Solving a trig equation.

**aeroflix** · August 24th 2010, 01:02 AM

solve for x

cot-1 (4x) + tan-1 (x) = 60 degrees

its so hard

**mathaddict** · August 24th 2010, 02:32 AM

Originally Posted by aeroflix

solve for x

cot-1 (4x) + tan-1 (x) = 60 degrees

its so hard

Let cot^(-1) (4x)= w

4x = cot w

tan w = 1/4x

Let also tan^(-1) x =p

x= tan p

so w + p=60

tan (w+p)= tan 60

(tan w+tan p)/(1-tan p tan w)= tan 60

Can you take it from here?

**aeroflix** · August 24th 2010, 04:16 AM

its kinda confusing. whats your point

**Soroban** · August 24th 2010, 08:17 AM

Hello, aeroflix!

The answers are truly ugly . . . but they check out!

I'll baby-talk through this for you . . .

$\text{Solve for }x\!:\;\;\cot^{-1}(4x) + \tan^{-1}(x) \:=\: 60^o$ .[1]

Let $\theta \,=\,\cot^{-1}(4x)$

. . Then: . $\cot\theta \,=\,4x \,=\,\frac{adj}{opp}$

So $\theta$ is in a right triangle with: . $opp = 1,\;adj = 4x$

. . $\tan\theta \,=\,\frac{1}{4x} \quad\Rightarrow\quad \theta \:=\:\tan^{-1}\left(\frac{1}{4x}\right)$

Hence: . $\cot^{-1}(4x) \:=\:\tan^{-1}\left(\frac{1}{4x}\right)$

Then [1] becomes: . $\tan^{-1}\left(\frac{1}{4x}\right) + \tan^{-1}(x) \;=\;60^o$ .[2]

We need this identity: . $\tan(A + B) \;=\;\dfrac{\tan A + \tan B}{1 - \tan A\tan B}$

Take the tangent of both sides of [2]:

. . $\tan\bigg[\tan^{-1}(\frac{1}{4x}) + \tan^{-1}(x)\bigg] \;=\; \tan(60^o)$

. . $\dfrac{\overbrace{\tan\left[\tan^{-1}(\tfrac{1}{4x})\right]}^{\text{This is }\frac{1}{4x}} + \overbrace{\tan\left[\tan^{-1}(x)\right]}^{\text{This is }x}}{1 - \underbrace{\tan\left[\tan^{-1}(\tfrac{1}{4x})\right]}_{\text{This is }\frac{1}{4x}} \underbrace{\tan\left[\tan^{-1}(x)\right]}_{\text{This is }x}} \;=\;\sqrt{3}$

So we have: . $\dfrac{\frac{1}{4x} + x}{1 - \frac{1}{4x}\cdot x} \;=\;\sqrt{3}$

. . which simplifies to: . $4x^2 - 3\sqrt{3}\,x + 1 \;=\;0$

$\text{Quadratic Formula: }\;x \;=\;\dfrac{3\sqrt{3} \pm\sqrt{11}}{8} \;\approx\;\begin{Bmatrix}1.064097152 \\ 0.234940954 \end{array}$

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