Trigonometry Proof from the Essential Textbook

[Alwin]

Hi guys, my friend recently asked me about this part of the theory from the Essential [3/4] textbook page 273.



Clearly,

But that's all I get, unsure of how to prove: 

My alternate approach was:

Spoiler

(copy and pasted from my word file):

Note that for this purpose, and can be interchanged.

Is it possible to prove:
Would I have to substitute in values to find both c₁ and c₂?

Or, can I prove it using trigonometry identities?
eg   and then perform some algebraic manipulation?

EDIT: tidied things up, and put stuff in the spoiler.

[brightsky]

i'm not sure what the issue is. you've already proven that sin^(-1)(blah) + cos^(-1)(blah) = pi/2.

[Alwin]

i'm not sure what the issue is. you've already proven that sin^(-1)(blah) + cos^(-1)(blah) = pi/2.

Sorry! I wasn't to clear, was I - too much preamble. My q was about other methods of proof, such as the calculus one the Essential book starts but doesn't finish, or if it's possible to use trig identities?

Is it possible to prove using calculus: ... Would I have to substitute in values to find both c₁ and c₂?

Or, can I prove it using trigonometry identities? eg   and then perform some algebraic manipulation?

[b^3]

EDIT: nvm, did something wrong, ignore this.

[lzxnl]

Well...if you REALLY want one...
You could expand arcsin x and arccos x into Taylor series about x=0
Then you'd see that pretty much all of the terms just cancel out when you add arcsin x and arccos x
And you'd be left with arccos 0 = pi/2.

Or, you can do this:
sin x = cos (pi/2-x)
arccos (sin x) = pi/2 - x
But arcsin (sin x) = x
Therefore arcsin (sin x) + arccos (sin x) = pi/2

assuming of course that x is within the necessary domain

Even easier, you already have arccos x + arcsin x = constant as the derivative of both sides is zero.
Sub in random value of x, say x=0, to find this constant, as this equation is true for all x^2<=1

I suppose the only problem with geometry is the confusing issue of signs of side lengths which would be a pain to address.

[SocialRhubarb]

has the range , so we can let , where

If we substitute these values into the first equation:







I think that works?

[BubbleWrapMan]

Do it from their graphs - one is just a vertical transformation of the other.

[Alwin]

Thanks guys!

Just one comment, @SocialRhubarb: your method is sound (although you said arcsin twice rather than arcsin + arccos haha), if you started with nliu's proof that arcsin(x)+arccos(x) is a constant function first. I actually really like that calculus method. Maybe because I was shown it ages ago but didn't realise it was possible to apply it here