What does this vector equal to?

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Homework Statement


Prove that |\vec{A}|2+|\vec{B}|2 equals to \frac{1}{2}(|\vec{A}+\vec {B}|2+|\vec{A}-\vec{B}|2)

Homework Equations


Perhaps the triangle inequality? I am not sure.

The Attempt at a Solution


I have no idea how to approach this question. Please help me with clear explanations. Thank you.
 
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Hi Cuisine123! :smile:

Use the definition of modulus: |A + B|2 = (A + B).(A + B) etc :wink:
 
Perhaps not the triangle inequality. That is an inequality. You are looking for an equality.

Have you tried expanding the right hand side?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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