Cosine of an inverse tangent

Geometry Level 1

In the right triangle above, 0 < b < a 0<b<a . One of the angle measures in the triangle is tan ⁡ − 1 ( a b ) \tan^{-1} \left( \dfrac ab\right) . What is cos ⁡ [ tan ⁡ − 1 ( a b ) ] \cos \left[ \tan^{-1} \left( \dfrac ab\right) \right ] ?

a a 2 + b 2 \frac a{\sqrt{a^2+b^2}} a 2 + b 2 a \frac {\sqrt{a^2+b^2}}a b a 2 + b 2 \frac b{\sqrt{a^2+b^2}} a b \frac ab b a \frac ba

Hi Hi by Hi Hi , 18, USA

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1 solution

Hi Hi
Jun 30, 2016

If you remember your trigonometry rules, you know that tan − 1 a b \frac{a}{b} is the same as saying tanΘ = a b \frac{a}{b} . Knowing our mnemonic device SOH, CAH, TOA, we know that tan Θ = opposite/adjacent. If a is our opposite and b is our adjacent, this means that Θ will be our right-most angle.

Knowing that, we can find the cos of Θ as well. The cosine will be the adjacent over the hypotenuse. the adjacent still being b and the hypotenuse being √a2+b2. So cos[tan − 1 a b \frac{a}{b} ] will be:

b √ a 2 + b 2 \frac{b}{√a2+b2}

Our final answer is... b √ a 2 + b 2 \frac{b}{√a2+b2}

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