Trigonometry

     
The standard proof of the identity $sin^2x + cos^2x = 1$ (the one that is taught in schools) is as follows: from pythagoras theorem, we have (where $h$ is hypotenuse, $b$ is base and $p$ is perpendicular)$$h^2 = p^2 + b^2$$dividing by $h^2$ on both sides:$$1 = fracp^2h^2+fracb^2h^2$$since $sin x = frac ph$ and $cos x = frac bh$, $$1 = sin^2x+cos^2x$$

Are there any more innovative ways of proving this common identity?


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Here"s a couple of different methods:

Proof using the cosine angel sum formula:$$1=cos(0) =cos(x + (-x)) =cos(x)cos(-x) - sin(x)sin(-x) = cos^2(x) + sin^2(x)$$Proof using differential equations:Recall the differential equation definitions of $sin$ and $cos$: They are the solutions to $f"" = - f$ with the appropriate initial conditions $f(0) = 0$, $f"(0) = 1$ for $sin(x)$ & $f(0) = 1$, $f"(0) = 0$ for $cos(x)$. Since solutions to differential equations are quality given the initial conditions, we immediately get $sin"(x) = cos(x)$ under this definition. Then:egineqnarrayfracddx left(sin^2(x) + cos^2(x) ight) &=& fracddx left(sin^2(x) + left(sin"(x) ight)^2 ight)\&=& 2sin(x)sin"(x) + 2sin"(x)sin""(x)\&=& 2sin(x)sin"(x) +2sin"(x)(-sin(x)) = 0endeqnarrayhence $sin^2(x) + cos^2(x)$ is constant, since its derivative is 0. Plugging in $x=0$ we see that it must equal 1.

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Proof using Euler"s formula: $$sin^2(x) + cos^2(x) = left(cos x + i sin x ight)left(cos x - i sin x ight) =left(cos x + i sin x ight)left(cos (-x) + i sin (-x) ight)= e^i x e^-i x = 1$$Proof using Taylor series:Using the Taylor series:egineqnarraysin(x) = sum_n=0^infty frac(-1)^n x^2n+1(2n+1)!\cos(x) = sum_n=0^infty frac(-1)^n x^2n(2n)!endeqnarraywe have $$sin^2(x) = left(sum_n=0^infty frac(-1)^n x^2n+1(2n+1)! ight)^2 = sum_n=0^infty sum_m=0^infty frac(-1)^n+m x^2n + 2m + 2(2n+1)!(2m+1)! = sum_k=0^infty x^2k+2(-1)^k sum_j=0^k frac1(2j+1)!(2(k-j)+1)!= sum_k=0^infty fracx^2k+2(-1)^k(2k+2)!sum_j=0^k inom2k+22j+1 = -sum_k=1^infty fracx^2k(-1)^k(2k)!sum_j=0^k-1 inom2k2j+1$$and $$cos^2(x) =left(sum_n=0^infty frac(-1)^n x^2n(2n)! ight)^2 = sum_n=0^infty sum_m=0^infty frac(-1)^n+m x^2n + 2m(2n)!(2m)! = sum_k=0^infty x^2k(-1)^k sum_j=0^k frac1(2j)!(2(k-j))! = sum_k=0^infty fracx^2k(-1)^k(2k)!sum_j=0^k inom2k2j = 1 + sum_k=1^infty fracx^2k(-1)^k(2k)!sum_j=0^k inom2k2j$$Adding these together:$$sin^2(x) + cos^2(x) = 1+sum_k=1^inftyfracx^2k(-1)^k(2k)!left(sum_j=0^k inom2k2j-sum_j=0^k-1 inom2k2j+1 ight)=1+sum_k=1^inftyfracx^2k(-1)^k(2k)!left(sum_j=0^2k (-1)^jinom2kj ight) = 1+sum_k=1^inftyfracx^2k(-1)^l(2k)! (1 + (-1))^2k = 1$$We use the binomial theorem on the second last step.

Carlson"s Theorem: Major overkill khổng lồ use this one, but it still works.

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Notice $cos(x)$, & $sin(x)$ have exponential type 1. Thus $$f(x) = cos^2(fracpi4x) + sin^2(fracpi4x) - 1$$has exponential type at most $fracpi2 . By inspection, you can see that $f(x) = 0$ for $xin dongan-group.com.vnbbN$. Hence Carlson"s theorem implies $f(x) = 0$ identically.