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Question Number 207638 by hardmath last updated on 21/May/24

Commented by TonyCWX08 last updated on 22/May/24

$$arcsin\left(sin5\right)=5$$$$arccis\left(cos6\right)=6$$$$arctan\left(tan2\right)=2$$$$5+6+2=13$$

Commented by mr W last updated on 22/May/24

$$arcsin\left(\sin 5°\right)=5°\left(indegree\right)$$$$arcsin\left(\sin 5\right)\ne 5\left(inrad\right)$$$$thereforepleasemakeclearwhat$$$$youmeanwith‘‘5^{″}.5°or5rad?$$$$usually5means5rad.$$

Commented by mr W last updated on 22/May/24

Commented by mr W last updated on 22/May/24

Answered by Simurdiera last updated on 21/May/24

$$=13$$

Commented by hardmath last updated on 21/May/24

$$\mathrm{How}$$

Answered by mr W last updated on 22/May/24

$$assumetheunitisrad.$$$$5=2\pi -(2\pi -5)$$$${\sin }^{-1}\left(\sin 5\right)=-(2\pi -5)=5-2\pi$$$$6=2\pi -(2\pi -6)$$$${\cos }^{-1}\left(\cos 6\right)=2\pi -6$$$$2=\pi -(\pi -2)$$$${\tan }^{-1}\left(\tan 2\right)=-(\pi -2)=2-\pi$$$$\Rightarrow {\sin }^{-1}\left(\sin 5\right)+{\cos }^{-1}\left(\cos 6\right)+{\tan }^{-1}\left(\tan 2\right)$$$$=5-2\pi +2\pi -6+2-\pi =1-\pi$$$$\approx -2.14159$$

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