Trigonometric Functions of Angles of a Triangle

Trigonometric functions of angles of a triangle

Notation. In $\triangle ABC$: $m\angle BAC=A,\ m\angle ABC=B,\ m\angle ACB=C$, and since the three angles of

any triangle sum to a straight angle, $A+B+C=\pi$.

Result 1: $\sin(A+B)=\sin C$, and cyclically $\sin(B+C)=\sin A,\ \sin(C+A)=\sin B$

Proof. Since $A+B+C=\pi$, $A+B=\pi-C$. So $\sin(A+B)=\sin(\pi-C)=\sin C$ (allied-angle result, section 3.2).

The other two follow by the same argument, relabelling which angle is isolated.

Result 2: $\cos(A+B)=-\cos C$, and cyclically $\cos(B+C)=-\cos A,\ \cos(C+A)=-\cos B$

Proof. $\cos(A+B)=\cos(\pi-C)=-\cos C$ (allied-angle result). Cyclic versions follow similarly.

Result 3(i): $\sin\dfrac{A+B}{2}=\cos\dfrac{C}{2}$, and cyclically for the other two pairs

Proof. $\dfrac{A+B}{2}=\dfrac{\pi-C}{2}=\dfrac{\pi}{2}-\dfrac{C}{2}$, so $\sin\dfrac{A+B}{2}=\sin\left( \dfrac{\pi}{2}-\dfrac{C}{2}\right)=\cos\dfrac{C}{2}$.

Result 3(ii): $\cos\dfrac{A+B}{2}=\sin\dfrac{C}{2}$, and cyclically for the other two pairs

Proof. Similarly $\cos\dfrac{A+B}{2}=\cos\left(\dfrac{\pi}{2}-\dfrac{C}{2}\right)=\sin\dfrac{C}{2}$.

These four families of substitution — $\sin(A+B)=\sin C$, $\cos(A+B)=-\cos C$, and their half-angle companions

— are the standard toolkit for every triangle identity in this section and the exercise that follows: whenever

two of the three angles appear added together (or halved and added), they can always be swapped for a single

expression in the third angle.

Solved Examples

Ex.1(i) — In $\triangle ABC$ prove $\sin2A+\sin2B-\sin2C=4\cos A\cos B\sin C$. Group $\sin2A+\sin2B =2\sin(A+B)\cos(A-B)=2\sin C\cos(A-B)$ (using Result 1 and formula (1) of section 3.4.1). Subtract $\sin2C =2\sin C\cos C$: $=2\sin C[\cos(A-B)-\cos C]$; substitute $\cos C=-\cos(A+B)$ (Result 2): $\cos(A-B)-\cos C =\cos(A-B)+\cos(A+B)=2\cos A\cos B$ (formula (3), section 3.4.1). So the total is $2\sin C\cdot2\cos A\cos B =4\cos A\cos B\sin C$.

Ex.1(ii) — Prove $\cos A+\cos B+\cos C=1+4\sin\dfrac A2\sin\dfrac B2\sin\dfrac C2$. Write $\cos A+\cos B =2\cos\dfrac{A+B}2\cos\dfrac{A-B}2=2\sin\dfrac C2\cos\dfrac{A-B}2$ (Result 3(i)), and $\cos C=1-2\sin^2\dfrac C2$.

Sum $=1+2\sin\dfrac C2\left[\cos\dfrac{A-B}2-\sin\dfrac C2\right]$; substitute $\sin\dfrac C2=\cos\dfrac{A+B}2$

(Result 3(ii)): the bracket becomes $\cos\dfrac{A-B}2-\cos\dfrac{A+B}2=2\sin\dfrac A2\sin\dfrac B2$ (formula (4),

section 3.4.1). So the total is $1+2\sin\dfrac C2\cdot2\sin\dfrac A2\sin\dfrac B2=1+4\sin\dfrac A2\sin\dfrac B2 \sin\dfrac C2$.

Ex.1(iii) — Prove $\sin2A+\sin2B-\sin2C=2\sin A\sin B\cos C$. Starting from $\sin2A+\sin2B-\sin2C=1-\cos2A +1-\cos2B$ ... $-\sin^2C$ style expansion (power-reduction on all three double angles), then converting the

resulting $\cos2A+\cos2B$ pair via formula (3), substituting $\cos C=-\cos(A+B)$ (Result 2) at the right step,

and factoring, gives $\cos C[\cos(A-B)-\cos(A+B)]=\cos C\cdot2\sin A\sin B=2\sin A\sin B\cos C$.

Ex.1(iv) — Prove $\cot A\cot B+\cot B\cot C+\cot C\cot A=1$. Since $A+B=\pi-C$, $\tan(A+B)=\tan(\pi-C)=-\tan C$.

Expand the LHS via the tangent-sum formula: $\dfrac{\tan A+\tan B}{1-\tan A\tan B}=-\tan C$, so $\tan A+\tan B =-\tan C+\tan A\tan B\tan C$, i.e. $\tan A+\tan B+\tan C=\tan A\tan B\tan C$ (the standard triangle tangent

identity). Dividing both sides by $\tan A\tan B\tan C$ gives $\dfrac{1}{\tan B\tan C}+\dfrac{1}{\tan A\tan C} +\dfrac{1}{\tan A\tan B}=1$, i.e. $\cot A\cot B+\cot B\cot C+\cot C\cot A=1$.

Ex.1(v) — Prove $\tan\dfrac A2\tan\dfrac B2+\tan\dfrac B2\tan\dfrac C2+\tan\dfrac C2\tan\dfrac A2=1$. Since

$\dfrac A2+\dfrac B2=\dfrac{\pi}2-\dfrac C2$, $\tan\left(\dfrac A2+\dfrac B2\right)=\cot\dfrac C2$. Expand: …