# Conditional Trigonometric Identities

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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}

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