# How do you prove sin^2x + cos^2x = 1?

Feb 13, 2017

See explanation...

#### Explanation:

Consider a right angled triangle with an internal angle $\theta$: Then:

$\sin \theta = \frac{a}{c}$

$\cos \theta = \frac{b}{c}$

So:

${\sin}^{2} \theta + {\cos}^{2} \theta = {a}^{2} / {c}^{2} + {b}^{2} / {c}^{2} = \frac{{a}^{2} + {b}^{2}}{c} ^ 2$

By Pythagoras ${a}^{2} + {b}^{2} = {c}^{2}$, so $\frac{{a}^{2} + {b}^{2}}{c} ^ 2 = 1$

So given Pythagoras, that proves the identity for $\theta \in \left(0 , \frac{\pi}{2}\right)$

For angles outside that range we can use:

$\sin \left(\theta + \pi\right) = - \sin \left(\theta\right)$

$\cos \left(\theta + \pi\right) = - \cos \left(\theta\right)$

$\sin \left(- \theta\right) = - \sin \left(\theta\right)$

$\cos \left(- \theta\right) = \cos \left(\theta\right)$

So for example:

${\sin}^{2} \left(\theta + \pi\right) + {\cos}^{2} \left(\theta + \pi\right) = {\left(- \sin \theta\right)}^{2} + {\left(- \cos \theta\right)}^{2} = {\sin}^{2} \theta + {\cos}^{2} \theta = 1$

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Pythagoras theorem

Given a right angled triangle with sides $a$, $b$ and $c$ consider the following diagram: The area of the large square is ${\left(a + b\right)}^{2}$

The area of the small, tilted square is ${c}^{2}$

The area of each triangle is $\frac{1}{2} a b$

So we have:

${\left(a + b\right)}^{2} = {c}^{2} + 4 \cdot \frac{1}{2} a b$

That is:

${a}^{2} + 2 a b + {b}^{2} = {c}^{2} + 2 a b$

Subtract $2 a b$ from both sides to get:

${a}^{2} + {b}^{2} = {c}^{2}$

Feb 13, 2017

Use the formula for a circle $\left({x}^{2} + {y}^{2} = {r}^{2}\right)$, and substitute $x = r \cos \theta$ and $y = r \sin \theta$.

#### 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 $\left(x , y\right)$ on the circle is the radius $r$ of the circle.

Picture a circle of radius $r$ centred at the origin, and pick a point $\left(x , y\right)$ 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 $\text{opposite side"/"hypotenuse}$. In our case, the length of the side opposite of $\theta$ is the $y$-coordinate of our point $\left(x , y\right)$, and the hypotenuse is our radius $r$. So:

$\sin \theta = \text{opp"/"hyp" = y/r" "<=>" } y = r \sin \theta$

Similarly, $\cos \theta$ is the ratio of the $x$-coordinate in $\left(x , y\right)$ to the radius $r$:

$\cos \theta = \text{adj"/"hyp"=x/r" "<=>" } x = r \cos \theta$

So we have $x = r \cos \theta$ and $y = r \sin \theta$. Substituting these into the circle formula gives

$\text{ "x^2" "+" "y^2" } = {r}^{2}$
${\left(r \cos \theta\right)}^{2} + {\left(r \sin \theta\right)}^{2} = {r}^{2}$
${r}^{2} {\cos}^{2} \theta + {r}^{2} {\sin}^{2} \theta = {r}^{2}$

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 $\left(x , y\right)$ 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 $\theta$.