. A single wrong sign on cos⁡α\cos\alpha or sin⁡β\sin\beta ruins the whole answer.

Going Deeper: Advanced Formula Techniques

Concept
Telescoping products of cosines

Multiplying a chain of angle-doubling cosines by 2sin⁡θ2\sin\theta collapses it, because each step obeys 2sin⁡ϕcos⁡ϕ=sin⁡2ϕ2\sin\phi\cos\phi=\sin 2\phi. The result is the beautiful identity

∏k=0n−1cos⁡ ⁣(2kθ)=cos⁡θ cos⁡2θ cos⁡4θ⋯cos⁡ ⁣(2n−1θ)=sin⁡ ⁣(2nθ)2nsin⁡θ.\prod_{k=0}^{n-1}\cos\!\left(2^{k}\theta\right) = \cos\theta\,\cos 2\theta\,\cos 4\theta\cdots\cos\!\left(2^{n-1}\theta\right) = \frac{\sin\!\left(2^{n}\theta\right)}{2^{n}\sin\theta}.

The whole product depends only on the first and last angles --- everything in between telescopes away.

Example
A striking exact value: cos⁡π7cos⁡2π7cos⁡4π7=−18\cos\dfrac{\pi}{7}\cos\dfrac{2\pi}{7}\cos\dfrac{4\pi}{7}=-\dfrac18

Take θ=π7\theta=\dfrac{\pi}{7} and n=3n=3, so the angles are θ,2θ,4θ\theta,2\theta,4\theta. Apply the telescoping identity:

cos⁡π7cos⁡2π7cos⁡4π7=sin⁡ ⁣(23⋅π7)23sin⁡π7=sin⁡8π78sin⁡π7=sin⁡ ⁣(π+π7)8sin⁡π7=−sin⁡π78sin⁡π7=−18\begin{aligned} \cos\frac{\pi}{7}\cos\frac{2\pi}{7}\cos\frac{4\pi}{7} &= \frac{\sin\!\left(2^{3}\cdot\tfrac{\pi}{7}\right)}{2^{3}\sin\tfrac{\pi}{7}} = \frac{\sin\tfrac{8\pi}{7}}{8\sin\tfrac{\pi}{7}}\\ &= \frac{\sin\!\left(\pi+\tfrac{\pi}{7}\right)}{8\sin\tfrac{\pi}{7}} = \frac{-\sin\tfrac{\pi}{7}}{8\sin\tfrac{\pi}{7}} = \boxed{-\frac18} \end{aligned}

The key step is sin⁡ ⁣(π+π7)=−sin⁡π7\sin\!\left(\pi+\tfrac{\pi}{7}\right)=-\sin\tfrac{\pi}{7}, which cancels the leftover sine exactly.

Concept
Triple-angle formulas

Chaining a double-angle with a sum formula (3A=2A+A3A=2A+A) gives the triple-angle identities:

sin⁡3A=3sin⁡A−4sin⁡3A,cos⁡3A=4cos⁡3A−3cos⁡A,\sin 3A = 3\sin A - 4\sin^3 A,\qquad \cos 3A = 4\cos^3 A - 3\cos A,
tan⁡3A=3tan⁡A−tan⁡3A1−3tan⁡2A.\tan 3A = \frac{3\tan A - \tan^3 A}{1-3\tan^2 A}.

Read as cubics in sin⁡A\sin A or cos⁡A\cos A, they turn angle problems into polynomial equations --- the doorway to exact values like cos⁡36∘\cos 36^\circ.

Example
sin⁡18∘\sin 18^\circ and cos⁡36∘\cos 36^\circ from a cubic

Let θ=18∘\theta=18^\circ, so 5θ=90∘5\theta=90^\circ and hence 3θ=90∘−2θ3\theta = 90^\circ-2\theta. Taking sines,

sin⁡3θ=sin⁡(90∘−2θ)=cos⁡2θ.\sin 3\theta = \sin(90^\circ-2\theta) = \cos 2\theta.

Now substitute the triple-angle and double-angle forms with s=sin⁡18∘s=\sin 18^\circ:

3s−4s3=1−2s24s3−2s2−3s+1=0(s−1)(4s2+2s−1)=0.\begin{aligned} 3s - 4s^3 &= 1 - 2s^2\\ 4s^3 - 2s^2 - 3s + 1 &= 0\\ (s-1)\left(4s^2 + 2s - 1\right) &= 0. \end{aligned}

Since s=sin⁡18∘≠1s=\sin18^\circ\neq 1, solve 4s2+2s−1=04s^2+2s-1=0 and keep the positive root:

s=−2+4+168=−2+258=sin⁡18∘=5−14.s = \frac{-2+\sqrt{4+16}}{8} = \frac{-2+2\sqrt5}{8} = \boxed{\sin 18^\circ = \frac{\sqrt5-1}{4}}.

Then cos⁡36∘=1−2sin⁡218∘=1−2⋅3−58=cos⁡36∘=1+54\cos 36^\circ = 1-2\sin^2 18^\circ = 1-2\cdot\dfrac{3-\sqrt5}{8} = \boxed{\cos 36^\circ = \dfrac{1+\sqrt5}{4}}, the golden ratio in disguise.

