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Find the value of $$\arccos(1/\sqrt2) + \arcsin (1/3) + 2 \arctan(1/\sqrt2).$$ Give your answer as a multiple of $\pi$.

This was the least well answered question on Edexcel's Advanced Extension Award annual paper in 2016. The next paper is tomorrow and I will be sitting it.

The examiner reports tells us that, out of the 7 marks that this question is worth, almost everyone got 2 marks.

  1. $\arccos (1/\sqrt2) = \pi/4$

  2. $\sin x = 1/3$

However, the rest of the mark scheme's solution is almost incomprehensible and doesn't explain anything at all well. There are no worked solutions online, so could someone please tell me how to come to the correct solution of $3\pi/4$?

user
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2 Answers2

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$$\sin x=\frac13\implies \tan x=\frac{\frac13}{\sqrt{1-(\frac13)^2}}=\frac1{2\sqrt2}\implies x=\arctan\frac1{2\sqrt2}\tag1$$

$$\tan(y/2)=\frac1{\sqrt2}\implies\tan y=\frac{2\frac1{\sqrt2}}{1-(\frac1{\sqrt2})^2}=2\sqrt2\implies y=\arctan(2\sqrt2)\tag2$$

$$(1)\& (2) \implies x+y=\frac\pi2.$$

Somewhat shorter aproach: $$ \cos y=\frac {1-\tan^2\frac y2}{1+\tan^2\frac y2}=\frac {1-\frac 12}{1+\frac 12}=\frac13.$$

$$\arcsin\frac13+\arccos\frac13=\frac\pi2.$$

user
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1

Using $\arcsin x=\arctan\dfrac x{\sqrt{1-x^2}}$ for $-1<x<1$

$$(2\sqrt2+i)(\sqrt2+i)^2=(2\sqrt2+i)(1+2\sqrt2i)=i(1+8)$$

Take argument in both sides

See also: atan2

lab bhattacharjee
  • 274,582