How might you go about proving that sin(x)2 + cos(x)2 = 1?

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- Apr 11th 2010, 04:48 AM #1

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- Apr 11th 2010, 04:56 AM #2

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Hi andy054,

to prove

$\displaystyle (sinx)^2+(cosx)^2=1$

one easy way is to draw a right-angled triangle and apply Pythagoras' theorem.

$\displaystyle sinx=\frac{opp}{hyp}\ \Rightarrow\ opp=(hyp)sinx$

$\displaystyle cosx=\frac{adj}{hyp}\ \Rightarrow\ adj=(hyp)cosx$

Now complete it using

$\displaystyle adj^2+opp^2=hyp^2$

As this only proves the identity for acute angles, you could expand to prove for all angles.

- Apr 11th 2010, 09:10 AM #3

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Archie Meade's answer is correct for that definition of sine and cosine. There are other ways of defining sine and cosine. And how you prove that depends upon exactly what your definitions of sine and cosine are.

For example, a more general definition of sine and cosine is this: construct the unit circle, that is, the circle with center at (0,0) and radius 1, on a coordinate system. If you start at (1, 0) and measure around the circumference of the circle a distance "t", the coordinates of the point are (cos(t), sin(t))- that is, whatever values the x and y coordinates of that point are, they are, by definition, the values of cos(x) and sin(x). Then "$\displaystyle sin^2(t)+ cos^2(t)= 1$" follows immediately from the fact that the equation of the unit circle is $\displaystyle x^2+ y^2= 1$.

That definition itself has some technical problems with giving a mathematical definition to "measuring" around the circumference. Other ways to define cos(x) and sin(x) are:

"y= cos(x) is the unique function, y, satisfying the intiial value problem $\displaystyle \frac{d^2y}{dx^2}= -y$ with y(0)= 1, y'(0)= 0 and y= sin(x) is the unique function, y, satisfying the intiial value problem $\displaystyle \frac{d^2y}{dx^2}= -y$ with y(0)= 0, y'(0)= 1."

Using that definition, you can show that the derivative of sin(x) is cos(x) and that the derivative of cos(x) is -sin(x). Then the derivative of $\displaystyle y= sin^2(x)+ cos^2(x)$ is $\displaystyle y'= 2sin(x)cos(x)+ 2cos(x)(-sin(x))= 0$ for all x. That proves y is a constant and since at x= 0, $\displaystyle y(0)= (sin(0))^2+ (cos(0))^2= 0^2+ 1^2= 1$, that constant is 1: $\displaystyle sin^2(x)+ cos^2(x)= 1$.

Yet another way to define sine and cosine is by their power series:**define**sin(x) to be $\displaystyle \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}x^{2n+1}$ and define cos(x) to be $\displaystyle \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}x^{2n}$.

You can show that those series converge uniformly and so are "term by term" differentiable. From that, you can show that the derivative of sin(x) is cos(x) and the derivative of cos(x) is -sin(x) and from that that both sin(x) and cos(x) satisfy y"= -y. Then use the proof above.