Write the given expression in algebraic form. tan(arccos x/3)?

Question

please show work / explain. i have problems like this and i need to learn how to do it. thank youuu.

Answer 1

tan [arccos(x/3)] =

let [arccos(x/3)] = y

thus

cosy = (x/3)

now you have to express tany in terms of cosy:

tany = siny/cosy = [±√(1 - cos²y)]/cosy

note that, due to arccosine function range [0, π] and thus y itself ranging from 0 to π, it ends either in the 1st or i the 2nd quadrant and siny is then positive; therefore you can rewrite the previous expression taking the plus sign:

tany =  siny/cosy = [√(1 - cos²y)]/cosy

hence, being cosy = (x/3), you get:

tany = {√[1 - (x/3)²]} /(x/3) =  {√[1 - (x²/9)]} /(x/3) = {√[(9 - x²)/9]} /(x/3) =

(1/3){√[(9 - x²)} (3/x) = {√[(9 - x²)} / x

therefore:

tan [arccos(x/3)] = tany = {√[(9 - x²)} / x

I hope it helps..

Bye!

Answer 2

Let  y = arccos(x/3), so we want tan(y).

Arccos(x/3) means the angle whose cosine is x/3. So draw a right triangle and call one angle y. Since y=arccos(x/3) this means that the cosine of y is x/3. The cosine is the adjacent over the hypotenuse, so the side adjacent to y equals x, and the hypotenuse of the right triangle equals 3. Now by the Pythagorean Thm, the side opposite y must equal  sqrt(9-x^2). Now we want tan(y). Tangent is the opposite over the adjacent, so tan(y) will be  sqrt(9-x^2) / x.

All these problems are done this way