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

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Feb 13, 2017

Answer:

Use the formula for a circle #(x^2+y^2=r^2)#, and substitute #x=rcostheta# and #y=rsintheta#.

Explanation:

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 #(x,y)# on the circle is the radius #r# of the circle.

Picture a circle of radius #r# centred at the origin, and pick a point #(x,y)# on the circle:

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 #r#. We can also draw a triangle for that point as follows:

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 (#theta#).

Now for the trigonometry.

For an angle #theta# in a right triangle, the trig function #sin theta# is the ratio #"opposite side"/"hypotenuse"#. In our case, the length of the side opposite of #theta# is the #y#-coordinate of our point #(x,y)#, and the hypotenuse is our radius #r#. So:

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

Similarly, #cos theta# is the ratio of the #x#-coordinate in #(x,y)# to the radius #r#:

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

So we have #x=rcostheta# and #y=rsintheta#. Substituting these into the circle formula gives

#"      "x^2"     "+"      "y^2"     "=r^2#
#(rcostheta)^2+(rsintheta) ^2 = r^2#
#r^2cos^2theta + r^2 sin^2 theta = r^2#

The #r^2#'s all cancel, leaving us with

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

This is often rewritten with the #sin^2# term in front, like this:

#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 #(x,y)# on a circle must be the circle's radius, the sum of the squared values for #sin theta# and #cos theta# must be 1 for any angle #theta#.

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Sep 17, 2017

Answer:

See explanation...

Explanation:

Consider a right angled triangle with an internal angle #theta#:

enter image source here

Then:

#sin theta = a/c#

#cos theta = b/c#

So:

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

By Pythagoras #a^2+b^2 = c^2#, so #(a^2+b^2)/c^2 = 1#

So given Pythagoras, that proves the identity for #theta in (0, pi/2)#

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#

#color(white)()#
Pythagoras theorem

Given a right angled triangle with sides #a#, #b# and #c# consider the following diagram:

enter image source here

The area of the large square is #(a+b)^2#

The area of the small, tilted square is #c^2#

The area of each triangle is #1/2ab#

So we have:

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

That is:

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

Subtract #2ab# from both sides to get:

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

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