Proving $\cos(x)^2+\sin(x)^2=1$

Question

I need to prove that $\cos(x)^2+\sin(x)^2=1$

Here's how I started (using the Cauchy product):

\begin{align} \cos(x)^2+\sin(x)^2 &=\sum_{k=0}^{\infty}\sum_{l=0}^k(-1)^l\frac{x^{2l}}{(2l)!}(-1)^{k-l}\frac{x^{2(k-l)}}{(2(k-l))!}+(-1)^l\frac{x^{2(l+1)}}{(2(l+1))!}(-1)^{k-l}\frac{x^{2(k-l)+1}}{(2(k-l)+1)!}\\ &=\sum_{k=0}^{\infty}\sum_{l=0}^k(-1)^k\frac{x^{2l}}{(2l)!}\frac{x^{2k}}{x^{2l}(2(k-l))!}\\ &\quad +(-1)^k\frac{x^{2l+1}}{(2l+1)!}\frac{x^{2k+1}}{x^{2l}(2(k-l)+1)!}\\ &=\sum_{k=0}^{\infty}\sum_{l=0}^k(-1)^k\frac{x^{2k}}{(2l)!(2k-2l))!}\\ &\quad +(-1)^k\frac{x^{2(k+1)}}{(2l+1)!(2k-2l+1)!} \end{align}

Here, I was told I had to use the equality $\sum_{k=0}^n(-1)^k\binom{n}{k}=0$

But I can't since the $x$ term is in the way. How do I proceed?

EDIT:

I have to prove this specifically using the power series and the binomial coefficient. Unfortunately I'm not allowed to do the proof any other way.

I managed to do a few more steps:

\begin{align} \hphantom{\cos^2 x + \sin^2 x} &=\sum_{k=0}^{\infty}\sum_{l=0}^k(-1)^k\frac{x^{2k}}{(2l)!(2k-2l))!}\\ &\quad +\sum_{k=0}^{\infty}\sum_{l=0}^k(-1)^k\frac{x^{2(k+1)}}{(2l+1)!(2k-2l+1)!}\\ &=\sum_{k=0}^{\infty}\frac{x^{2k}}{(2k)!}\sum_{l=0}^k(-1)^k\binom{2k}{2l}\\ &\quad +\sum_{k=0}^{\infty}\sum_{l=0}^k(-1)^k\binom{2k}{2l}\frac{x^{2k+2}}{(2l+1)(2k-2l+1)(2k)!}\\ &=\sum_{k=0}^{\infty}0\\ &\quad +\sum_{k=0}^{\infty}\sum_{l=0}^k(-1)^k\binom{2k}{2l}\frac{x^{2k+2}}{(2l+1)(2k-2l+1)(2k)!} \end{align}

Now I'm stuck again. Also, I'm not sure the last step is correct, since that would make $\cos(x)^2$ zero.

Some more hints would be nice.

PS: Unfortunately, the linked question didn't help.

PPS: I am also not allowed to do this using derivatives.

Answer 1

well, you can start with a more simpler solution: $$ \sin(x) = 1/2i\ *\ (e^{ix} - e^{-ix}) \\ \cos(x) = 1/2\ *\ (e^{ix} + e^{-ix}) \\ \sin^2(x) = -1/2\ *\ (e^{2ix} + e^{-2ix} -2)\\ \cos^2(x) = 1/2\ *\ (e^{2ix} + e^{-2ix} +2)\\ \sin^2(x) + \cos^2(x) = -1/2\ *\ (e^{2ix} + e^{-2ix} -2) + 1/2\ *\ (e^{2ix} + e^{-2ix} +2) = 1 $$

Answer 2

You could also integrate (if you're allowed to do that) and note that \begin{align} \int \cos^2(x) +\sin^2(x)\,dx &= \underbrace{\int \cos^2(x)\,dx}_{=\frac{1}{2}x+\frac{1}{2}\sin(x)\cos(x)+c_1} + \underbrace{\int \sin^2(x)\,dx}_{\frac{1}{2}x-\frac{1}{2}\sin(x)\cos(x)+c_2} \\ &= \frac{1}{2}x + c_1 + \frac{1}{2}x + c_2 \\ &= x+c_3. \end{align}

Answer 3

Let's start with a circle. We have that $r^2 = x^2 + y^2$, and then we draw a right triangle to that point. We then have the $\sin \theta = \frac{y}{r}$ and $\cos \theta = \frac{x}{r}$. We then divide the first equation to get $1 = \frac{x^2}{r^2} + \frac{y^2}{r^2}$. Substituting in the equations for $\cos \theta$ and $\sin \theta,$ we get $1 = \cos^2 \theta + \sin^2 \theta$

