\displaystyle \text{Question 1: Evaluate:}  \cos \Big( \sin^{-1} \frac{3}{5}+ \sin^{-1} \frac{5}{13} \Big)      

\displaystyle  \text{Answer:}  

\displaystyle  \cos \Big( \sin^{-1} \frac{3}{5}+ \sin^{-1} \frac{5}{13} \Big)

\displaystyle  = \cos \Bigg\{ \sin^{-1} \Bigg[ \frac{3}{5} \sqrt{1 - \Big(\frac{5}{13}\Big)^2} + \frac{5}{13} \sqrt{1 - \Big(\frac{3}{5}\Big)^2 } \Bigg]  \Bigg\}

\displaystyle  = \cos \Bigg\{ \sin^{-1} \Big( \frac{3}{5} \times \frac{12}{13} + \frac{5}{13} \times \frac{4}{5} \Big) \Bigg\}

\displaystyle  = \cos \Bigg\{ \sin^{-1} \Big( \frac{36}{65} + \frac{4}{13} \Big) \Bigg\}

\displaystyle  = \cos \Bigg\{ \cos^{-1} \sqrt{ 1 - \Big( \frac{56}{65} \Big)^2  } \Bigg\}

\displaystyle  = \cos \Bigg\{ \cos^{-1} \frac{33}{65} \Bigg\}

\displaystyle  = \frac{33}{65}

\\

\displaystyle \text{Question 2: Prove the following results:}   

\displaystyle \text{(i) }  \sin^{-1} \Big( \frac{63}{65} \Big) = \sin^{-1} \Big( \frac{5}{13} \Big) + \cos^{-1} \Big( \frac{3}{5} \Big) { \hspace{5.0cm} \text{[CBSE 2012]} }  

\displaystyle \text{(ii) }  \sin^{-1} \Big( \frac{5}{13} \Big) + \cos^{-1} \Big( \frac{3}{5} \Big) = \tan^{-1} \Big( \frac{63}{16} \Big)  

\displaystyle \text{(iii) }  \frac{9\pi}{8} - \frac{9}{4} \sin^{-1}\Big( \frac{1}{3} \Big) = \frac{9}{4} \sin^{-1} \Big( \frac{2\sqrt{2}}{3} \Big)    

\displaystyle  \text{Answer:}  

\displaystyle  \text{(i) }  

\displaystyle \text{RHS } = \sin^{-1} \Big( \frac{5}{13} \Big) + \cos^{-1} \Big( \frac{3}{5} \Big)

\displaystyle = \sin^{-1} \Big( \frac{5}{13} \Big) + \sin^{-1} \Big( \frac{4}{5} \Big) \ \ \ \ \ [\because \cos^{-1} x = \sin^{-1} \sqrt{1 - x^2} ]

\displaystyle = \sin^{-1} \Bigg\{ \frac{5}{13} \times \frac{3}{5} + \frac{4}{5} \times \frac{12}{13} \Bigg\}

\displaystyle = \sin^{-1} \Bigg\{ \frac{15}{65} + \frac{48}{65}\Bigg\}

\displaystyle = \sin^{-1}\frac{63}{65} = \text{LHS }

\displaystyle  \text{(ii) }  

\displaystyle \text{LHS } = \sin^{-1} \Big( \frac{5}{13} \Big) + \cos^{-1} \Big( \frac{3}{5} \Big)

\displaystyle = \sin^{-1} \Big( \frac{5}{13} \Big) + \sin^{-1} \Bigg( \sqrt{ 1 - \Big(\frac{3}{5} \Big)^2 } \Bigg) \ \ \ \ \ [\because \sin^{-1} x = \cos^{-1} \sqrt{1 - x^2} ]

\displaystyle = \sin^{-1} \Big( \frac{5}{13} \Big) + \sin^{-1} \Big( \frac{4}{5} \Big)

\displaystyle = \sin^{-1} \Bigg[ \frac{5}{13} \sqrt{1 - \Big(\frac{4}{5}\Big)^2} + \frac{4}{5} \sqrt{1 - \Big(\frac{5}{13}\Big)^2 } \Bigg] \\ \\ \ \ \ \ \ [\because \sin^{-1} x + \sin^{-1} y = \sin^{-1}( x  \sqrt{1 - y^2} + y \sqrt{1 - x^2} ]

\displaystyle =  \sin^{-1} \Big( \frac{5}{13} \times \frac{3}{5} + \frac{4}{5} \times \frac{12}{13} \Big)

\displaystyle = \sin^{-1} \Big( \frac{3}{13} + \frac{48}{65} \Big)

\displaystyle = \sin^{-1} \Big( \frac{63}{65} \Big)

\displaystyle = \tan^{-1} \Bigg( \frac{ \frac{63}{65}  }{\sqrt{1 - \Big(\frac{63}{55}\Big)^2}} \Bigg) \ \ \ \ \ [\because \sin^{-1} x =  \tan^{-1} \Big(  \frac{x}{\sqrt{1 - x^2}}  \Big) ] 

\displaystyle = \tan^{-1} \Bigg( \frac{ \frac{63}{65}  }{\frac{16}{65}} \Bigg)

\displaystyle = \tan^{-1} \Big( \frac{63}{16} \Big) = \text{RHS }

\displaystyle  \text{(iii) }  

\displaystyle \text{LHS } = \frac{9\pi}{8} - \frac{9}{4} \sin^{-1}\Big( \frac{1}{3} \Big)

