# Examples on Conditional Trigonometric Identities Set 1

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Example - 26

If  $$A + B + C = \pi$$, prove that

${\sin ^2}\frac{A}{2} + {\sin ^2}\frac{B}{2} - {\sin ^2}\frac{C}{2} = 1 - 2\cos \frac{A}{2}\cos \frac{B}{2}\sin \frac{C}{2}$

Solution:

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

Example - 27

If  $$A + B + C = \pi$$, prove that

${\sin ^3}A\cos (B - C) + {\sin ^3}B\cos (C - A) + {\sin ^3}C\cos (A - B) = 3\sin A\sin B\sin C$

Solution: Lets concentrate on one term of the LHS:

\begin{align}&{\sin ^3}A\cos (B - C) = \frac{1}{2}{\sin ^2}A\left\{ {2\sin (B + C)\cos (B - C)} \right\} \\ \,\,\,\,\, &\qquad\qquad\qquad\;\quad= \frac{1}{2}{\sin ^2}A\left\{ {\sin 2B + \sin 2C} \right\} \\ \,\,\,\,\, &\qquad\qquad\qquad\;\quad= \underbrace {{{\sin }^2}A\{ \sin B\cos B}_{} + \sin C\cos C\} \\ \end{align}

From the term  $${\sin ^3}B\cos (C - A)$$, we’ll have a term of the form $${\sin ^2}B\sin A\cos A$$  generated. Adding this to the highlighted term above gives

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

This term is generated from one pair. We have three pairs and so the total sum is $$3\sin A\sin B\sin C$$.

Example - 28

If $$A + B + C = \pi$$, show that

(a) $$\tan A + \tan B + \tan C = \tan A\tan B\tan C$$

(b) \begin{align}\tan \frac{A}{2}\tan \frac{B}{2} + \tan \frac{B}{2}\tan \frac{C}{2} + \tan \frac{C}{2}\tan \frac{A}{2} = 1\end{align}

Solution: (a) Since  \begin{align}&A + B + C = \pi , A + B = \pi - C\\ \Rightarrow\qquad &\tan (A + B) = - \tan C \\ \Rightarrow\qquad &\frac{{\tan A + \tan B}}{{1 - \tan A\tan B}} = - \tan C \\ \end{align}

Cross-multiplying and rearranging generates the required identity.

(b)\begin{align}tan \left( {\frac{A}{2} + \frac{B}{2}} \right) = \tan \left( {\frac{\pi }{2} - \frac{C}{2}} \right) = \cot \frac{C}{2}\end{align}

$\Rightarrow \frac{{\tan \frac{A}{2} + \tan \frac{B}{2}}}{{1 - \tan \frac{A}{2}\tan \frac{B}{2}}} = \frac{1}{{\tan \frac{C}{2}}}$

Again, cross-multiplying and rearranging yields the required identify.