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Trigonometry> root 2 cosx+ root6 sinx= max (sintheta + ...

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root 2 cosx+ root6 sinx= max (sintheta + costheta) where theta belongs to R find the general value ofx

keerthi priya , 10 Years ago

Grade 11

2 Answers

Jitender Singh

Last Activity: 10 Years ago

Ans:Hello student, please find answer to your question
$\sqrt{2}cosx + \sqrt{6}sinx = max(sin\theta +cos\theta )$
$f(\theta ) = \sin\theta +cos\theta$
$f'(\theta ) = cos\theta -sin\theta = 0$
$\theta = n\pi + \frac{\pi }{4}$
$[f(\theta )]_{max} = \sqrt{2}$
$\sqrt{2}cosx + \sqrt{6}sinx = \sqrt{2}$
$cosx + \sqrt{3}sinx = 1$
$\frac{1}{2}.cosx + \frac{\sqrt{3}}{2}sinx = \frac{1}{2}$
$sin(x + \frac{\pi }{6}) = sinx.\frac{\sqrt{3}}{2} + cosx.\frac{1}{2}$
$sin(x + \frac{\pi }{6}) = \frac{1}{2}$
$x + \frac{\pi }{6} = 2n\pi + \frac{\pi }{6}$
$x = 2n\pi$

keerthi priya

Last Activity: 10 Years ago

thnk u very much

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