Q.If $\sin x = \dfrac{15}{17}$ and $\cos y = \dfrac{12}{13}$, $0 < x < \dfrac{\pi}{2}$, $0 < y < \dfrac{\pi}{2}$, find the value of

Sum and Difference (Compound Angle) Identities

A compound angle is an angle expressed as an algebraic sum or difference of two (or more) angles, e.g. $\alpha+\beta$ or $\alpha-\beta$. Because $\sin,\cos,\tan$ are not linear functions, a trig ratio of a compound angle can not be found by applying the function to each piece separately and combining ($\cos(\alpha+\beta)\ne\cos\alpha+\cos\beta$, in general). Instead, six standard identities give the exact relationship, and all six can be derived from a single geometric fact about chord lengths on the unit circle.

The six identities

Where they come from

Identity (1) is proved first and everything else is a short substitution away. Place four points on the unit circle centred at $O$: the fixed point $P=(1,0)$, and $Q=(\cos\alpha,\sin\alpha)$, $R=(\cos(\alpha+\beta),\sin(\alpha+\beta))$, $S=(\cos(-\beta),\sin(-\beta))$, chosen so $\angle POR=\alpha+\beta$ and $\angle QOS=\alpha-(-\beta)=\alpha+\beta$ — the same central angle. Equal central angles on a circle of the same radius cut off equal chords, so $PR=SQ$, i.e. $PR^2=SQ^2$. Writing both squared lengths with the distance formula and simplifying with $\cos^2+\sin^2=1$ collapses directly to Identity (1). (The argument is carried out for $0\le\alpha,\beta<2\pi$, but periodicity extends it to every real $\alpha,\beta$.)

Every other identity is then a two-line substitution, never a fresh geometric argument:

• (2) from (1): write $\alpha-\beta=\alpha+(-\beta)$ and use $\cos(-\beta)=\cos\beta,\ \sin(-\beta)=-\sin\beta$.
• (3) from (2): write $\sin\theta=\cos(\tfrac\pi2-\theta)$ and expand $\cos\big[(\tfrac\pi2-\alpha)-\beta\big]$.
• (4) from (3): write $\alpha-\beta=\alpha+(-\beta)$.
• (5) from (3) and (1): divide $\sin(\alpha+\beta)$ by $\cos(\alpha+\beta)$, then divide every term top and bottom by $\cos\alpha\cos\beta$.
• (6) from (5): write $\alpha-\beta=\alpha+(-\beta)$ and use $\tan(-\beta)=-\tan\beta$.

Special cases worth remembering

• Setting $\alpha=\beta$ in (2): $\cos(\alpha-\alpha)=\cos^2\alpha+\sin^2\alpha$, i.e. $1=\cos^2\alpha+\sin^2\alpha$ — the Pythagorean identity re-emerges as a consistency check.
• Setting $\alpha=0,\ \beta=x$ in (2): $\cos(-x)=\cos x$, i.e. cosine is an even function.
• Setting $\alpha=\tfrac\pi2,\ \beta=\theta$ in (4): $\sin(\tfrac\pi2-\theta)=\cos\theta$, the co-function relation used to derive (3) in the first place.
• Setting $\alpha+\beta=\tfrac\pi2$ in (3): reduces again to $\cos^2\alpha+\sin^2\alpha=1$.

Why they matter

Any angle decomposable into a sum or difference of the standard special angles ($0^\circ,30^\circ,45^\circ,60^\circ,90^\circ,\dots$) now has an exact trig value — e.g. $75^\circ=45^\circ+30^\circ$, $105^\circ=60^\circ+45^\circ$, $165^\circ=120^\circ+45^\circ$. The identities are also the algebraic engine behind three-angle expansions such as $\sin(A+B+C)$ and $\tan(A+B+C)$ (grouped as $A+(B+C)$, then the two-angle identities applied twice), and behind the well-known triangle fact $\tan A+\tan B+\tan C=\tan A\tan B\tan C$ whenever $A+B+C=\pi$.

Multiple-Angle and Sub-multiple (Half) Angle Identities

If $A$ is an angle, its multiples are $2A,3A,4A,\ldots$ and its sub-multiples are $\dfrac A2,\dfrac A3,\ldots$. This part of the concept collects the identities that rewrite the sine/cosine/tangent of a multiple or sub-multiple angle purely in terms of the ratios of the original angle.

Double-angle identities

All follow immediately from the sum identities $\sin(\alpha+\beta)$, $\cos(\alpha+\beta)$, $\tan(\alpha+\beta)$ by setting $\beta=\alpha$ (and, for the last two, dividing through by $\cos^2A+\sin^2A=1$).

Power-reducing identities

Solving the all-cosine double-angle forms for the squared ratio:

Triple-angle identities

Writing $3A=2A+A$ and expanding with the sum + double-angle identities:

Half-angle (sub-multiple) identities

Putting $2A=\theta$ (so $A=\theta/2$) in every double-angle identity converts it into a half-angle identity:

and solving the middle pair for the half-angle ratio itself:

(sign fixed by the quadrant $\theta/2$ lies in).

Two workhorse techniques for the exercise problems

• The $A+(45^\circ-A)$ pairing. If $A+B=45^\circ$, then $\tan(A+B)=1$ gives $\tan A+\tan B=1-\tan A\tan B$, and substituting into $(1+\tan A)(1+\tan B)=1+\tan A+\tan B+\tan A\tan B$ collapses it to exactly $2$ -- the building block behind products like $(1+\tan1^\circ)(1+\tan2^\circ)\cdots(1+\tan44^\circ)$, which pairs off into $22$ such factors of $2$.
• Telescoping via $1+\sec2A=\dfrac{\tan2A}{\tan A}$. Since $1+\sec2A=\dfrac{1+\cos2A}{\cos2A}=\dfrac{2\cos^2A}{\cos2A}=\dfrac{\tan2A}{\tan A}$ (multiply top and bottom of the middle fraction by $\sin A/\cos A$), a chain $(1+\sec2\theta)(1+\sec4\theta)\cdots(1+\sec2^n\theta)$ collapses ratio-by-ratio to $\dfrac{\tan2^n\theta}{\tan\theta}=\tan2^n\theta\cot\theta$. The same telescoping idea, run on $2\sin A\cos A=\sin2A$ instead, collapses a chain of cosines $\cos A\cos2A\cos4A\cdots$ down to a single sine over a power of $2$.

Worked illustration -- the golden-ratio exact values ($\sin18^\circ$)

Let $\theta=18^\circ$, so $5\theta=90^\circ$, i.e. $2\theta=90^\circ-3\theta$, giving $\sin2\theta=\cos3\theta$: $2\sin\theta\cos\theta=4\cos^3\theta-3\cos\theta$. Dividing by $\cos\theta\ne0$ and writing $\cos^2\theta=1-\sin^2\theta$ turns this into a quadratic in $\sin\theta$:

Taking the positive root (first quadrant): $\sin18^\circ=\dfrac{\sqrt5-1}4$. From this, $\cos18^\circ=\dfrac14\sqrt{10+2\sqrt5}$, and the complementary-angle relations give $\sin72^\circ=\cos18^\circ$, $\cos36^\circ=1-2\sin^218^\circ=\dfrac{\sqrt5+1}4$, and $\sin54^\circ=\cos36^\circ$ -- five exact values, all built from one small quadratic.