Free Study Material

NCERT Solutions

NCERT Books

Revision Notes

Practice Problems

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

Product-to-sum formulas Quiz 2

mathematics

  $\sin (5{{x}^{2}}+3{{y}^{4}}-y-z)\cos (8{{x}^{3}}-2y+99z)$ is equal to:

 Use the product-to-sum formula $ \sin \alpha \cos \beta =\dfrac{1}{2}[\sin (\alpha +\beta )+\sin (\alpha -\beta )]$ .

$  \sin (5{{x}^{2}}+3{{y}^{4}}-y-z)\cos (8{{x}^{3}}-2y+99z) \\ =\dfrac{1}{2}\left[ \sin \left( 5{{x}^{2}}+3{{y}^{4}}-y-z+8{{x}^{3}}-2y+99z \right)+\sin \left( 5{{x}^{2}}+3{{y}^{4}}-y-z-8{{x}^{3}}+2y-99z \right) \right] \\ =\dfrac{1}{2}\left[ \sin \left( 3{{y}^{4}}+8{{x}^{3}}+5{{x}^{2}}-3y+98z \right)+\sin \left( 3{{y}^{4}}-8{{x}^{3}}+5{{x}^{2}}+y-100z \right) \right]  $

  Evaluate: $ \left[ \sin \left( 7x \right) \right]\left[ \sin 2x \right]$  at $x=\dfrac{\pi }{2}$.

 Use the formula, $ \sin \alpha \sin \beta =\dfrac{1}{2}\left[ \cos \left( \alpha -\beta  \right)-\cos \left( \alpha +\beta  \right) \right]$  then put the trigonometric values for the exact value.

 $  \left[ \sin \left( 7x \right) \right]\left[ \sin 2x \right] \\ =\dfrac{1}{2}\left[ \cos \left( 7x-2x \right)-\cos \left( 7x+2x \right) \right] \\ =\dfrac{1}{2}\left( \cos 5x-\cos 9x \right)\Rightarrow \dfrac{1}{2}\left[ \cos \left( 5\times 90{}^\circ  \right)-\cos \left( 9\times 90{}^\circ  \right) \right] \\ =\dfrac{1}{2}\left[ \cos \left( 360{}^\circ +90{}^\circ  \right)-\cos \left( 810{}^\circ -720{}^\circ  \right) \right] \\ =\dfrac{1}{2}\left( \cos 90{}^\circ -\cos 90{}^\circ  \right)=\dfrac{1}{2}\left( 0-0 \right)=0 $

 Evaluate: $ f+g\text{, if }f=\cos 75{}^\circ \sin 45{}^\circ \text{ and }g=\cos 150{}^\circ \sin 90{}^\circ $

 Use the product-to-sum formula $ \cos \alpha \sin \beta =\dfrac{1}{2}[\sin (\alpha +\beta )-\sin (\alpha -\beta )]$  for both the functions given and then add the values after putting the trigonometric ratios.

 $  f=\cos 75{}^\circ \sin 45{}^\circ  \\ =\dfrac{1}{2}\left[ \sin \left( 75{}^\circ +45{}^\circ  \right)-\sin \left( 75{}^\circ -45{}^\circ  \right) \right]=\dfrac{1}{2}\left( \sin 120{}^\circ -\sin 30{}^\circ  \right)=\dfrac{1}{2}\times \left( \dfrac{\sqrt{3}}{2}-\dfrac{1}{2} \right) \\ g=\cos 150{}^\circ \sin 90{}^\circ  \\ =\dfrac{1}{2}\left[ \sin \left( 150{}^\circ +90{}^\circ  \right)-\sin \left( 150{}^\circ -90{}^\circ  \right) \right]=\dfrac{1}{2}\left( \sin 240{}^\circ -\sin 60{}^\circ  \right)=\dfrac{1}{2}\times \left( -\dfrac{\sqrt{3}}{2}-\dfrac{1}{2} \right) \\ \therefore f+g=-\dfrac{1}{2} $

 Find the value of $ \cos 2\theta {{\sin }^{-2}}\theta $ if it is given that$ a=\dfrac{1}{\sqrt{7}},\text{ }\dfrac{1}{2}\left( \sin A+\sin B \right)=a\cos \left( \dfrac{A-B}{2} \right)\text{ and }\theta $  is the average of $A\text{ and }B$.