\displaystyle = \frac{9}{4} \Big( \frac{\pi}{2} - \sin^{-1} \frac{1}{3} \Big)

\displaystyle = \frac{9}{4} \Big( \cos^{-1} \frac{1}{3} \Big)

\displaystyle = \frac{9}{4} \Bigg(\sin^{-1} \sqrt{1 - \frac{1}{9}} \Bigg)

\displaystyle = \frac{9}{4} \Bigg( \sin^{-1} \frac{2\sqrt{2}}{3} \Bigg) = \text{RHS }

\\

\displaystyle \text{Question 3: Solve the following:}   

\displaystyle \text{(i)}  \sin^{-1}  x + \sin^{-1}  2x = \frac{\pi}{3} 

\displaystyle \text{(ii)}  \cos^{-1} x + \sin^{-1}  \frac{x}{2} - \frac{\pi}{6} = 0 { \hspace{7.0cm} \text{[CBSE 2012]} }  

\displaystyle  \text{Answer:}  

\displaystyle  \text{(i) }  

\displaystyle \text{Given, }  \sin^{-1}  x + \sin^{-1}  2x = \frac{\pi}{3}

\displaystyle \Rightarrow \sin^{-1}  x + \sin^{-1}  2x = \sin^{-1} \Big( \frac{\sqrt{3}}{2} \Big)

\displaystyle \Rightarrow \sin^{-1}  x  -  \sin^{-1} \Big( \frac{\sqrt{3}}{2} \Big) =  - \sin^{-1}  2x

\displaystyle \Rightarrow \sin^{-1}  \Big[ x \sqrt{1 - \frac{3}{4}} + \frac{\sqrt{3}}{2} \sqrt{1 - x^2} \Big] = - \sin^{-1} 2x

\displaystyle \Rightarrow \sin^{-1}  \Big[ \frac{x}{2} + \frac{\sqrt{3}}{2} \sqrt{1 - x^2} \Big] =  \sin^{-1} (-2x)

\displaystyle \Rightarrow \frac{x}{2} + \frac{\sqrt{3}}{2} \sqrt{1 - x^2} = -2x

\displaystyle \Rightarrow  x + \sqrt{3} \sqrt{1=x^2} = -4x

\displaystyle \Rightarrow 5x = - \sqrt{3} \sqrt{1=x^2}

\displaystyle \Rightarrow 25x^2 = 3 - 3x^2

\displaystyle \Rightarrow 28x^2 = 3

\displaystyle \Rightarrow x = \pm \frac{1}{2} \sqrt{\frac{3}{7}}

\displaystyle  \text{(ii) }  

\displaystyle \text{Given, }  \cos^{-1} x + \sin^{-1}  \frac{x}{2} - \frac{\pi}{6} = 0

\displaystyle \Rightarrow  \cos^{-1} x + \sin^{-1}  \frac{x}{2} =  \frac{\pi}{6}

\displaystyle \Rightarrow  \Big( \frac{\pi}{2} - \sin^{-1} \Big) + \sin^{-1}  \frac{x}{2} =  \frac{\pi}{6}

\displaystyle \Rightarrow  \frac{\pi}{2} - \frac{\pi}{6} =  \sin^{-1} x - \sin^{-1} \frac{x}{2}

\displaystyle \Rightarrow \sin^{-1} x - \sin^{-1} \frac{x}{2} = \frac{\pi}{3}

\displaystyle \Rightarrow \sin^{-1} x - \sin^{-1} \frac{x}{2} = \sin^{-1} \frac{\sqrt{3}}{2}

\displaystyle \Rightarrow \sin^{-1} x  = \sin^{-1} \frac{\sqrt{3}}{2} + \sin^{-1} \frac{x}{2}

\displaystyle \Rightarrow \sin^{-1} x  = \sin^{-1}\Bigg( \frac{\sqrt{3}}{2} \sqrt{1 - \frac{x^2}{4}} + \frac{x}{2} \sqrt{1 - \frac{3}{4}} \Bigg)  \\ \\ \ \ \ \ \ [\because \sin^{-1} x + \sin^{-1} y = \sin^{-1}( x  \sqrt{1 - y^2} + y \sqrt{1 - x^2} ] 

\displaystyle \Rightarrow \sin^{-1} x  = \sin^{-1}\Bigg[ \Bigg(  \frac{\sqrt{3}}{2} \frac{\sqrt{4-x^2}}{2} \Bigg) + \frac{x}{2} \cdot \frac{1}{2} \Bigg]

\displaystyle \Rightarrow x = \frac{\sqrt{3}}{2} \frac{\sqrt{4-x^2}}{2} + \frac{x}{4}

\displaystyle \Rightarrow x - \frac{x}{4} = \frac{\sqrt{3}}{2} \frac{\sqrt{4-x^2}}{2}

\displaystyle \Rightarrow  \frac{3x}{4} = \frac{\sqrt{3}}{2} \frac{\sqrt{4-x^2}}{2}

\displaystyle \Rightarrow  9x^2 = 3( 4 - x^2)

\displaystyle \Rightarrow 3x^2 =4- x^2

\displaystyle \Rightarrow 4x^2 = 4

\displaystyle \Rightarrow x^2 = 1

\displaystyle \Rightarrow x = \pm 1

\\

Discover more from ICSE / ISC / CBSE Mathematics Portal for K12 Students

Subscribe to get the latest posts sent to your email.