How do you prove ${sin}^{2} x + {cos}^{2} x = 1$ $sin^2x + cos^2x = 1$ ?

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How do you prove ${sin}^{2} x + {cos}^{2} x = 1$ $sin^2x + cos^2x = 1$ ?

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Answer 1 · 2017-02-13T19:35:39

See explanation...

Explanation:

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

enter image source here

Then:

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

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

So:

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

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

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

For angles outside that range we can use:

$sin (θ + π) = - sin (θ)$ $sin (theta + pi) = -sin (theta)$

$cos (θ + π) = - cos (θ)$ $cos (theta + pi) = -cos (theta)$

$sin (- θ) = - sin (θ)$ $sin (- theta) = - sin(theta)$

$cos (- θ) = cos (θ)$ $cos (- theta) = cos(theta)$

So for example:

${sin}^{2} (θ + π) + {cos}^{2} (θ + π) = {(- sin θ)}^{2} + {(- cos θ)}^{2} = {sin}^{2} θ + {cos}^{2} θ = 1$ $sin^2 (theta + pi) + cos^2 (theta + pi) = (-sin theta)^2 + (-cos theta)^2 = sin^2 theta + cos^2 theta = 1$

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Pythagoras theorem

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

enter image source here

The area of the large square is ${(a + b)}^{2}$ $(a+b)^2$

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

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

So we have:

${(a + b)}^{2} = c^{2} + 4 \cdot \frac{1}{2} a b$ $(a+b)^2 = c^2 + 4 * 1/2ab$

That is:

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

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

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

Answer 2 · 2017-02-13T20:13:40

Use the formula for a circle $(x^{2} + y^{2} = r^{2})$ $(x^2+y^2=r^2)$ , and substitute $x = r cos θ$ $x=rcostheta$ and $y = r sin θ$ $y=rsintheta$ .

Explanation:

The formula for a circle centred at the origin is

$x^{2} + y^{2} = r^{2}$ $x^2+y^2=r^2$

That is, the distance from the origin to any point $(x, y)$ $(x,y)$ on the circle is the radius $r$ $r$ of the circle.

Picture a circle of radius $r$ $r$ centred at the origin, and pick a point $(x, y)$ $(x,y)$ 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 $r$ $r$ . 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 ( $θ$ $theta$ ).

Now for the trigonometry.

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

$sin θ = \frac{opp}{hyp} = \frac{y}{r} \Leftrightarrow y = r sin θ$ $sin theta = "opp"/"hyp" = y/r" "<=>" "y=rsintheta$

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

$cos θ = \frac{adj}{hyp} = \frac{x}{r} \Leftrightarrow x = r cos θ$ $cos theta = "adj"/"hyp"=x/r" "<=>" "x = rcostheta$

So we have $x = r cos θ$ $x=rcostheta$ and $y = r sin θ$ $y=rsintheta$ . Substituting these into the circle formula gives

$x^{2} + y^{2} = r^{2}$ $" "x^2" "+" "y^2" "=r^2$
${(r cos θ)}^{2} + {(r sin θ)}^{2} = r^{2}$ $(rcostheta)^2+(rsintheta) ^2 = r^2$
$r^{2} {cos}^{2} θ + r^{2} {sin}^{2} θ = r^{2}$ $r^2cos^2theta + r^2 sin^2 theta = r^2$

The $r^{2}$ $r^2$ 's all cancel, leaving us with

${cos}^{2} θ + {sin}^{2} θ = 1$ $cos^2 theta + sin^2 theta = 1$

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

${sin}^{2} θ + {cos}^{2} θ = 1$ $sin^2 theta + cos^2 theta = 1$

And that's it. That's really all there is to it. Just as the distance between the origin and any point $(x, y)$ $(x,y)$ on a circle must be the circle's radius, the sum of the squared values for $sin θ$ $sin theta$ and $cos θ$ $cos theta$ must be 1 for any angle $θ$ $theta$ .

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