Question

solve cos(cos^-1X + sin^-1X)

Answer 1

http://www.wolframalpha.com/input/?i=cos%28cos^-1X+%2B+sin^-1X%29

Answer 2

remember cos(x+y)=cosxcosy-sinxsiny
so we have
cos(cos^-1(x))cos(sin^-1(x))-sin(cos^-1(x))sin(sin^-1(x))
=xcos(sin^-1(x))-xsin(cos^-1(x))
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let sin^-1(x)=@
then sin@=x=x/1
so if we draw a right triangle and we label one of the angles @(not the right angle) then
opposite side of @ is x
and hyp of that triangle is 1 since sin@=opp/hyp
so the other can be found using pythagorean thm
opp^2+adj^2=hyp^2
\[x^2+adj^2=1^2\]
\[adj=\sqrt{1-x^2}\]
so cos@=sqrt{1-x^2}/1=sqrt{1-x^2}
so
\[\cos(\sin^{-1}x)=\cos(@)=\sqrt{1-x^2}\]
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now let cos^-1(x)=&
so cos&=x=x/1
remember cos&=adj/hyp
again assume we have a right traingle we can find the oppsite side of &
by using pythagorean thm
so we have
opp^2+adj^2=hyp^2
\[opp^2+x^2=1^2\]
\[opp=\sqrt{1-x^2}\]
so \[\sin\&=\sqrt{1-x^2}\]
\[\sin(\cos^{-1}x)=\sin(\$)=\sqrt{1-x^2}\]
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finally
\[\cos(\cos^{-1}x+\sin^{-1}x)=\cos(\cos^{-1}x)\cos(\sin^{-1}x)-\sin(\sin^{-1}x)\sin(\cos^{-1}x)\]
\[=x*\sqrt{1-x^2}-x*\sqrt{1-x^2}=0\]

Answer 3

Isn't cos(cos^-1 x) equal to x? (inverse just undoes the operation.

Answer 4

No that's true but rubbish, I can't split it up like that (time for bed).

Answer 5

estudier are u here.. can u take a look at my lin algebra problem??

Answer 6

If we let Cos a = x and Sin b = x (ie a = Cos^-1 x, b = Sin^-1 y) then we want

Cos (a+b) = Cos^2 a - Sin^2 b

= (Cos a + Sin b)(Cos a - Sin b) = 2x * 0 = 0.

Answer 7

yes i have cos(cos^(-1)x)=x
and sin(sin^(-1)x)=x

Answer 8

Typo, "b = Sin^-1 y" in my answer should read "b = Sin^-1 x"