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Glockmeister · « **on:** March 19, 2008, 11:02:55 pm »

Simplify $arcsin x + arccos x = 1$

Collin Li · « **Reply #1 on:** March 20, 2008, 01:14:32 am »

What does it mean by simplify? I believe that the LHS should actually equal to $\frac{\pi}{2}$ for all values of $x$ .

Glockmeister · « **Reply #2 on:** March 20, 2008, 01:36:40 am »

Argh.. bleh, that = 1 shouldn't be there.

The actual equation is $\sin^{-1}{x} + \cos^{-1}{x}$

Collin Li · « **Reply #3 on:** March 20, 2008, 01:52:38 am »

Quote from: Glockmeister on March 20, 2008, 01:36:40 am

$\sin^{-1}{x} + \cos^{-1}{x}$

Let $\arcsin{x} = \alpha$ and $\arccos{x} = \beta$

Since inverse trigonometric functions represent angles, realise that this is the sum of two angles:

$\arcsin{x} + \arccos{x} = \alpha + \beta$

$= \arcsin [\sin (\alpha + \beta) ]$

$= \arcsin ( \sin\alpha \cdot \cos\beta + \cos\alpha \cdot \sin\beta )$

Using triangles, or the trigonometric identity:

$\cos\alpha = \sqrt{1 - \sin^2\alpha} = \sqrt{1-x^2}$ , and

$\sin\beta = \sqrt{1 - \cos^2\beta} = \sqrt{1-x^2}$

$\therefore\; \arcsin ( \sin\alpha \cdot \cos\beta + \cos\alpha \cdot \sin\beta ) = \arcsin \left( x^2 + \left(\sqrt{1-x^2}\right)^2 \right)$

$= \arcsin\left( x^2 + (1-x^2) \right) = \arcsin 1 = \frac{\pi}{2}$

AppleXY · « **Reply #4 on:** March 20, 2008, 08:11:46 am »

I can prove $\arctan{x}$ + $\arctan\frac{1}{x}$ = $\frac{\pi}{2}$ (for x>0) using calculus. But how would you prove $\arctan\frac{1}{x} =$ $\frac{\pi}{2}$ - $\arctan{x}$ using trig?

Collin Li · « **Reply #5 on:** March 20, 2008, 08:28:10 am »

Quote

$\arctan\frac{1}{x} = \frac{\pi}{2} - \arctan{x}$

To prove this with algebra and trigonometry, start from the RHS, realise it is an addition of two angles (or subtraction), so your next step should be: $\mbox{R.H.S.} = \arcsin [\sin (\mbox{R.H.S.}) ]$ of it. Note I did not use $\arctan [\tan (\cdots) ]$ , because then the expanded version of the compound angle formula for $\tan$ would result in $\tan \frac{\pi}{2}$ (undefined) coming up.

Since this is boring, and I demonstrated it above, I will explain the geometric meaning of this proof, by proving it geometrically:

Let $\arctan \frac{1}{x} = \theta$ , and draw a right-angled triangle of angle theta. The opposite is length 1 and the adjacent is length $x$ so that $\tan\theta = \frac{1}{x}$ .

Now, let the remaining angle be $\phi = \frac{\pi}{2} - \theta$ (by definition, using the fact that the interior angles of a triangle add up to $\pi$ ).

From triangle (reading the opposite and adjacent relative to $\phi$ ):
$\tan\phi = \left(\frac{1}{x}\right)^{-1} = x$

$\implies \arctan x = \phi$

$\implies \arctan x = \frac{\pi}{2} - \theta$

$\implies \arctan x = \frac{\pi}{2} - \arctan \frac{1}{x}$

$\implies \arctan \frac{1}{x} = \frac{\pi}{2} - \arctan x$ (as required)

I'm not sure whether the first proof ( $\arcsin x + \arccos x = \frac{\pi}{2}$ ) can be done as easily.

AppleXY · « **Reply #6 on:** March 20, 2008, 08:44:37 am »

WOW. That's awesome man. Thanks and yeah

Glockmeister · « **Reply #7 on:** March 20, 2008, 08:10:45 pm »

Quote from: coblin on March 20, 2008, 01:52:38 am

Quote from: Glockmeister on March 20, 2008, 01:36:40 am
$\sin^{-1}{x} + \cos^{-1}{x}$

Let $\arcsin{x} = \alpha$ and $\arccos{x} = \beta$

Since inverse trigonometric functions represent angles, realise that this is the sum of two angles:

$\arcsin{x} + \arccos{x} = \alpha + \beta$

$= \arcsin [\sin (\alpha + \beta) ]$

$= \arcsin ( \sin\alpha \cdot \cos\beta + \cos\alpha \cdot \sin\beta )$

Using triangles, or the trigonometric identity:

$\cos\alpha = \sqrt{1 - \sin^2\alpha} = \sqrt{1-x^2}$ , and

$\sin\beta = \sqrt{1 - \cos^2\beta} = \sqrt{1-x^2}$

$\therefore\; \arcsin ( \sin\alpha \cdot \cos\beta + \cos\alpha \cdot \sin\beta ) = \arcsin \left( x^2 + \left(\sqrt{1-x^2}\right)^2 \right)$

$= \arcsin\left( x^2 + (1-x^2) \right) = \arcsin 1 = \frac{\pi}{2}$

Thanks for that man, knew it had to have something to do with the inverse.

