Free Study Material

NCERT Solutions

NCERT Books

Revision Notes

Practice Problems

This is one of our many quizzes on Sum-to-product formulas. Click on this to access more such competitive quizzes on Trigonometry.

Sum-to-product formulas Quiz 2

mathematics

 Evaluate the value of$\dfrac{{\left( {\cos 10x + \cos 2x} \right) + \left( {\cos 7x + \cos 5x} \right)}}{{\left( {\sin 10x + \sin 2x} \right) + \left( {\sin 7x + \sin 5x} \right)}}$.

 Apply the formula of $\cos C + \cos D = 2\cos \dfrac{{C + D}}{2}\cos \dfrac{{C - D}}{2}$ and $\sin C + \sin D = 2\sin \dfrac{{C + D}}{2}\cos \dfrac{{C - D}}{2}$ to solve the trigonometric expressions. 

  $  \dfrac{{\left( {\cos 10x + \cos 2x} \right) + \left( {\cos 7x + \cos 5x} \right)}}{{\left( {\sin 10x + \sin 2x} \right) + \left( {\sin 7x + \sin 5x} \right)}}   \\   \Rightarrow \cos C + \cos D = 2\cos \dfrac{{C + D}}{2}\cos \dfrac{{C - D}}{2}   \\   \Rightarrow \sin C + \sin D = 2\sin \dfrac{{C + D}}{2}\cos \dfrac{{C - D}}{2}   \\   \Rightarrow \dfrac{{2\cos 6x\cos 4x + 2\cos 6x\cos x}}{{2\sin 6x\cos 4x + 2\sin 6x\cos x}}   \\   \Rightarrow \dfrac{{2\cos 6x\left( {\cos 4x + \cos x} \right)}}{{2\sin 6x\left( {\cos 4x + \cos x} \right)}}   \\   \Rightarrow \cot 6x   \\  $

 Evaluate the value of $\dfrac{{\sin 5x + \sin x}}{{\cos 5x + \cos x}}.$

 Apply the formula of $\sin C + \sin D = 2\sin \dfrac{{C + D}}{2}\cos \dfrac{{C - D}}{2}$ and $\cos C + \cos D = 2\cos \dfrac{{C + D}}{2}\cos \dfrac{{C - D}}{2}$ to solve the trigonometric expressions. 

  $  \dfrac{{\sin 5x + \sin x}}{{\cos 5x + \cos x}}   \\   \Rightarrow \sin C + \sin D = 2\sin \dfrac{{C + D}}{2}\cos \dfrac{{C - D}}{2}   \\   \Rightarrow \cos C + \cos D = 2\cos \dfrac{{C + D}}{2}\cos \dfrac{{C - D}}{2}   \\   \Rightarrow \dfrac{{2\sin 3x + \cos 2x}}{{2\cos 3x + \cos 2x}}   \\   \Rightarrow \dfrac{{\sin 3x}}{{\cos 3x}}   \\   \Rightarrow \tan 3x   \\  $

 Value of trigonometric expression $\dfrac{{\sin {{53}^ \circ } + \sin {{37}^ \circ }}}{{\cos {{53}^ \circ } + \cos {{37}^ \circ }}}.$

 Apply the formula of $\sin C + \sin D = 2\sin \dfrac{{C + D}}{2}\cos \dfrac{{C - D}}{2}$ and $\cos C + \cos D = 2\cos \dfrac{{C + D}}{2}\cos \dfrac{{C - D}}{2}$ , then put the values of trigonometric values of solve the trigonometric expressions. 

$  \dfrac{{\sin {{53}^ \circ } + \sin {{37}^ \circ }}}{{\cos {{53}^ \circ } + \cos {{37}^ \circ }}}   \\   \Rightarrow \sin C + \sin D = 2\sin \dfrac{{C + D}}{2}\cos \dfrac{{C - D}}{2}   \\   \Rightarrow \cos C + \cos D = 2\cos \dfrac{{C + D}}{2}\cos \dfrac{{C - D}}{2}   \\   \Rightarrow \dfrac{{2\sin {{45}^\circ }\cos {8^\circ }}}{{2\cos {{45}^\circ }\cos {8^\circ }}}   \\   \Rightarrow \dfrac{{\sin {{45}^\circ }}}{{\cos {{45}^\circ }}}   \\   \Rightarrow \dfrac{{\dfrac{1}{{\sqrt 2 }}}}{{\dfrac{1}{{\sqrt 2 }}}}  \Rightarrow 1   \\  $

 Evaluate the value of$\dfrac{{{{\left( {\cos 3\theta  + i\sin 3\theta } \right)}^4}{{(\cos 5\theta  - i\sin 5\theta )}^{\dfrac{4}{5}}}}}{{{{\left( {\cos \dfrac{3}{2}\theta  + i\sin \dfrac{3}{2}\theta } \right)}^{\dfrac{2}{9}}}{{\left( {\cos \dfrac{4}{5}\theta  - i\sin \dfrac{4}{5}\theta } \right)}^{10}}}}$.

