1. ## 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

2. Can it be done this way? - I put x = any real number, x = -1, x = 1 and check validity of equation.

3. Hello, fardeen_gen!

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

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 $(B)$

4. I have a doubt.
In , 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?