# Math Help - tan

1. ## tan

Without the aid of a calculator, what is the value of this expresion, tan(arcsin 1/3 + arccos 1/2)?

2. Use formula for tan(a+b)

Note tan(arcsin(1/3)) = 1/sqrt(8) Draw a rt triangle with an angle whose sine is 1/3 you should get a 1-sqrt(8)- 3 triangle

tan(arcos(1/2)) = sqrt(3) Draw a rt triangle with an angle whose cosine is 1/2 you should get a 1 - sqrt(3) - 2 triangle

3. Originally Posted by Calculus26

Note tan(arcsin(1/3)) = 1/sqrt(8) Draw a rt triangle with an angle whose sine is 1/3 you should get a 1-sqrt(8)- 3 triangle

tan(arcos(1/2)) = sqrt(3) Draw a rt triangle with an angle whose cosine is 1/2 you should get a 1 - sqrt(3) - 2 triangle
$\arcsin x=\alpha\Rightarrow\sin\alpha=x, \alpha\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$

$\tan\alpha=\frac{\sin\alpha}{\cos\alpha}=\frac{\si n\alpha}{\sqrt{1-\sin^2\alpha}}$

Then $\tan(\arcsin x)=\frac{x}{\sqrt{1-x^2}}$

$\arccos x=\beta\Rightarrow \cos\beta=x, \ \beta\in[0,\pi]$

$\tan\beta=\frac{\sqrt{1-\cos^2\beta}}{\cos\beta}$

Then $\tan(\arccos x)=\frac{\sqrt{1-x^2}}{x}$

4. Red_dog, your personal message is awesome, the solution was step by step. thanks