trig help please

[FrogyJil]

tan(cos inverse (1/x))

i dont understand how to do this i got this far thought

cos inverse = (1/x) so cos = 1/x x=1 r=x

using x^2 + y^2 = r^2

would give me 1^2 + y^2 = x^2

i dont know were to go with this please help. would the ending x be 1 also but then y would = 0

 

[stapel]

Try not to be so casual with your notation, as this can create its own difficulties.

Using "ß" for "angle beta", we have:

. . . . .cos^-1(1/x) = ß

...for some angle ß. So draw a right triangle. Don't worry about the dimensions of the sides; this triangle is just a schematic we'll use to hold our values and variables. Label the angle as ß.

By definition of "inverse cosine", we then know that the adjacent side (to ß) has length "1" and the hypotenuse has length "x". Label the triangle.

Now use the Pythagorean Theorem to find an expression for the opposite side, in terms of x.

Since cos^-1(1/x) = ß, then tan(cos^-1(1/x)) = tan(ß).

Read the value (in terms of x) for tan(ß) from your triangle.

Eliz.