sin2x – cos2x = 1 for all values of x

Prove the identity, sin^2x + cos^2x = 1 ?

Unit Circle’s equation is x² + y² = 1

All the points on the circle contains coordinates which make the equation x² + y² = 1, true!

On the unit circle, from any arbitrary point (x, y), the representation of coordinates can be given by (sin theta + cos theta), where, theta is rotation’s degree from the x-axis which is positive.

On substitution, sin theta = x & cos theta = y into the unit circle’s equation, thus, we can observe that sin theta ² + cos theta ² = 1.

Generally, the identity given is just proven with the help of the Pythagoras theorem and the unit circle, either range of different methods or calculus.

Although if one proves it with the help of different methods, it’s important to remember that we can’t prove all the trigonometric identities with the use of alternative trigonometric identities, as that would be depending on a circular reasoning, in addition to that even if we prove it using different trigonometric identities, we should be using different methods in some or the other way.

It so appears that sin²(x)+cos²(x)=1 is known to be one of the simpler identities to verify with the use of alternative methods, and therefore, it’s usually done in this way.

Nevertheless, let’s now switch on to the proof with the formula of angle addition use for cosine:

cos(α + β)= cos(α)cos(β)−sin(α)sin(β)

Further, it’s important to note that odd function is sine and the even function is cosine, which means,

sin(−x) = −sin(x);

cos(−x) = cos(x)

Now, proceeding for the proof:

Let α = x; β = −x

Cos ( x + (-x)) = Cos (x) Cos (-x) – Sin (x) Sin (-x) ⇒ Cos (0) = cos (x) cos ( x – (-sin (x) sin (x)) …………………………eq.1

Considering the above equation

sin(−θ)= −sin(θ)

cos(−θ) = cos(θ)

Therefore, 1 = cos²(x) + sin²(x)

Another simplest method to prove this identity is:


Let’s consider the right angled triangle, having θ as the internal angle:

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Then: sinθ= a / c,

cosθ =b / c


sin²θ + cos²θ = a² / c² + b² / c² = a² + b² / c²

By the Pythagoras theorem:

As a² + b² = c²,

Accordingly, a² + b² / c² = 1 ; (a² + b² – c² = 0)

So, by the given Pythagoras, which proves this identity for

θ∈ (0, π / 2) or (0, 90º)

For angles that are present outside this range then, we can make use of:

  1. sin(−θ)= −sin(θ)
  2. Sin (θ+π) = −sin(θ)
  3. cos(−θ) = cos(θ)
  4. cos(θ+π) = −cos(θ)

Hence, for eg:

sin²(θ+π) + cos²(θ+π) = (−sinθ)² + (−cosθ)² = sin²θ + cos²θ = 1


sin²(θ+180º) + cos²(θ+180º) = (−sinθ)² + (−cosθ)² = sin²θ + cos²θ = 1

Considering Pythagoras theorem

In a right angle triangle given having a, b, and c as their sides

So, consider this below diagram:

enter image source here

Area of larger square = (a+b)²

Area of smaller square is = c²

Area of each triangle = 1 / 2 a.b

Therefore, we have:

(a+b)² = c² + 4 ⋅ 1 / 2 a*b ; i.e.

a² + 2ab + b² = c² + 2ab


From both sides, subtract “2a” to get:

a²+ b² = c2

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