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Evaluate $\cos(2\cot^{-1}(32/49))$

What I did for this problem was:

  • Turn $\cot$ into $\tan(49/32)$
  • And I am not sure what I do with the $2*$

I tried multiplying $49$ by $2$ and plugging the values into a reference triangle.

After I had all three sides I just found the $\cos$ of the triangle and used that as the answer but it was incorrect. I also knew i couldn't use any of the double angle formulas because the two is in front of the identity.

The cot is inverse

Where did I go wrong?

bjcolby15
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Jeremy Hernandez
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2 Answers2

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By a well-known formula you have $$\sin\left(\cot^{-1}x\right)=\frac1{\sqrt{1+x^2}}$$ And, of course, the double angle formula $$\cos2x=1-2\sin^2x$$ Now your solution is straightforward $$\begin{align}\cos\left(\cot^{-1}\frac{32}{49}\right)&=1-2\sin^2\left(\cot^{-1}\frac{32}{49}\right)\\[10pt] &=1-2\left(\frac1{\sqrt{1+\frac{32^2}{49^2}}}\right)^2\\[10pt] &=-\frac{1377}{3425}\end{align}$$

Michael Hardy
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  • Instead of a "well known formual" $\sin\left(\cot^{-1}x\right)=\dfrac1{\sqrt{1+x^2}}$ I'd say $\cot = \dfrac{\text{adjacent}}{\text{opposite}} = \dfrac{32}{49},$ so $\sin = \dfrac{\text{opposite}}{\text{hypotenuse}} = \dfrac{49}{\sqrt{32^2 + 49^2}}.$ – Michael Hardy Nov 25 '17 at 00:05
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Use Are $\mathrm{arccot}(x)$ and $\arctan(1/x)$ the same function?

Now if $\tan^{-1}x=y,\tan y=x$

$$\cos\left(2\tan^{-1}x\right)=\cos2y=\dfrac{1-\tan^2y}{1+\tan^2y}=?$$

lab bhattacharjee
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