 Apply Demoivre’s theorem, in mathematical form it states that ${\left( {\cos x + i\sin x} \right)^n}\; = \cos \left( {nx} \right) + i\sin \left( {nx} \right)$.

 $\dfrac{{{{\left( {\cos 3\theta  + i\sin 3\theta } \right)}^4}{{(\cos 5\theta  - i\sin 5\theta )}^{\dfrac{4}{5}}}}}{{{{\left( {\cos \dfrac{3}{2}\theta  + i\sin \dfrac{3}{2}\theta } \right)}^{\dfrac{2}{9}}}{{\left( {\cos \dfrac{4}{5}\theta  - i\sin \dfrac{4}{5}\theta } \right)}^{10}}}} \\ $ By De Moivre’s theorem, $ \\ \dfrac{{{{\left( {\cos 3\theta  + i\sin 3\theta } \right)}^4}{{(\cos 5\theta  - i\sin 5\theta )}^{\dfrac{4}{5}}}}}{{{{\left( {\cos \dfrac{3}{2}\theta  + i\sin \dfrac{3}{2}\theta } \right)}^{\dfrac{2}{9}}}{{\left( {\cos \dfrac{4}{5}\theta  - i\sin \dfrac{4}{5}\theta } \right)}^{10}}}}   \\  {(\cos 3\theta  + i\sin 3\theta )^4} = \cos 12\theta  + i\sin 12\theta    \\  {(\cos 5\theta  - i\sin 5\theta )^{\dfrac{4}{5}}} = \cos 4\theta  - i\sin 4\theta    \\   {\left( {\cos \dfrac{3}{2}\theta  + i\sin \dfrac{3}{2}\theta } \right)^{\dfrac{2}{9}}}   \\   = \cos \left[ {\dfrac{2}{9}\left( {\dfrac{3}{2}\theta } \right)} \right] + i\sin \left[ {\dfrac{2}{9}\left( {\dfrac{3}{2}\theta } \right)} \right]   \\   = \cos \left( {\dfrac{\theta }{3}} \right) + i\sin \left( {\dfrac{\theta }{3}} \right)   \\ \text{ Now, }  \\   {\left( {\cos \dfrac{4}{5}\theta  - i\sin \dfrac{4}{5}\theta } \right)^{10}}   \\   = \cos \left[ {10\left( {\dfrac{4}{5}\theta } \right)} \right] + i\sin \left[ {10\left( {\dfrac{4}{5}\theta } \right)} \right]   \\   = \cos (8\theta ) - i\sin (8\theta )   \\   \dfrac{{{{(\cos 3\theta  + i\sin 3\theta )}^4}{{(\cos 5\theta  - i\sin 5\theta )}^{\dfrac{4}{5}}}}}{{{{\left( {\cos \dfrac{3}{2}\theta  + i\sin \dfrac{3}{2}\theta } \right)}^{\dfrac{2}{9}}}{{\left( {\cos \dfrac{4}{5}\theta  - i\sin \dfrac{4}{5}\theta } \right)}^{10}}}}   \\   = \dfrac{{\left( {\cos 12\theta  + i\sin 12\theta } \right)\left( {\cos 4\theta  - i\sin 4\theta } \right)}}{{\left( {\cos \dfrac{\theta }{3} + i\sin \dfrac{\theta }{3}} \right)\left( {\cos 8\theta  - i\sin 8\theta } \right)}}   \\   = {(\cos \theta  + i\sin \theta )^{12 - 4\dfrac{1}{3} + 8}}   \\   = {(\cos \theta  + i\sin \theta )^{\dfrac{{47}}{3}}}   \\   = \cos \dfrac{{47}}{3}\theta  + i\sin \dfrac{{47}}{3}\theta    \\  $

 Value of $\dfrac{{\sin 3x + \sin x + \sin 2x}}{{\cos 3x + \cos x + \cos 2x}}$ is

 $ \dfrac{{\sin 3x + \sin x + \sin 2x}}{{\cos 3x + \cos x + \cos 2x}}   \\   \Rightarrow \sin C + \sin D = 2\sin \dfrac{{C + D}}{2}\cos \dfrac{{C - D}}{2}   \\   \Rightarrow \cos C + \cos D = 2\cos \dfrac{{C + D}}{2}\cos \dfrac{{C - D}}{2}   \\   \Rightarrow \dfrac{{2\sin 2x\cos x + \sin 2x}}{{2\cos 2x\cos x + \cos 2x}}   \\   \Rightarrow \dfrac{{\sin 2x\left( {2\cos x + 1} \right)}}{{\cos 2x\left( {2\cos x + 1} \right)}}   \\   \Rightarrow \tan 2x   \\  $