what-is-the-maximum-and-minimum-value-of-sinx-cosx-sinxcosx-

Question Number 10458 by paonky last updated on 10/Feb/17

what is the maximum and minimum

value of \sin x + \cos x + \sin x \cos x

Answered by mrW1 last updated on 10/Feb/17

\sin x + \cos x

= \sqrt{2} (\sin x \times \frac{\sqrt{2}}{2} + \cos x \times \frac{\sqrt{2}}{2})

= \sqrt{2} (\sin x \times \cos \frac{π}{4} + \cos x \times \sin \frac{π}{4})

= \sqrt{2} \sin (x + \frac{π}{4})

\sin x \cos x = \frac{1}{2} \sin 2 x = - \frac{1}{2} \cos (2 x + \frac{π}{2})

= - \frac{1}{2} \cos 2 (x + \frac{π}{4}) = - \frac{1}{2} [1 - {2 \sin}^{2} (x + \frac{π}{4})]

= \sin^{2} (x + \frac{π}{4}) - \frac{1}{2}

f = \sin x + \cos x + \sin x \cos x

= \sqrt{2} \sin (x + \frac{π}{4}) + \sin^{2} (x + \frac{π}{4}) - \frac{1}{2}

= {(\frac{\sqrt{2}}{2})}^{2} + 2 \times \frac{\sqrt{2}}{2} \sin (x + \frac{π}{4}) + \sin^{2} (x + \frac{π}{4}) - 1

= {[\frac{\sqrt{2}}{2} + \sin (x + \frac{π}{4})]}^{2} - 1

w h e n \sin (x + \frac{π}{4}) = - \frac{\sqrt{2}}{2}, i . e . x = π + 2 i π o r x = \frac{3 π}{2} + 2 i π

\Rightarrow f_{m i n} = - 1

w h e n \sin (x + \frac{π}{4}) = 1, i . e . x = \frac{π}{4} + 2 i π

\Rightarrow f_{m a x} = {(\frac{\sqrt{2}}{2} + 1)}^{2} - 1 = \frac{1}{2} + \sqrt{2} \approx 1.914

Commented by paonky last updated on 10/Feb/17

thank you sir

Facebook Tweet Pin