Concept
Summing equally spaced sines and cosines

For an arithmetic progression of angles a, a+d, a+2d,…a,\,a+d,\,a+2d,\dots, multiply by 2sin⁡d22\sin\tfrac{d}{2} and telescope with product-to-sum. The closed forms are

∑k=0n−1cos⁡(a+kd)=sin⁡ ⁣(nd2)sin⁡ ⁣(d2) cos⁡ ⁣(a+(n−1)d2),∑k=0n−1sin⁡(a+kd)=sin⁡ ⁣(nd2)sin⁡ ⁣(d2) sin⁡ ⁣(a+(n−1)d2).\begin{aligned} \sum_{k=0}^{n-1}\cos(a+kd) &= \frac{\sin\!\left(\tfrac{nd}{2}\right)}{\sin\!\left(\tfrac{d}{2}\right)}\, \cos\!\left(a+\tfrac{(n-1)d}{2}\right),\\[4pt] \sum_{k=0}^{n-1}\sin(a+kd) &= \frac{\sin\!\left(\tfrac{nd}{2}\right)}{\sin\!\left(\tfrac{d}{2}\right)}\, \sin\!\left(a+\tfrac{(n-1)d}{2}\right). \end{aligned}

When the angles wrap evenly around the circle (so ndnd is a multiple of 2π2\pi), the factor sin⁡nd2=0\sin\tfrac{nd}{2}=0 and both sums vanish --- e.g. ∑k=0n−1cos⁡2πkn=0\displaystyle\sum_{k=0}^{n-1}\cos\tfrac{2\pi k}{n}=0 for n≥2n\ge 2.

Concept
Nested half-angles for unusual exact values

Applying the half-angle cosine formula repeatedly to cos⁡π4=22\cos\tfrac{\pi}{4}=\tfrac{\sqrt2}{2} builds a tower of nested radicals:

cos⁡π8=1+cos⁡π42=122+2,cos⁡π16=122+2+2,\cos\frac{\pi}{8} = \sqrt{\frac{1+\cos\tfrac{\pi}{4}}{2}} = \frac12\sqrt{2+\sqrt2}, \qquad \cos\frac{\pi}{16} = \frac12\sqrt{2+\sqrt{2+\sqrt2}},

and in general cos⁡π2n=122+2+⋯+2⏟n−1 radicals\cos\dfrac{\pi}{2^{n}} = \dfrac12\underbrace{\sqrt{2+\sqrt{2+\cdots+\sqrt2}}}_{n-1\ \text{radicals}}. Each extra half-angle nests one more square root; the ++ sign is used throughout because every π2n\tfrac{\pi}{2^n} lands in Quadrant I.

Example
Given quadrants, find sin⁡(α−β)\sin(\alpha-\beta) and tan⁡2α\tan 2\alpha

Let sin⁡α=45\sin\alpha=\tfrac45 with α\alpha in Quadrant II, and cos⁡β=1213\cos\beta=\tfrac{12}{13} with β\beta in Quadrant IV. First recover the partners with correct signs:

α in QII: cos⁡α=−35;β in QIV: sin⁡β=−513.\alpha\ \text{in QII}:\ \cos\alpha=-\tfrac35; \qquad \beta\ \text{in QIV}:\ \sin\beta=-\tfrac{5}{13}.

Difference formula:

sin⁡(α−β)=sin⁡αcos⁡β−cos⁡αsin⁡β=45⋅1213−(−35)(−513)=4865−1565=3365.\begin{aligned} \sin(\alpha-\beta) &= \sin\alpha\cos\beta - \cos\alpha\sin\beta = \tfrac45\cdot\tfrac{12}{13} - \left(-\tfrac35\right)\left(-\tfrac{5}{13}\right)\\ &= \frac{48}{65} - \frac{15}{65} = \boxed{\dfrac{33}{65}}. \end{aligned}

Double-angle in tangent: with tan⁡α=sin⁡αcos⁡α=4/5−3/5=−43\tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}=\dfrac{4/5}{-3/5}=-\dfrac43,

tan⁡2α=2tan⁡α1−tan⁡2α=2(−43)1−169=−83−79=−83⋅9−7=247.\begin{aligned} \tan 2\alpha &= \frac{2\tan\alpha}{1-\tan^2\alpha} = \frac{2\left(-\tfrac43\right)}{1-\tfrac{16}{9}} = \frac{-\tfrac83}{-\tfrac79} = \frac{-8}{3}\cdot\frac{9}{-7} = \boxed{\dfrac{24}{7}}. \end{aligned}
Tip

Tip: Advanced exact values almost always come from one of three moves --- telescoping a product with 2sin⁡θ2\sin\theta, turning a triple-angle relation into a cubic, or nesting half-angles. Spot which structure your target angle fits, and the radicals fall out.

Formulas, Proofs & Tips

Tip
Sum, difference and double-angle formulas
sin⁡(A±B)=sin⁡Acos⁡B±cos⁡Asin⁡B,cos⁡(A±B)=cos⁡Acos⁡B∓sin⁡Asin⁡B,sin⁡2A=2sin⁡Acos⁡A\sin(A\pm B)=\sin A\cos B\pm\cos A\sin B,\quad \cos(A\pm B)=\cos A\cos B\mp\sin A\sin B,\quad \sin 2A=2\sin A\cos A

What it means. Angles do not distribute — these say what really happens.

Example. cos⁡2θ=1−2sin⁡2θ\cos 2\theta=1-2\sin^2\theta; if sin⁡θ=12\sin\theta=\tfrac12, then cos⁡2θ=12\cos2\theta=\tfrac12.

Why it works. The double-angle results are the sum formulas with B=AB=A: sin⁡2A=sin⁡Acos⁡A+cos⁡Asin⁡A=2sin⁡Acos⁡A\sin 2A=\sin A\cos A+\cos A\sin A=2\sin A\cos A, and cos⁡2A=cos⁡2A−sin⁡2A\cos 2A=\cos^2A-\sin^2A, which the Pythagorean identity rewrites as 2cos⁡2A−12\cos^2A-1 or 1−2sin⁡2A1-2\sin^2A.

Tip. sin⁡(A+B)≠sin⁡A+sin⁡B\sin(A+B)\neq\sin A+\sin B. Note the sign flip: cos⁡\cos of a sum takes a minus.