cos(sin^-1(5/x))

Hi CPhill,

when the argument in sin is positiv then,  cos can be positive (I.) or negativ(II.):

What needs to be done here?

cos(sin^-1(5/x))    ?

$$\cos{ ( \sin^{-1}( \frac{5}{x} ) ) } = \pm \sqrt{ 1- \left( \dfrac{5}{x} \right) ^2 }$$

I always do these with a triangle to aid me.

if x is positive then the angle is in the 1st or 2nd quads and cos of the angle could be pos or neg.

If xis neg then the angle is in the 3rd or 4th quads and cos  of the angle could be pos or neg.

now I'll look at the first quad.

$$cos\left(sin^{-1}\left(\frac{5}{x}\right)\right)$$

$$I let $\theta =sin^{-1}\left(\frac{5}{x}\right)$\\ I then found the third side using Pythagoras' Theorum.\\ Hence\\ $cos\left(sin^{-1}\left(\frac{5}{x}\right)\right)=\pm \frac{\sqrt{x^2-25}}{x}$$ $

$$ok, let me see if I can make this look like Heureka's answer\\\\ $\pm \frac{\sqrt{x^2-25}}{x}$\\\\ =$\pm \sqrt{\frac{{x^2-25}}{x^2}$\\\\ =$\pm \sqrt{1-\frac{{25}}{x^2}$\\\\ =$\pm \sqrt{1-\frac{{25}}{x^2}$\\\\ =$\pm \sqrt{1-\left(\frac{{5}}{x}\right)^2$\\\\$$

There you go 

cos(sin^-1(5/x))

$$\cos{(\sin^{-1}{( \frac{5}{x} )})} = \pm \sqrt{ 1 - \left( \frac{5}{x} \right)^2 }$$

or

$$\cos{(\sin^{-1}{( \dfrac{z}{1} )})} = \pm \sqrt{ 1 - \left(z \right)^2 }$$

Thanks Heureka,

Is your way the standard 'formula' presentation?

Hi Melody,

no

ok thanks 

It may not be clear as to why we need to take both roots in this answer......

Remember that....sin²(Θ) + cos²(Θ) = 1     ......so we have.....using Melody's triangle as a reference.......

(5/x)² + cos²(Θ) = 1

cos²(Θ) = 1 - (5/x)²

cos²(Θ) =  (x² - 25) / x²      ......take the square root of both sides......remember we need to use the square root property......!!!!

cos(Θ)  =  ±√ (x² - 25) / x

I did cover the positive and negative bit (it was the same as Heureka's  but I like your alternate way of looking at it Chris.  Thanks.