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edu89 · Aug 29, 2009

Guys can anyone do this with step-by-step

Prove
$\sin (\cos^{-1}(x))+3=\sqrt(-x^2+1)+3$

Drongoski · Aug 29, 2009

same as proving: sin(cos^-1(x)) = sqrt(1-x²)

Draw a right-angled triangle with hypotenuse = 1, one angle @ so that wrt this angle, the base is x and therefore the opp side by Pythag theorem = sqrt(1 - x²)

Then LHS = sin @ = opp/hyp = sqrt(1 - x²)/1 = sqrt(1 - x²)

Gussy Booo · Aug 29, 2009

I don't undertand how you worked with the LHS of the question.

sin(cos^-1(x)).......start from there please.

kaz1 · Aug 29, 2009

Gussy Booo said:
I don't undertand how you worked with the LHS of the question.

sin(cos^-1(x)).......start from there please.

You can't do it algebraically, you have to do it using the method Dragoski stated.

Thecorey0 · Aug 29, 2009

$LHS = sin(cos^{-1}x) +3\\ Let\ cos^{-1}x = \alpha\\ So\ cos\alpha = x\\ But\ Cos\alpha =\frac{Adjacent}{Hypotenuse}\\ Adjacent = x\ and\ Hypotenuse = 1\\ Using\ Pythagoras’\ theorm,\ Opposite = \sqrt{(1)^2-(x)^{2}}\\ So\ Sin\alpha = \frac{Opposite}{Hypotenuse}\\ = \sqrt{1-x^{2}}\\ \therefore\ Sin(cos^{-1}x) +3= \sqrt{1-x^2} + 3 =RHS$
Should be right.

Gussy Booo · Aug 30, 2009

I seeeee ! Cooool

Help! (1 Viewer)

edu89

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Drongoski

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Gussy Booo

Mathematics <3

kaz1

et tu

Thecorey0

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Gussy Booo

Mathematics <3

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