Mao · « **Reply #8 on:** March 20, 2008, 08:42:43 pm »

Quote from: coblin on March 20, 2008, 08:28:10 am

I'm not sure whether the first proof ( $\arcsin x + \arccos x = \frac{\pi}{2}$ ) can be done as easily.

it turns out a lot simpler

by the same token, if we assume a right angled triangle with hypotenuse of 1, angle of $\theta$ and opposite of $x$ :

$sin( \theta )=x$

$\implies \theta=sin^{-1}x$

let $\phi$ be the complementary angle

$cos( \phi ) = sin(\theta)$

$\implies cos \left( \frac{\pi}{2} - \theta \right) = x$

$\implies \frac{\pi}{2} - \theta=cos^{-1}x$

adding these two together:

$\theta +\frac{\pi}{2} - \theta=sin^{-1}x+cos^{-1}x$

$\therefore \arcsin x + \arccos x = \frac{\pi}{2}$

Ahmad · « **Reply #9 on:** March 21, 2008, 08:56:15 am »

This might be obvious, but another method:

Let $f(x) = \sin^{-1} x + \cos^{-1} x$

$f'(x) = \frac{1}{\sqrt{1-x^2}} + \frac{-1}{\sqrt{1-x^2}} = 0$

$\implies f(x) = C$ i.e. constant function

$f(0) = \sin^{-1} 0 + \cos^{-1} 0 = \frac{\pi}{2} = C$

$\implies \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$

Works the same with the other problems.

evaporade · « **Reply #10 on:** March 21, 2008, 09:14:14 am »

Since arsinx = pi/2 - arcosx for x in [-1,1] .: arsinx + arcosx = pi/2

Ahmad · « **Reply #11 on:** March 21, 2008, 10:56:20 am »

Quote from: Ahmad on March 21, 2008, 08:56:15 am

This might be obvious, but another method:

Let $f(x) = \sin^{-1} x + \cos^{-1} x$

$f'(x) = \frac{1}{\sqrt{1-x^2}} + \frac{-1}{\sqrt{1-x^2}} = 0$

$\implies f(x) = C$ i.e. constant function

$f(0) = \sin^{-1} 0 + \cos^{-1} 0 = \frac{\pi}{2} = C$

$\implies \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$

Works the same with the other problems.

Extension: Use the above technique to show that $\cos \theta + i \sin \theta = e^{i\theta}$

Collin Li · « **Reply #12 on:** March 21, 2008, 12:13:54 pm »

Sounds fun!

Let $y = \cos x + i\sin x$

$\implies \frac{dy}{dx} = -\sin x + i\cos x = i\left(\cos x - \frac{1}{i}\sin x\right)$

Since $-\frac{1}{i} = -\frac{i}{i^2} = i$ :

$\frac{dy}{dx} = i(\cos x + i\sin x) = iy$

$\implies \frac{dx}{dy} = \frac{1}{iy}$

$\implies x = \frac{1}{i}\int \frac{1}{y}\,dy = \frac{1}{i}\log_ey + C$

$\implies i(x-C) = \log_ey \implies y = e^{i(x-C)}$

Since we know that $y(0) = \cos 0 + i\sin 0 = 1$ :

$\implies 1 = e^{iC} \implies \log_e1 = iC \implies C = 0$

$\therefore\; y = e^{ix} = \cos x + i\sin x$

dcc · « **Reply #13 on:** March 21, 2008, 12:43:27 pm »

another way:
Let:

$f(x) = \dfrac{\cos \theta + i \sin \theta}{e^{i\theta}}$

$f'(x) = \dfrac{e^{i\theta}(- \sin \theta + i \cos \theta) - (\cos \theta + i \sin \theta)(i e^{i\theta})}{(e^{i\theta})^2}$

multiplying out the i in the second bracket:

$f'(x) = \dfrac{e^{i\theta}(- \sin \theta + i \cos \theta) - e^{i\theta}(- \sin \theta + i \cos \theta)}{(e^{i\theta})^2} = 0 \mbox{ therefore f(x) is a constant function}$

since f(x) is a constant function, we have:

$f(0) = \dfrac{\cos 0 + i \sin 0}{e^{0i}} = 1$

$\implies 1 = \dfrac{\cos \theta + i \sin \theta}{e^{i\theta}}$

$\implies e^{i\theta} = \cos \theta + i \sin \theta$

AppleXY · « **Reply #14 on:** March 21, 2008, 10:21:41 pm »

Quote from: Ahmad on March 21, 2008, 10:56:20 am

Quote from: Ahmad on March 21, 2008, 08:56:15 am
This might be obvious, but another method:

Let $f(x) = \sin^{-1} x + \cos^{-1} x$

$f'(x) = \frac{1}{\sqrt{1-x^2}} + \frac{-1}{\sqrt{1-x^2}} = 0$

$\implies f(x) = C$ i.e. constant function

$f(0) = \sin^{-1} 0 + \cos^{-1} 0 = \frac{\pi}{2} = C$

$\implies \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$

Works the same with the other problems.

Extension: Use the above technique to show that $\cos \theta + i \sin \theta = e^{i\theta}$

Yeah, thats exactly what I did with my problem. Calculus really rocks.

Also applicable to the polar cis form complex number.

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