Exercises — Sum and difference formulas — sin(A±B), cos(A±B), tan(A±B) — proofs

$105°=60°+45°$ (both standard). Use the sum form of sine: $\sin 105°=\sin (60°+45°)=\sin 60°\cos 45°+\cos 60°\sin 45°.$ (Any valid split works, e.g. $105°=45°+60°$; the point is to land on standard angles.)

Cosine is contrary, so $\cos (A-B)=\cos A\cos B+\sin A\sin B$ (the minus inside flips to a plus). Sine keeps company, so $\sin (A-B)=\sin A\cos B-\cos A\sin B$ (the sign inside matches the sign in the middle).

With $\tan A=\tan B=0$: $\tan (A+B)=\frac{0+0}{1-0}=0.$ Here $\tan A+\tan B=0$ too, so they happen to agree — but that is a coincidence of the zero case, not a rule.

$\cos (60°+45°)=\cos 60°\cos 45°-\sin 60°\sin 45°$ $=\frac{1}{2}\cdot \frac{\sqrt{2}}{2}-\frac{\sqrt{3}}{2}\cdot \frac{\sqrt{2}}{2}=\frac{\sqrt{2}-\sqrt{6}}{4}.$ This is negative ($\sqrt{6}>\sqrt{2}$), which makes sense: $105°$ is in the second quadrant where cosine is negative.

$\tan (45°-30°)=\frac{\tan 45°-\tan 30°}{1+\tan 45°\tan 30°}=\frac{1-\frac{1}{\sqrt{3}}}{1+1\cdot \frac{1}{\sqrt{3}}}=\frac{\sqrt{3}-1}{\sqrt{3}+1}.$ Rationalise by multiplying top and bottom by $\sqrt{3}-1$: $=\frac{(\sqrt{3}-1{)}^2}{(\sqrt{3}+1)(\sqrt{3}-1)}=\frac{3-2\sqrt{3}+1}{3-1}=\frac{4-2\sqrt{3}}{2}=2-\sqrt{3}.$

$\cos \left(x-\frac{\pi }{2}\right)=\cos x\cos \frac{\pi }{2}+\sin x\sin \frac{\pi }{2}=\cos x\cdot 0+\sin x\cdot 1=\sin x.$ So shifting cosine right by $90°$ produces sine — a cofunction fact falling straight out of the formula.

Find $\cos A$. From ${\sin }^{2}A+{\cos }^{2}A=1$: ${\cos }^{2}A=1-\frac{9}{25}=\frac{16}{25}$, so $\cos A=\pm \frac{4}{5}$. In Quadrant II cosine is negative, so $\cos A=-\frac{4}{5}$ (look at the left-pointing blue arrow in the figure). Find $\sin B$. ${\sin }^{2}B=1-\frac{25}{169}=\frac{144}{169}$, so $\sin B=\pm \frac{12}{13}$. Quadrant I ⇒ positive, $\sin B=\frac{12}{13}$. Assemble: $\sin (A+B)=\sin A\cos B+\cos A\sin B=\frac{3}{5}\cdot \frac{5}{13}+\left(-\frac{4}{5}\right)\cdot \frac{12}{13}=\frac{15}{65}-\frac{48}{65}=-\frac{33}{65}.$ $\cos (A+B)=\cos A\cos B-\sin A\sin B=\left(-\frac{4}{5}\right)\frac{5}{13}-\frac{3}{5}\cdot \frac{12}{13}=-\frac{20}{65}-\frac{36}{65}=-\frac{56}{65}.$ Both negative ⇒ $A+B$ lands in Quadrant III, consistent with $A\approx 143°$, $B\approx 67°$, sum $\approx 210°$.

$\tan (45°+45°)=\frac{\tan 45°+\tan 45°}{1-\tan 45°\tan 45°}=\frac{1+1}{1-1}=\frac{2}{0}.$ The denominator hits zero, so the expression is undefined. Geometrically $A+B=90°$, and $\tan 90°$ is undefined because the unit-circle point is at $(0,1)$: $\tan =\frac{\sin }{\cos }=\frac{1}{0}$. The algebra's $0$ in the denominator is that vertical line. The condition $1-\tan A\tan B=0$ (i.e. $\tan A\tan B=1$) is exactly the signature of a right-angle sum.

