Prove that sinx=x for x is a real number has only one solution at x=0
on the same grid, graph both
y = x, which is a straight line, with slope 1
y = sinx
http://www.wolframalpha.com/input/?i=plot+y+%3D+x,+y+%3D+sinx
What do you notice?
Where do they intersect ?
slope of y = sinx
= dy/dx = cosx
when x = 0, cos0 = 1
for any value of -π/2 < x < p/2 , cosx < 1
so the graphs rise at different rates and will never meet again.
To prove that the equation sin(x) = x has only one solution at x = 0, we need to show that the two functions are equal at this specific value and also show that there are no other values of x where they intersect.
Let's start by substituting x = 0 into the equation sin(x) = x:
sin(0) = 0
The sine of 0 is indeed equal to 0. Therefore, x = 0 satisfies the equation sin(x) = x.
To show that there are no other solutions besides x = 0, we can use properties of the sine function and the slope of the line y = x.
The sine function is periodic and oscillates between -1 and 1 for all real values of x. On the other hand, the linear equation y = x represents a straight line with a slope of 1.
At x = 0, the slope of the line is 1, while the slope of the sine function at this point is 1 also because the derivative of sin(x) with respect to x at x = 0 is cos(0) = 1.
Since the slope of the line y = x at x = 0 is equal to the slope of the sine function at the same point, and the two functions intersect at this point, there can be no other intersection points. This means that x = 0 is the only solution to the equation sin(x) = x for real values of x.
Hence, it is proven that sin(x) = x has only one solution at x = 0.