How to prove algebraically that \(cos^{-1} (\frac{1}{2} \)) is \(\frac{\pi}{3} \)?

How to prove algebraically that \(cos^{-1} (\frac{1}{2} \)) is \(\frac{\pi}{3} \)?

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TopNewestMaybe first start with cos 3x = 4 \(\cos^{3} \) x - 3 cos x? – Noel Lo · 2 years, 2 months ago

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cos 3x = 0 means the smallest positive value for 3x is \(\frac{\pi}{2} \) since cos \(\frac{\pi}{2} \) = 0 which follows that x= \(\frac{\pi}{6} \).

We can also prove here that cos \(\frac{\pi}{6} \) = \(\frac{\sqrt{3}}{2}\). Then \(\cos^{-1} (\frac{\sqrt{3}}{2}\) ) = \(\frac{\pi}{6} \). – Noel Lo · 2 years, 2 months ago

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Did you know that \(cos^{-1} cos x = x \forall x \in [0,\pi] \) ?

I have provided you with the graph of \(y=cos^{-1} cos x\) – Azhaghu Roopesh M · 2 years, 2 months ago

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– Noel Lo · 2 years, 2 months ago

And just like what we did for \(\frac{\pi}{3} \), we can also show that \(\sin^{-1} (\frac{1}{2} \)) = \(\frac{\pi}{6} \).Log in to reply

– Noel Lo · 2 years, 2 months ago

Or perhaps sin 3x = 3 sin x - 4 \(\sin^{3} x \). Equate 3 sin x - 4 \(\sin^{3} x \) to 0 and we have sin x= 0 or \(\frac{\sqrt{3}} {2} \). Similarly, for sin 3x =0, the basic value for x to satisfy would be \(\frac{\pi}{3} \). This means sin \(\frac{\pi}{3} \) = \(\frac{\sqrt{3}} {2} \) and \(\sin^{-1} \) (\(\frac{\sqrt{3}}{2} \)) =\(\frac{\pi}{3} \) . Now, cos \(\frac{\pi}{3} \) = \(\frac{1}{2} \) since \(\sin^2 x\) + \(\cos^2 x\) =1. It follows that \(cos^{-1} (\frac{1}{2} \))= \(\frac{\pi}{3} \)Log in to reply

– Noel Lo · 2 years, 2 months ago

Someone think of a shorter way?Log in to reply

– Noel Lo · 2 years, 2 months ago

By the way, to further prove the identity for sin 3x and cos 3x, we can express 3x as 2x + x and use the double angle formula.Log in to reply