Math, asked by gurpreetsingh22961, 3 months ago

find the minimum and maximum value of cos(x+π/3)+sinx

Answers

Answered by mathdude500
1

  \large \underline{\tt \:  \red{ According  \: to  \: statement }}

\tt \longmapsto\:f(x) = cos(x + \dfrac{\pi}{3} ) + sinx

\tt \longmapsto\:f(x) = cosx \: cos\dfrac{\pi}{3}  - sinx \: sin\dfrac{\pi}{3}  + sinx

\tt \longmapsto\:f(x) = \dfrac{1}{2}cosx -  \dfrac{ \sqrt{3} }{2} sinx  +  sinx

\tt \longmapsto\:f(x) = \dfrac{1}{2}cosx +  \bigg(1 - \dfrac{ \sqrt{3} }{2}  \bigg) sinx

\tt \longmapsto\:f(x) = \dfrac{1}{2}cosx +  \bigg( \dfrac{ 2 - \sqrt{3} }{2}  \bigg) sinx

Now,

We know that

  • If f(x) = a sinx + b cosx, then

\tt \longmapsto\:minimum \: value \: of \: f(x) =  -  \sqrt{ {a}^{2} +  {b}^{2}  }

And

\tt \longmapsto\:maximum \: value \: of \: f(x) =\sqrt{ {a}^{2} +  {b}^{2}  }

So,

  • Minimum value is given by

 \tt \:  =  -  \:  \sqrt{ {\bigg( \dfrac{1}{2}\bigg)  }^{2}  + {\bigg(\dfrac{2 -  \sqrt{3} }{2} \bigg) }^{2}  }

 \tt \:  =  -   \: \sqrt{\dfrac{1}{4} + \dfrac{4 + 3 - 4 \sqrt{3} }{4}  }

 \tt \:  =   - \:  \sqrt{\dfrac{1 + 7 - 4 \sqrt{3} }{4} }

 \tt \:  =  \:  -  \:  \sqrt{\dfrac{8 - 4 \sqrt{3} }{4} }

 \tt \:  =  \:  -  \:  \sqrt{2  -  \sqrt{3} }

And

  • Maximum value is given by

 \tt \:  =    \:  \sqrt{ {\bigg( \dfrac{1}{2}\bigg)  }^{2}  + {\bigg(\dfrac{2 -  \sqrt{3} }{2} \bigg) }^{2}  }

 \tt \:  =    \: \sqrt{\dfrac{1}{4} + \dfrac{4 + 3 - 4 \sqrt{3} }{4}  }

 \tt \:  =  \:  \sqrt{\dfrac{1 + 7 - 4 \sqrt{3} }{4} }

 \tt \:  =  \:  \sqrt{\dfrac{8 - 4 \sqrt{3} }{4} }

 \tt \:  =  \:  \sqrt{2 -  \sqrt{3} }

Hence,

 \tt \:  -  \sqrt{2 -  \sqrt{3} }  \leqslant cos(x + \dfrac{\pi}{3}) + sinx \leqslant  \sqrt{2 -  \sqrt{3} }

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