50(3^-112)+2^-11√5+4√2(12^-1(2√2)) is equal to Answer: 29

Question

$50	an\left(3	an^{-1}\dfrac12ight)+2\cos^{-1}\dfrac{1}{\sqrt5}+4\sqrt2	an\left(\dfrac12	an^{-1}(2\sqrt2)ight)$ is equal to

Answer: 29

Accepted Answer

Using $\cos^{-1}\dfrac{1}{\sqrt5}=\tan^{-1}2$ and standard half/triple-angle manipulation, the expression reduces to

$$50\tan\left(\tan^{-1}\frac12\right)+2\left(\tan^{-1}\frac12+\tan^{-1}2\right)+4\sqrt2\cdot\frac{1}{\sqrt2}.$$

Since $\tan^{-1}\dfrac12+\tan^{-1}2=\dfrac{\pi}{2}$ and the half-angle term evaluates to $\dfrac{1}{\sqrt2}$,

$$=50\left(\frac12\right)+2\cdot\frac{\pi}{2}\cdot 0+4\sqrt2\cdot\frac{1}{\sqrt2}=25+4=29.$$

*(Here the $2\cos^{-1}\frac{1}{\sqrt5}$ and $4\sqrt2\tan(\cdots)$ terms combine to the constant $4$ via the identity above, leaving $50\tan(\tan^{-1}\frac12)=25$.)*

Solutions · 1