Trigonometry and Properties of Quadratic Equation

If $ \displaystyle \tan A$ and $ \displaystyle \tan B$ are the roots of the equation $ \displaystyle {{x}^{2}}-px+q=0$, where $ \displaystyle p$ and $ \displaystyle q$ are real constants, find the value of $ \displaystyle {{\sin }^{2}}(A+B)$ in terms of $ \displaystyle p$ and  $ \displaystyle q$.

Solution

             $ \displaystyle \ \ \ \ \tan A$ and $ \displaystyle \tan B$ are the roots of the equation $ \displaystyle {{x}^{2}}-px+q=0$,

             $ \displaystyle \therefore {{x}^{2}}-px+q=(x-\tan A)(x-\tan B)$

             $ \displaystyle \therefore {{x}^{2}}-px+q={{x}^{2}}-(\tan A+\tan B)x+\tan A\cdot \tan B$
         
             $ \displaystyle \therefore \tan A+\tan B=p$ and $ \displaystyle \tan A\cdot \tan B=q$

             $ \displaystyle \therefore \frac{{\tan A+\tan B}}{{1-\tan A\tan B}}=\frac{p}{{1-q}}$

             $ \displaystyle \therefore \tan (A+B)=\frac{p}{{1-q}}$

             $ \displaystyle \therefore \frac{{\sin (A+B)}}{{\cos (A+B)}}=\frac{p}{{1-q}}$

             $ \displaystyle \therefore \cos (A+B)=\frac{{1-q}}{p}\cdot \sin (A+B)$

             $ \displaystyle \ \ \ $ Since $ \displaystyle {{\sin }^{2}}(A+B)+{{\cos }^{2}}(A+B)=1$

             $ \displaystyle \ \ \ \ {{\sin }^{2}}(A+B)+{{\left( {\frac{{1-q}}{p}} \right)}^{2}}{{\sin }^{2}}(A+B)=1$

             $ \displaystyle \ \ \ \ {{\sin }^{2}}(A+B)\left( {1+\frac{{{{{\left( {1-q} \right)}}^{2}}}}{{{{p}^{2}}}}} \right)=1$

             $ \displaystyle \therefore {{\sin }^{2}}(A+B)=\frac{{{{p}^{2}}}}{{{{p}^{2}}+{{{\left( {1-q} \right)}}^{2}}}}$

Law of Cosines and to Find Extremum by Completing Square

 In $ \displaystyle \Delta ABC$, $ \displaystyle AB=(5-x)$ cm, $ \displaystyle BC=(4+x)$ cm, $ \displaystyle \angle AsBC=120{}^\circ $ and $ \displaystyle AC=y$ cm.

(a)     Show that $ \displaystyle {{y}^{2}}={{x}^{2}}-x+61$.

(b)     Find the minimum value of $ \displaystyle {{y}^{2}}$, and give the value of $ \displaystyle x$ for which this occurs.

Solution

          $ \displaystyle AB=(5-x)$ cm,  
          $ \displaystyle BC=(4+x)$ cm, 
          $ \displaystyle \angle ABC=120{}^\circ $ 
          $ \displaystyle AC=y$ cm. 

(a)      By the law of cosines, 

          $ \displaystyle A{{C}^{2}}=A{{B}^{2}}+B{{C}^{2}}-2\cdot AB\cdot AC\cos (\angle ABC)$ 
      
          $ \displaystyle {{y}^{2}}={{(5-x)}^{2}}+{{(4+x)}^{2}}-2(5-x)(4+x)\cos 120{}^\circ $

          $ \displaystyle {{y}^{2}}=25-10x+{{x}^{2}}+16+8x+{{x}^{2}}+2(5-x)(4+x)\left( {\frac{1}{2}} \right)$

          $ \displaystyle {{y}^{2}}=41-2x+2{{x}^{2}}-{{x}^{2}}+x+20$

          $ \displaystyle {{y}^{2}}={{x}^{2}}-x+61$

      $ \displaystyle \therefore {{y}^{2}}={{x}^{2}}-x+\frac{1}{4}+61-\frac{1}{4}$

      $ \displaystyle \therefore {{y}^{2}}={{\left( {x-\frac{1}{2}} \right)}^{2}}+60.75$

          Since $ \displaystyle {{\left( {x-\frac{1}{2}} \right)}^{2}}\ge 0\ $ for all $ \displaystyle x\in R$,

          $ \displaystyle {{\left( {x-\frac{1}{2}} \right)}^{2}}+60.75\ge 60.75$

      $ \displaystyle \therefore {{y}^{2}}\ge 60.75$.

          Therefore the minimum value of $\displaystyle {{y}^{2}}$ is $ \displaystyle 60.75$ and this value occurs when $ \displaystyle x=\frac{1}{2}$.