Question

prove that cos^-1(x) + cos^-1 { x/2 + √3-3x^2/2} = π/3​

Answer 1

we have to prove that

$\displaystyle\sf cos^{-1}(x) + cos^{-1}\left\{ \dfrac{x}{2} + \dfrac{\sqrt{3-3x^2}}{2} \right\} = \dfrac{\pi}{3}$

$\displaystyle\sf$

consider

$\displaystyle\sf LHS = cos^{-1}(x) + cos^{-1}\left\{ \dfrac{x}{2} + \dfrac{\sqrt{3-3x^2}}{2} \right\}$

$\displaystyle\sf$

Let $\displaystyle\sf cos^{-1} x = \alpha$

$\displaystyle\sf \implies x = cos \ \alpha$

$\displaystyle\sf$

$\displaystyle\sf \therefore LHS = \alpha + cos^{-1} \left[ cos\ \alpha \cdot cos\dfrac{\pi}{3} + \dfrac{\sqrt{3}}{2}\sqrt{1-cos^2\alpha}\right]$

$\displaystyle\sf$

$\displaystyle\sf = \alpha + cos^{-1}\left[ cos\dfrac{\pi}{3} \: cos\:\alpha + sin\dfrac{\pi}{3}\:sin\:\alpha\right] \bf \left[\because sin\:\boldsymbol\alpha = \sqrt{1-cos^2\boldsymbol\alpha}, sin\dfrac{\boldsymbol\pi}{3} = \dfrac{\sqrt{3}}{2}\right]$

$\displaystyle\sf$

$\displaystyle\sf = \alpha + cos^{-1} \left[ cos\left(\dfrac{\pi}{3}-\alpha\right)\right]$

$\displaystyle\sf$

$\displaystyle\sf = \alpha + \dfrac{\pi}{3} - \alpha$

$\displaystyle\sf$

$\displaystyle\sf = \dfrac{\pi}{3} = \: RHS$