Sum and difference identities or compound angles formulas

Identity 3.1 — $\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta$.

This is proved first; every other identity in this sub-section is then a short substitution away from it. Work on the unit circle centred at $O$, fix $P=(1,0)$, and mark three more points at the standard positions given by the angles $\alpha$, $\alpha+\beta$, and $-\beta$ (all measured from $OP$):

The angle $\angle QOS=\alpha-(-\beta)=\alpha+\beta$ is exactly the same central angle as $\angle POR=\alpha+\beta$. Since triangles $POR$ and $SOQ$ share two radii of equal length and the same included angle (SAS), they are congruent, so the chord $PR$ equals the chord $SQ$ in length — i.e. $PR^2=SQ^2$.

Compute both squared distances with the ordinary distance formula:

Expand every square and repeatedly use $\cos^2\theta+\sin^2\theta=1$; all the squared single-ratio terms collapse to $1$'s, the $2$'s cancel, and (using $\cos(-\beta)=\cos\beta,\ \sin(-\beta)=-\sin\beta$) what remains is

The underlying geometric fact is simply that the straight-line distance between two points on a circle depends only on the radius and the central angle between them — that is exactly why $PR=SQ$ once the two central angles agree. The argument is carried out for $0\le\alpha,\beta<2\pi$, but periodicity of sine/cosine then extends the identity to every real $\alpha,\beta$.

Identity 3.2 — $\cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta$.

Rather than repeat the geometric argument, write $\alpha-\beta=\alpha+(-\beta)$ and reuse Identity 3.1:

Two quick special cases: setting $\beta=\alpha$ collapses this to $\cos^2\alpha+\sin^2\alpha=1$ (a consistency check, not new information); setting $\alpha=0,\beta=x$ gives $\cos(-x)=\cos x$ — i.e. cosine is an even function.

Identity 3.3 — $\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta$.

Write sine as a shifted cosine, $\sin\theta=\cos\!\left(\tfrac\pi2-\theta\right)$, applied at $\theta=\alpha+\beta$:

Expand the right side as a difference of the angles $\tfrac\pi2-\alpha$ and $\beta$ using Identity 3.2:

using the co-function identities $\cos(\tfrac\pi2-\alpha)=\sin\alpha$, $\sin(\tfrac\pi2-\alpha)=\cos\alpha$. As a check, if $\alpha+\beta=\tfrac\pi2$ the identity again reduces to $\cos^2\alpha+\sin^2\alpha=1$.

Identity 3.4 — $\sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta$.

Write $\alpha-\beta=\alpha+(-\beta)$ and apply Identity 3.3:

Identity 3.5 — $\tan(\alpha+\beta)=\dfrac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}$.

Write tangent as sine over cosine and substitute the sum formulas:

Divide every term, top and bottom, by $\cos\alpha\cos\beta$ (valid when $\cos\alpha,\cos\beta\ne0$):

Identity 3.6 — $\tan(\alpha-\beta)=\dfrac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}$.

Write $\alpha-\beta=\alpha+(-\beta)$, substitute into Identity 3.5, and use $\tan(-\beta)=-\tan\beta$:

A few consequences and reading notes worth keeping in mind:

• $\cos(\alpha\pm\beta)\ne\cos\alpha\pm\cos\beta$ in general (and likewise for sine/tangent) — the identities above are the actual relationship, never a term-by-term split.
• Setting $\alpha=\beta$ in Identity 3.4 gives $\sin(\alpha-\alpha)=\sin\alpha\cos\alpha-\cos\alpha\sin\alpha$, i.e. $\sin0=0$.
• Setting $\alpha=\tfrac\pi2,\beta=\theta$ in Identity 3.4 gives $\sin(\tfrac\pi2-\theta)=\cos\theta$ — the same co-function fact used above to derive Identity 3.3.
• Practical payoff. Any angle expressible as a sum/difference of the "special" angles ($0^\circ,30^\circ,45^\circ,60^\circ,90^\circ,\dots$) now has an exact value: e.g. $\tan75^\circ=\tan(45^\circ+30^\circ)$, $\cos135^\circ=\cos(180^\circ-45^\circ)$. Exercise 3.4 leans heavily on this trick.

Worked applications (the textbook's Examples 3.15–3.20). All six use only the identities above:

• Splitting $15^\circ=45^\circ-30^\circ$ and $165^\circ=120^\circ+45^\circ$ (with $\tan120^\circ=\tan(90^\circ+30^\circ)=-\cot30^\circ=-\sqrt3$ found along the way) gives exact values for $\cos15^\circ$ and $\tan165^\circ$ via Identities 3.2 and 3.5.
• Given $\sin x$ in one quadrant and $\cos y$ in another, first fix the sign of the missing ratio in each quadrant using the Pythagorean identity, then substitute both pairs into Identities 3.2 and 3.4 to get exact fractions for $\sin(x-y)$ and $\cos(x-y)$.
• An identity like $\cos\!\left(\tfrac{3\pi}4+x\right)-\cos\!\left(\tfrac{3\pi}4-x\right)=-\sqrt2\sin x$ is proved by expanding both cosine terms with Identities 3.1/3.2, cancelling the matching $\cos\tfrac{3\pi}4\cos x$ pieces, and simplifying what remains — the same pattern that gives the general rule $\cos(A+x)-\cos(A-x)=-2\sin A\sin x$.
• A point rotated about the origin through a fixed angle has its new coordinates found by writing the original point as $(r\cos\theta,r\sin\theta)$ (with $r$ the distance from the origin), then applying Identities 3.1/3.3 to $(r\cos(\theta+\phi),\,r\sin(\theta+\phi))$ for the rotation angle $\phi$ — a genuine coordinate-geometry application. …