Conditional Trigonometric Identities

Suppose that  \(A + B + C = \pi \),  Consider the expression  \(\sin 2A + \sin 2B + \sin 2C\).

Let us convert this sum into a product using the constraint specified.

\[\begin{align}&\sin 2A + \sin 2B + \sin 2C = 2(A + B)\cos (A - B) + \sin 2C \\&\qquad\qquad\qquad\qquad\qquad\;\,= 2\sin C\cos (A - B) + 2\sin C\cos C\\ &\qquad\qquad\qquad\qquad\qquad\;\,= 2\sin C(\cos (A - B) - \cos (A + B)) \\&\qquad\qquad\qquad\qquad\qquad\;\,= 4\sin A\sin B\sin C \\ \end{align} \]

The identity

\[\sin 2A + \sin 2B + \sin 2C = 4\sin A\sin B\sin C \qquad{\text{for}}\;A + B + C = \pi \]

is an example of a conditional identity. It is an identity because it is satisfied for all values of the angles A, B, C which satisfy the constraint \(A + B + C = \pi.\)

Here are a few more examples, all having same constraint  \(A + B + C = \pi \).

(i)

\[\begin{align}&\sin 2A + \sin 2B - \sin 2C = 2\sin (A + B)\cos (A - B) - \sin 2C\\ \,\, &\qquad\qquad\qquad\qquad\qquad\;= 2\sin C\cos (A - B) - 2\sin C\cos C \\ \,\, &\qquad\qquad\qquad\qquad\qquad\;= 2\sin C(\cos (A - B) + \cos (A + B)) \\ \,\, &\qquad\qquad\qquad\qquad\qquad\;= 4\cos A\cos B\sin C  \\ \end{align} \]

(ii)

\[\begin{align}&\cos 2A + \cos 2B + \cos 2C = 2\cos (A + B)\cos (A - B) + \cos 2C  \\\,\,\,\, &\qquad\qquad\qquad\qquad\qquad\;\;=  - 2\cos C\cos (A - B) + 2{\cos ^2}C - 1  \\\,\,\,\, &\qquad\qquad\qquad\qquad\qquad\;\;= 2\cos C(\cos C - \cos (A - B)) - 1  \\\,\,\,\, &\qquad\qquad\qquad\qquad\qquad\;\;=  - 2\cos C(\cos (A + B) + \cos (A - B)) - 1 \\\,\,\,\, &\qquad\qquad\qquad\qquad\qquad\;\;=  - 1 - 4\cos A\cos B\cos C  \\ \end{align} \]

(iii)  \[\sin A + \sin B + \sin C = 2\sin \left( {\frac{{A + B}}{2}} \right)\cos \left( {\frac{{A - B}}{2}} \right) + \sin C\]

Now,  \(\begin{align}\frac{{A + B}}{2} = \frac{{\pi - C}}{2} = \frac{\pi }{2} - \frac{C}{2} \qquad \Rightarrow \qquad \sin \left( {\frac{{A + B}}{2}} \right) = \cos \frac{C}{2}\end{align}\)

Also,  \(\begin{align}\sin C = 2\sin \frac{C}{2}\cos \frac{C}{2}.\end{align}\)  Thus, the sum becomes

\[\begin{align}&2\cos \frac{C}{2}\left\{ {\sin \frac{C}{2} + \cos \left( {\frac{{A - B}}{2}} \right)} \right\}  \\ &= 2\cos \frac{C}{2}\left\{ {\sin \left( {\frac{{\pi  - (A + B)}}{2}} \right) + \cos \left( {\frac{{A - B}}{2}} \right)} \right\} \\ &= 2\cos \frac{C}{2}\left\{ {\cos \left( {\frac{{A + B}}{2}} \right) + \cos \left( {\frac{{A - B}}{2}} \right)} \right\}  \\   &= 4\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}  \\ \end{align} \]

(iv)

\[\begin{align}&\cos A + \cos B + \cos C   = 2\cos \left( {\frac{{A + B}}{2}} \right)\cos \left( {\frac{{A - B}}{2}} \right) + \cos C \\  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &\qquad\qquad\qquad\qquad\quad\;= 2\sin \frac{C}{2}\cos \left( {\frac{{A - B}}{2}} \right) + 1 - 2{\sin ^2}\frac{C}{2} \\  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &\qquad\qquad\qquad\qquad\quad\;= 1 + 2\sin \frac{C}{2}\left\{ {\cos \left( {\frac{{A - B}}{2}} \right) - \cos \left( {\frac{{A + B}}{2}} \right)} \right\} \\  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &\qquad\qquad\qquad\qquad\quad\;= 1 + 4\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}  \\ \end{align} \]

(v)

\[\begin{align}&{\sin ^2}A + {\sin ^2}B - {\sin ^2}C = {\sin ^2}A + \sin (B + C)\sin (B - C)  \\  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &\qquad\qquad\qquad\qquad\;\;\quad= {\sin ^2}A + \sin A\sin (B - C)  \\  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &\qquad\qquad\qquad\qquad\;\;\quad= \sin A(\sin (B + C) + \sin (B - C)) \\  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &\qquad\qquad\qquad\qquad\;\;\quad= 2\sin A\sin B\sin C\\ \end{align} \]