Problem (1) For what value of $x$ $x$ , $sin x + cos x = 2$ $sinx+cosx=2$ Problem (2) For what value of $x$ $x$ , $sin x cos x = 1$ $sinxcosx=1$ ?

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Answer 1 · 2017-03-14T14:44:15

$(sin x + cos x)$ $(sinx+cosx)$ can never be $2$ $2$

and $sin x cos x$ $sinxcosx$ can never be $1$ $1$ .

Problem (1)

$sin x + cos x$ $sinx+cosx$ can never be $2$ $2$ , the maximum value of $sin x + cos x$ $sinx+cosx$ can only be $\sqrt{2}$ $sqrt2$ as

$sin x + cos x = \sqrt{2} (sin x \times \frac{1}{\sqrt{2}} + cos x \times \frac{1}{\sqrt{2}})$ $sinx+cosx=sqrt2(sinx xx 1/sqrt2+cosx xx 1/sqrt2)$

= $\sqrt{2} (sin x cos (\frac{π}{4}) + cos x sin (\frac{π}{4}))$ $sqrt2(sinxcos(pi/4)+cosxsin(pi/4))$

= $\sqrt{2} sin (x + \frac{π}{4})$ $sqrt2sin(x+pi/4)$

as maximum value of any sine ratio can only be $1$ $1$ ,

maximum value of $\sqrt{2} sin (x + \frac{π}{4})$ $sqrt2sin(x+pi/4)$ or $sin x + cos x$ $sinx+cosx$ can only be $\sqrt{2}$ $sqrt2$ .

Hence $sin x + cos x$ $sinx+cosx$ can never be $2$ $2$ .

Problem (2)

As $sin x cos x = \frac{1}{2} (2 sin x cos x) = \frac{1}{2} \times sin 2 x$ $sinxcosx=1/2(2sinxcosx)=1/2xxsin2x$

and again as maximum value of any sine ratio can only be $1$ $1$ ,

maximum value of $sin x cos x$ $sinxcosx$ or $\frac{1}{2} sin 2 x$ $1/2sin2x$ can only be $\frac{1}{2}$ $1/2$

and it can never be more than $\frac{1}{2}$ $1/2$ and hence cannot be $1$ $1$ .

Problem (1) For what value of xx, sinx+cosx=2sinx+cosx=2 Problem (2) For what value of xx, sinxcosx=1sinxcosx=1?