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  • Help me understand! - The maximum value of for any real value of is : - Trigonometry - JEE Main

The maximum value of 3\cos \theta +5\sin \left ( \theta -\frac{\pi }{6} \right )  for any real value of \theta is :

  • Option 1)

    \sqrt{34}

     

     

     

  • Option 2)

    \sqrt{19}

  • Option 3)

    \sqrt{31}

  • Option 4)

    \frac{\sqrt{79}}{2}

Answers (1)

best_answer

 

Subtraction Formulae -

\sin \left ( A-B \right )= \sin A\cos B-\cos A\sin B

- wherein

A and B are two angles.

5(\sin \theta \cos \frac{\pi}{6}-\cos \theta \sin \frac{\pi}{6})+3\cos \theta

\frac{5\sqrt 3}{2}\sin \theta- \frac{5}{2}\cos \theta +3\cos \theta

Maximum value = \sqrt{(\frac{5\sqrt 3}{2})^{2}+(\frac{1}{2})^{2}}

                          =   \sqrt{\frac{76}{4}}

                          =\sqrt{19}

 

 


Option 1)

\sqrt{34}

 

 

 

Option 2)

\sqrt{19}

Option 3)

\sqrt{31}

Option 4)

\frac{\sqrt{79}}{2}

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