Conditional Trigonometric Identities

An identity in the usual sense (like $\sin^2\theta+\cos^2\theta=1$) is true for every admissible value of the angle, with no extra assumption needed. A conditional trigonometric identity, by contrast, is a relation between trigonometric functions of several angles that is true only when those angles satisfy some extra stated condition — most commonly that they are the three angles of a triangle, so $A+B+C=180^\circ=\pi$ (or, in radian problems phrased slightly differently, $A+B+C=\pi/2$, or a symmetric perimeter-style condition like $A+B+C=2s$). Take away the condition and the "identity" is simply false for generic $A,B,C$.

The one trick that proves almost all of them. Whenever $A+B+C=\pi$ (or any fixed constant), you can always eliminate one angle in favour of the other two — e.g. $C=\pi-A-B$, so $\cos C=-\cos(A+B)$ and $\sin C=\sin(A+B)$; or, working with half-angles, $\frac{A+B}2=\frac{\pi}2-\frac C2$, so $\sin\frac C2=\cos\frac{A+B}2$ and $\cos\frac C2=\sin\frac{A+B}2$. Substituting one of these relations into the sum-to-product identities of §3.5.3 is what turns a three-angle expression into a compact product — this section is really an extended workout of that one substitution idea.

Cosine sum for a triangle. For $A+B+C=\pi$, group $\cos A+\cos B$ first via the sum-to-product rule, giving $2\cos\frac{A+B}2\cos\frac{A-B}2$; since $\frac{A+B}2=\frac\pi2-\frac C2$, this is $2\sin\frac C2\cos\frac{A-B}2$. Writing $\cos C=1-2\sin^2\frac C2$ (the double-angle form) turns the whole sum $\cos A+\cos B+\cos C$ into $1+2\sin\frac C2\big[\cos\frac{A-B}2-\sin\frac C2\big]$; replacing $\sin\frac C2$ once more by $\cos\frac{A+B}2$ and combining the bracket with another sum-to-product step collapses everything to

Two useful corollaries follow immediately. First, setting $u=\sin\frac A2\sin\frac B2\sin\frac C2$ and re-expressing it (again via a product-to-sum step, this time treating it as a quadratic in $\cos\frac{A+B}2$) shows the quadratic's discriminant must be non-negative for a real solution to exist, which forces $u\le\frac18$ — i.e. $\sin\frac A2\sin\frac B2\sin\frac C2\le\frac18$. Second, combining $u\le\frac18$ with the identity above (and the fact that $u>0$ for a genuine triangle, since every half-angle lies strictly between $0$ and $\frac\pi2$) sandwiches the cosine sum: $1<\cos A+\cos B+\cos C\le\frac32$.

Sine of the half-angles. A parallel computation — this time starting from $\sin\frac A2+\sin\frac B2+\sin\frac C2$, rewritten as a sum of cosines via the complementary-angle swap $\sin\frac A2=\cos\big(\frac\pi2-\frac A2\big)$, then grouped and reduced exactly as above but at the quarter-angle level — shows

The mechanics are identical to the cosine-sum proof, just carried out one level down (working with quarter-angles rather than half-angles), which is why the same "convert to a sum-to-product pattern, then substitute the triangle condition" strategy reappears.

Sum of squared cosines. For $\cos^2A+\cos^2B+\cos^2C$, first double every term using $\cos^2\theta=\frac{1+\cos2\theta}2$, so the sum becomes $\frac32+\frac12\big[(\cos2A+\cos2B)+\cos2C\big]$. Converting $\cos2A+\cos2B$ to a product gives $2\cos(A+B)\cos(A-B)$, and since $A+B=\pi-C$ we have $\cos(A+B)=-\cos C$; substituting and simplifying (using $\cos2C=2\cos^2C-1$ once more) collapses the whole expression to …