# Can anyone Prove #(sin theta)^2+(cos theta)^2=1# ?

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Use the formula for a circle

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The formula for a circle centred at the origin is

#x^2+y^2=r^2#

That is, the distance from the origin to any point

Picture a circle of radius

graph{(x^2+y^2-1)((x-sqrt(3)/2)^2+(y-0.5)^2-0.003)=0 [-2.5, 2.5, -1.25, 1.25]}

If we draw a line from that point to the origin, its length is

graph{(x^2+y^2-1)(y-sqrt(3)x/3)((y-0.25)^4/0.18+(x-sqrt(3)/2)^4/0.000001-0.02)(y^4/0.00001+(x-sqrt(3)/4)^4/2.7-0.01)=0 [-2.5, 2.5, -1.25, 1.25]}

Let the angle at the origin be theta (

Now for the trigonometry.

For an angle

#sin theta = "opp"/"hyp" = y/r" "<=>" "y=rsintheta#

Similarly,

#cos theta = "adj"/"hyp"=x/r" "<=>" "x = rcostheta#

So we have

#" "x^2" "+" "y^2" "=r^2#

#(rcostheta)^2+(rsintheta) ^2 = r^2#

#r^2cos^2theta + r^2 sin^2 theta = r^2#

The

#cos^2 theta + sin^2 theta = 1#

This is often rewritten with the

#sin^2 theta + cos^2 theta = 1#

And that's it. That's really all there is to it. Just as the distance between the origin and any point

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#### Explanation

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See explanation...

#### Explanation:

Consider a right angled triangle with an internal angle

Then:

#sin theta = a/c#

#cos theta = b/c#

So:

#sin^2 theta + cos^ theta = a^2/c^2+b^2/c^2 = (a^2+b^2)/c^2#

By Pythagoras

So given Pythagoras, that proves the identity for

For angles outside that range we can use:

#sin (theta + pi) = -sin (theta)#

#cos (theta + pi) = -cos (theta)#

#sin (- theta) = - sin(theta)#

#cos (- theta) = cos(theta)#

So for example:

#sin^2 (theta + pi) + cos^2 (theta + pi) = (-sin theta)^2 + (-cos theta)^2 = sin^2 theta + cos^2 theta = 1#

**Pythagoras theorem**

Given a right angled triangle with sides

The area of the large square is

The area of the small, tilted square is

The area of each triangle is

So we have:

#(a+b)^2 = c^2 + 4 * 1/2ab#

That is:

#a^2+2ab+b^2 = c^2+2ab#

Subtract

#a^2+b^2 = c^2#

Describe your changes (optional) 200