# How to prove the identity `sin^2x + cos^2x = 1` ?

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Another method is knowing to take the derivative of

f(x) = sin^2(x) + cos^2(x)

f '(x) = 2 sin(x) cos(x) + 2 cos(x) (-sin(x))

= 2 sin(x) cos(x) - 2 cos(x) sin(x)

= 0

Since the derivative is zero everywhere the function must be a constant.

Take f(0) = sin^2(0) + cos^2(0) = 0 + 1 = 1

So

sin^2(x) + cos^2(x) = 1 everywhere.

An alternate approach to proving this identity involves using the "unit circle" (radius = 1). Since the radius is also the hypotenuse of the right triangle formed by the angle "x" within the circle, the sine is y and the cosine is x. By the Pythagorean Theorem, x^2 + y^2 = 1^2 ... or x^2 + y^2 = 1

The equation of the unit circle is x^2+y^2=1.

All points on this circle have coordinates that make this equation true.

For any random point (x, y) on the unit circle, the coordinates can be represented by (cos `theta` , sin `theta` ) where `theta` is the degrees of rotation from the positive x-axis (see attached image).

By substituting cos `theta` = x and sin `theta` = y into the equation of the unit circle, we can see that (cos `theta` )^2 + (sin `theta` )^2 = 1.

sin^2X + cos^2X = 1

L.H.S

=sin^2X + cos^2X

=sinXsinX+cosXcosX

=cos(X-X)

=cos0

=1

R.H.S

Keep in mind that sin t = y/r, cos t = x/r, and r^2 = x^2+y^2. By substitution, sin^2x = y^2/r^2 and cos^2x = x^2/r^2. So the left hand side of the identitiy becomes (y^2/r^2) + (x^2/r^2). Since these already have a common denominator, we can write them as one fraction: (y^2+x^2)/r^2 = r^2/r^2 = 1.

from pythagoras theorem.....

**r^2=x^2+y^2**

divide through by **r^2**..

**i.e..r^2/r^2=x^2/r^2+y^2/r^2**

:. it implies that.. **1=x^2/r^2+y^2/r^2**

**1=(x/r)^2+(y/r)^2**

but **sinΘ=y/r,cosΘ=x/r**

by substitution...

**1=sin^2Θ+cos^2Θ**

You can prove it by using pythagoras' theorem using the following equation:

r^2=y^2+x^2

Then divide throughout by r^2

y^2/r^2 + x^2/y^2 = 1

then subsituting **sin Θ = y/r, cos**** Θ = x/r**

sin^2x + Cos^2x = 1 (shown)

In a right triangle, let the opposite side of angle x is a, the other right-angle side is b, and the hypontenuse is r.

So, we can get: sin^2x+cos^2x=(a^2)/r^2+(b^2)/r^2=(a^2+b^2)/r^2=r^2/r^2=1

Remember: sin is y/r & cos is x/r

So:

LHS= sin ²x + cos ²x

= (y/r ) ² (x/r) ²

=y²/r² + x²/r²

= (y² + x²)/ r² r² = x² + y² (theorem of Pythagoras)

= r²/r²

=1

= RHS

cosΘ = adjacent / hypotenuse (b/c) We know from the Pythagorean theorum that a² + b² = c² Divide both sides by c² 1/c² (a² + b²) = 1/c² ( c²) Then a²/c² + b²/c² = c²/c² (a/c)² + (b/c)² = 1 sin²Θ + cos²Θ = 1

sin theta=p/h

cos theta=b/h

Therefore,

sin^2 theta=p^2/h^2

cos^2 theta=b^2/h^2

Now,

sin^2 theta + cos^2 theta=(p^2/h^2) + (b^2/h^2)

=(p^2 + b^2)/h^2

By pythagoras' theorem,

p^2 + b^2=h^2

therefore,

(p^2 + b^2)/h^2 = h^2/h^2

=1

Hence, proved

This is not a proof, but it sure is compelling evidence:

Enter y=(sin(x))^2+(cos(x))^2 into a graphing calculator and look at the result for -2pi < x < 2pi. Yep, it's the constant function y=1.

Or think about it without technology. What would the graph of f(x)=(sin(x))^2 look like? All points with y-values of 0 or 1 would not change, and points with y=-1 would keep their x-values but get y-values of +1. So this graph would be zero at even multiples of pi/2 and 1 at odd multiples of pi/2. A similar analysis of g(x)=(cos(x))^2 gives a graph that is 1 where f(x) is 0, and 0 where f(x) is 1. So y=f(x)+g(x) is clearly 1 in all of those places.

Again, it's not a proof, but it's good to have a graphical look at this important trig identity.

**sin^2X + cos^2X = 1**

**L.H.S**

**=sin^2X + cos^2X**

**=sinXsinX+cosXcosX**

**=cos(X-X)**

**=cos0**

**=1= R.H.S hence proved!!**

sin^2X + cos^2X = 1

L.H.S

=sin^2X + cos^2X

=sinXsinX+cosXcosX

=cos(X-X)

=cos0

=1

R.H.S

In a right angle triangle,

sinx=perpendicular(p)/hypotenuse(h)

cosx=base(b)/hypotenuse(h)

Now,

L.H.S=Sin^2x+Cos^2x

=P^2/h^2 + b^2/h^2

=(p^2+b^2)/h^2

=h^2/h^2 (p^2+b^2 =h^2)

=1

proved