 Use the product-to-sum formula $ \left( \sin A+\sin B \right)=2\sin \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)$  and the trigonometric identity $\cos 2x={{\cos }^{2}}x-{{\sin }^{2}}x$, then put the value of $\cos ec\theta $.

 $   \dfrac{1}{2}\left( \sin A+\sin B \right)=a\cos \left( \dfrac{A-B}{2} \right)\text{ } \\ \text{Now, }\left( \sin A+\sin B \right)=2\sin \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)=2\sin \theta \cos \left( \dfrac{A-B}{2} \right)\text{           }\left[ \because \dfrac{A+B}{2}=\theta  \right] \\ \Rightarrow 2a\cos \left( \dfrac{A-B}{2} \right)=2\sin \theta \cos \left( \dfrac{A-B}{2} \right)\Rightarrow a=\sin \theta =\dfrac{1}{\sqrt{7}}\text{    } \\ \text{Here, }\cos 2\theta {{\sin }^{-2}}\theta =\dfrac{\cos 2\theta }{{{\sin }^{2}}\theta } \\ \Rightarrow \cos 2\theta {{\sin }^{-2}}\theta =\dfrac{{{\cos }^{2}}\theta -{{\sin }^{2}}\theta }{{{\sin }^{2}}\theta }=\dfrac{1-2{{\sin }^{2}}\theta }{{{\sin }^{2}}\theta }=\dfrac{1}{{{\sin }^{2}}\theta }-\dfrac{2{{\sin }^{2}}\theta }{{{\sin }^{2}}\theta } \\ \Rightarrow \cos 2\theta {{\sin }^{-2}}\theta =\cos e{{c}^{2}}\theta -2 \\ \therefore \cos 2\theta {{\sin }^{-2}}\theta =7-2=5\text{         }\left[ \because \sin \theta =\dfrac{1}{\sqrt{7}}\Rightarrow \text{cosec}\theta =\sqrt{7}\text{  } \right] $

 Find the value of $'xy'$ if $\dfrac{x}{2}=\cos \left( \dfrac{11\theta }{2} \right)\cos \left( \dfrac{5\theta }{2} \right)\text{ and }\dfrac{y}{2}=\sin \left( \dfrac{11\theta }{2} \right)\sin \left( \dfrac{5\theta }{2} \right)$

 Use the product-to-sum formula, $ \cos \alpha \cos \beta =\dfrac{1}{2}[\cos (\alpha -\beta )+\cos (\alpha +\beta )]\text{ and sin}\alpha \sin \beta =\dfrac{1}{2}[\cos (\alpha -\beta )-\cos (\alpha +\beta )]$  to find values of $''x''$ and $''y''$. Now, find their product.

 $\dfrac{x}{2}=\cos \left( \dfrac{11\theta }{2} \right)\cos \left( \dfrac{5\theta }{2} \right)=2\cos \left( \dfrac{11\theta }{2} \right)\cos \left( \dfrac{5\theta }{2} \right) \\ \Rightarrow x=\cos \left( \dfrac{11\theta }{2}-\dfrac{5\theta }{2} \right)+\cos \left( \dfrac{11\theta }{2}+\dfrac{5\theta }{2} \right) \\ \Rightarrow x=\cos \left( \dfrac{6\theta }{2} \right)+\cos \left( \dfrac{16\theta }{2} \right)=\cos 3\theta +\cos 8\theta  \\ \dfrac{y}{2}=\sin \left( \dfrac{11\theta }{2} \right)\sin \left( \dfrac{5\theta }{2} \right) \\ \Rightarrow y=\cos \left( \dfrac{11\theta }{2}-\dfrac{5\theta }{2} \right)-\cos \left( \dfrac{11\theta }{2}+\dfrac{5\theta }{2} \right) \\ \Rightarrow y=\cos 3\theta -\cos 8\theta  \\ \therefore xy=\left( \cos 3\theta +\cos 8\theta  \right)\left( \cos 3\theta -\cos 8\theta  \right)={{\cos }^{2}}3\theta -{{\cos }^{2}}8\theta  $