$A=\arctan 2\approx 63.4°$, $B=\arctan 3\approx 71.6°$, so $A+B\approx 135°$ — Quadrant II, where tangent is negative, matching $\tan (A+B)=-1$. Indeed $\tan 135°=-1$. The negative value is not an error: $\tan A\tan B=6>1$ forced the denominator negative, signalling the sum crossed past $90°$.

$\sin (A+A)=\sin A\cos A+\cos A\sin A=2\sin A\cos A,$ so $\sin 2A=2\sin A\cos A$ — see Double Angle Formulas. With $A=15°$: $\sin 30°=2\sin 15°\cos 15°.$ Using $\sin 15°=\frac{\sqrt{6}-\sqrt{2}}{4}$ and $\cos 15°=\frac{\sqrt{6}+\sqrt{2}}{4}$: $2\cdot \frac{(\sqrt{6}-\sqrt{2})(\sqrt{6}+\sqrt{2})}{16}=2\cdot \frac{6-2}{16}=2\cdot \frac{4}{16}=\frac{1}{2}.$ And $\sin 30°=\frac{1}{2}$ ✓.

$\sin (A+B)+\sin (A-B)=(\sin A\cos B+\cos A\sin B)+(\sin A\cos B-\cos A\sin B)=2\sin A\cos B.$ So $2\sin A\cos B=\sin (A+B)+\sin (A-B)$. With $A=75°,B=15°$: $2\sin 75°\cos 15°=\sin 90°+\sin 60°=1+\frac{\sqrt{3}}{2}=\frac{2+\sqrt{3}}{2}.$

Divide numerator and denominator of the left side by $\cos A\cos B$ (the same trick as the tangent proof — WHY? to turn every term into a tangent):

Write $3\theta =2\theta +\theta$ and apply sine's sum formula: $\sin 3\theta =\sin 2\theta \cos \theta +\cos 2\theta \sin \theta .$ Insert $\sin 2\theta =2\sin \theta \cos \theta$ and $\cos 2\theta =1-2{\sin }^2\theta$ (both from Double Angle Formulas): $=2\sin \theta \cos \theta \cdot \cos \theta +(1-2{\sin }^2\theta )\sin \theta =2\sin \theta {\cos }^2\theta +\sin \theta -2{\sin }^3\theta .$ Replace ${\cos }^2\theta =1-{\sin }^2\theta$ using the Pythagorean Identity: $=2\sin \theta (1-{\sin }^2\theta )+\sin \theta -2{\sin }^3\theta =2\sin \theta -2{\sin }^3\theta +\sin \theta -2{\sin }^3\theta =3\sin \theta -4{\sin }^3\theta .■$

Expand the target with the sum formula: $R\sin (x+\varphi )=R\cos \varphi \sin x+R\sin \varphi \cos x$. Match coefficients of $\sin x$ and $\cos x$: $R\cos \varphi =\sqrt{3},R\sin \varphi =1.$ Square and add (why? to eliminate $\varphi$ via ${\cos }^2\varphi +{\sin }^2\varphi =1$): $R^2=(\sqrt{3}{)}^2+1^2=4\Rightarrow R=2.$ Divide (why? to isolate $\varphi$): $\tan \varphi =\frac{R\sin \varphi }{R\cos \varphi }=\frac{1}{\sqrt{3}}$, and since both $R\cos \varphi >0$ and $R\sin \varphi >0$, $\varphi$ is in Quadrant I: $\varphi =\frac{\pi }{6}$ ($30°$). $\boxed{\sqrt{3}\sin x+\cos x=2\sin \left(x+\frac{\pi }{6}\right).}$ The figure shows the two component waves (yellow, blue) adding to the single taller wave (pink) of amplitude $2$, shifted left by $\frac{\pi }{6}$.

$\frac{\sin (x+h)-\sin x}{h}=\frac{\sin x\cos h+\cos x\sin h-\sin x}{h}=\sin x\cdot \frac{\cos h-1}{h}+\cos x\cdot \frac{\sin h}{h}.$ As $h\to 0$, the standard small-angle limits give $\frac{\sin h}{h}\to 1$ and $\frac{\cos h-1}{h}\to 0$. Why these two limits? They isolate exactly how $\sin$ and $\cos$ behave near zero — the sum formula is what splits the shifted sine into those two known pieces. Hence: $\frac{d}{dx}\sin x=\sin x\cdot 0+\cos x\cdot 1=\cos x.$ This is precisely the bridge used in Derivatives of Trig Functions — the sum formula is what makes the derivative computable at all.