Graphically find $x$ $x$ for which $sin x = cos x$ $sinx=cosx$ ?

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Graphically find $x$ $x$ for which $sin x = cos x$ $sinx=cosx$ ?

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Answer 1 · 2016-09-18T05:00:36

Please see below.

Explanation:

As $sin (x + \frac{π}{2}) = cos x$ $sin(x+pi/2)=cosx$ , $sin x$ $sinx$ function moves with a gap of $\frac{π}{2}$ $pi/2$ with respect to $cos x$ $cosx$ . Between $x = 0$ $x=0$ and $x = \frac{π}{2}$ $x=pi/2$ , $cos x$ $cosx$ falls from $1$ $1$ to $0$ $0$ and $sin x$ $sinx$ rises from $0$ $0$ to $1$ $1$ .

Further $cos (- x) = cos x$ $cos(-x)=cosx$ and hence while $cos x$ $cosx$ is symmetric around $y$ $y$ -axis i.e. $x = 0$ $x=0$ , As $sin x$ $sinx$ appears with a lag of $\frac{π}{2}$ $pi/2$ , it is symmetric around $x = \frac{π}{2}$ $x=pi/2$ .

Hence, $cos x = sin x$ $cosx=sinx$ appears exactly at the midpoint between $0$ $0$ and $\frac{π}{2}$ $pi/2$ i.e. at $\frac{π}{4} = 0.7854$ $pi/4=0.7854$
graph{(y-sinx)(y-cosx)=0 [-1.365, 3.635, -1, 1.5]}

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