# arccos identity

• Sep 25th 2008, 06:38 AM
fardeen_gen
arccos identity
arccos [1/2.x^2 + √(1- x^2).√(1 - (x^2/4)) = arcos x/2 - arcos x, is true if:

A)|x| ≤ 1
B) x is a real number
C) 0 ≤ x ≤ 1
D)-1 ≤ x ≤ 0
• Sep 26th 2008, 05:20 AM
fardeen_gen
Can it be done this way? - I put x = any real number, x = -1, x = 1 and check validity of equation.
• Sep 26th 2008, 03:39 PM
Soroban
Hello, fardeen_gen!

Quote:

$\arccos\left(\frac{x^2}{2} + \sqrt{1- x^2}\cdot\sqrt{1 - \frac{x^2}{4}}\right) \;=\; \arccos\frac{x}{2} - \arccos x$

This is true if:

. . $(A)\;|x| \leq 1 \qquad(B)\;x\text{ is a real number} \qquad(C)\;0 \leq x \leq 1 \qquad(D)\;-1 \leq x \leq 0$

Let: $\alpha \:=\:\arccos\left(\frac{x^2}{2} + \frac{\sqrt{(1-x^2)(4-x^2)}}{2}\right) \qquad \beta \:=\:\arccos\frac{x}{2} \qquad \gamma \:=\:\arccos x$

$\text{Then: }\;\cos\alpha \:=\:\frac{x^2}{2} + \frac{\sqrt{(1-x^2)(4-x^2)}}{2}\qquad\begin{array}{ccccccc}\cos\beta &=& \frac{x}{2} & \Rightarrow & \sin\beta &=&\frac{\sqrt{4-x^2}}{2} \\ cos\gamma &=& x & \Rightarrow & \sin\gamma &=&\frac{\sqrt{1-x^2}}{1} \end{array}$ .[1]

We have: . $\alpha \;=\;\beta - \gamma$

Take cosines: . $\cos\alpha \;=\;\cos(\beta - \gamma) \;=\;\cos\beta\cos\gamma + \sin\beta\sin\gamma$

Substitute [1]: . $\frac{x^2}{2} + \frac{\sqrt{(1-x^2)(4-x^2)}}{2} \;=\;\frac{x}{2}\cdot x + \frac{\sqrt{4-x^2}}{2}\cdot\frac{\sqrt{1-x^2}}{1}$

And we have: . $\frac{x^2}{2} + \frac{\sqrt{(1-x^2)(4-x^2)}}{2} \;=\;\frac{x^2}{2} + \frac{\sqrt{(1-x^2)(4-x^2)}}{2}$

We have an identity . . . It is true for all values of $x$ . . . answer $(B)$

• Sep 26th 2008, 09:09 PM
fardeen_gen
I have a doubt.
In http://www.mathhelpforum.com/math-he...e5b506c2-1.gif, shouldn't we keep a restriction on x so that the terms under square root don't become negative? For all real values of x, eg. 5, the terms under the sq. root become negative. But is it ok, since this happens on both sides of the equality?