Answer 4

If you are to derive this directly from the definition $\cos(x) \equiv \sum_{n=0}^\infty\frac{x^{2n}(-1)^{n}}{(2n)!}$ and $\sin(x) \equiv \sum_{n=0}^\infty\frac{x^{2n+1}(-1)^{n}}{(2n+1)!}$ then we can do this without having to perform the product by first using term-by-term differentiation to get

$$\cos'(x) = -\sin(x),~~~~~\sin'(x) = \cos(x)$$

If we now take $f(x) = \cos^2(x) + \sin^2(x)$ then the result above gives

$$f'(x) = -2\sin(x)\cos(x) + 2\sin(x)\cos(x) = 0 \implies f(x) = f(0) = 1^2+0^2 = 1$$

Answer 5

Let $f(x)=\cos ^2 x+ \sin ^2x$ , then $f'(x)= -2 \sin x \cos x+2 \sin x\cos x=0$. As $f(0)=1, f(x)=1$

Answer 6

Let us suppose a=verical side,b=horizontal side be the sides of a right triangle.and c be hypotenuse so by pythagoras principle $c^2=a^2+b^2$ so let $x$ be the angle between a and c so by trigonometric ratios we have $sin^2(x)=\frac{b^2}{c^2}, cos^2(x)=\frac{a^2}{c^2}$ so adding we get $\frac{a^2+b^2}{c^2}$ but $c^2=a^2+b^2$ so it is equal to $1$. So proved $cos^2(x)+sin^2(x)=1$. Thats all than using binomials for it.

Answer 7

In situations like this, it is more convenient to write the coefficients of a power series as $\dfrac{c_n}{n!}$. Then

$$f(x) = \sum_{n = 0}^{\infty} \frac{a_n}{n!}x^n\quad\text{and}\quad g(x) = \sum_{n = 0}^{\infty} \frac{b_n}{n!}x^n$$

yields

$$f(x)g(x) = \sum_{m = 0}^{\infty} \biggl(\sum_{k = 0}^m \frac{a_k}{k!}\cdot \frac{b_{m-k}}{(m-k)!}\biggr)x^m = \sum_{m = 0}^{\infty} \frac{1}{m!}\biggl(\sum_{k = 0}^m \binom{m}{k}a_k b_{m-k}\biggr)x^m.$$

For the cosine, we have

$$c_n = \begin{cases} (-1)^{n/2} &, n \text{ even}\\ \quad 0 &, n \text{ odd},\end{cases}$$

so in the Cauchy product for $\cos^2 x$ the coefficients of odd powers of $x$ are all zero (since one of $c_k$ and $c_{2r+1-k}$ is zero for all $k$). The coefficient of $x^m$ for $m = 2r$ in $\cos^2 x$ is then

$$\frac{1}{(2r)!}\sum_{k = 0}^r \binom{2r}{2k}(-1)^k(-1)^{r-k} = \frac{(-1)^r}{(2r)!}\sum_{k = 0}^r \binom{2r}{2k}(-1)^{2k}.$$

Similarly, for the sine we have

$$c_n = \begin{cases}\quad 0 &, n \text{ even}\\ (-1)^{\lfloor n/2\rfloor} &, n \text{ odd}.\end{cases}$$

Again, the coefficients of odd powers of $x$ in the Cauchy product of $\sin^2 x$ are all zero, and the coefficient of $x^m$ for $m = 2r$ - with $r \geqslant 1$ - is

$$\sum_{k = 0}^{r-1} \frac{c_{2k+1}}{(2k+1)!}\frac{c_{2r-(2k+1)}}{(2r-2k-1)!} = \frac{1}{(2r)!} \sum_{k = 0}^{r-1}\binom{2r}{2k+1}(-1)^k(-1)^{r-k-1} = \frac{(-1)^r}{(2r)!}\sum_{k = 0}^{r-1}\binom{2r}{2k+1}(-1)^{2k+1}.$$

Adding these two then yields

\begin{align} \cos^2 x + \sin^2 x &= \sum_{r = 0}^{\infty} \frac{(-1)^r}{(2r)!}\biggl(\sum_{k = 0}^r \binom{2r}{2k}(-1)^{2k}\biggr)x^{2r} + \sum_{r = 0}^{\infty} \frac{(-1)^r}{(2r)!}\biggl(\sum_{k = 0}^r \binom{2r}{2k+1}(-1)^{2k+1}\biggr)x^{2r} \\ &= \sum_{r = 0}^{\infty} \frac{(-1)^r}{(2r)!}\biggl(\sum_{k = 0}^r \binom{2r}{2k}(-1)^{2k} + \sum_{k = 0}^{r-1}\binom{2r}{2k+1}(-1)^{2k+1}\biggr)x^{2r}\\ &= \sum_{r = 0}^{\infty} \frac{(-1)^r}{(2r)!}\biggl(\sum_{j = 0}^{2r} \binom{2r}{j}(-1)^j\biggr)x^{2r}. \end{align}

Now the equality

$$\sum_{k = 0}^n \binom{n}{k}(-1)^k = 0$$

for $n > 0$ shows that in the above all coefficients except for the coefficient of $x^0$ vanish, and the coefficient of $x^0$ is $1$, so indeed

$$\cos^2 x + \sin^2 x \equiv 1.$$