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# Prove special values for trigonometry

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

Note by Noel Lo
2 years, 7 months ago

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Maybe first start with cos 3x = 4 $$\cos^{3}$$ x - 3 cos x?

- 2 years, 7 months ago

For 4 $$\cos^{3}$$ x - 3 cos x=0, we similarly have cos x =0 or $$\frac{\sqrt{3}}{2}$$.

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 $$\cos^{-1} (\frac{\sqrt{3}}{2}$$ ) = $$\frac{\pi}{6}$$.

- 2 years, 7 months ago

I think instead of using the word "smallest positive value" , you should use : the smallest solution in the Principal Interval .

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

I have provided you with the graph of $$y=cos^{-1} cos x$$

- 2 years, 7 months ago

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

- 2 years, 7 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 $$cos^{-1} (\frac{1}{2}$$)= $$\frac{\pi}{3}$$

- 2 years, 7 months ago

Someone think of a shorter way?

- 2 years, 7 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.

- 2 years, 7 months ago