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Calculate the value of x if \cos ^{-1} \frac{1}{3}+\cos ^{-1} \frac{2}{3}=\cos ^{-1} x

Option: 1

\pi+\frac{1}{9} (2-2\sqrt{10})


Option: 2

\frac{1}{9} (2-2\sqrt{10})


Option: 3

\frac{1}{9} (2+2\sqrt{10})


Option: 4

2\pi+\frac{1}{9} (2-2\sqrt{10})


Answers (1)

best_answer

Sum and difference of angles in terms of arccos

cos ^{-1} x+\cos ^{-1} y=\cos ^{-1}\{x y-\sqrt{1-x^{2}} \sqrt{1-y^{2}}\} \,\,\text { if }0 < x, y \leq 1

 

Now,

\\\cos ^{-1} \frac{1}{3}+\cos ^{-1} \frac{2}{3}=\cos ^{-1} x\\ \\ \cos^{-1} \{ \frac{1}{3} \times \frac{2}{3}-\sqrt{1-(\frac{1}{3})^2}\sqrt{1-(\frac{2}{3})^2} \ \}=\cos^{-1} x\\ \\\cos^{-1} \{ \frac{1}{9} (2-2\sqrt{10})\}=\cos^{-1}x \\ x=\frac{1}{9} (2-2\sqrt{10})

Posted by

Suraj Bhandari

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