1. ## Trig. problems.

Hi,

Can someone help me to do these problems?

1).If tan(θ+α)-(3+2√2)tanθ = 0,
Prove that, sin(2θ+α )= √2sinα

2).Prove that, cos⁻¹x-sin⁻¹x= π/6

2. Hello, Malsha!

2) Prove that: cos⁻¹x-sin⁻¹x= π/6
Could that be: .$\displaystyle \cos^{\text{-}1}x - \sin^{\text{-}1}x \:=\:\frac{\pi}{6}\;?\quad\hdots$ This is not true!

$\displaystyle \text{We have: }\;\underbrace{\cos^{\text{-}1}x}_{\alpha} - \underbrace{\sin^{\text{-}1}x}_{\beta}$

That is: .$\displaystyle \begin{array}{ccc}\alpha \:=\:\cos^{-1}x & \Rightarrow & \cos\alpha \:=\:x \\ \beta \:=\:\sin^{-1}x & \Rightarrow & \sin\beta \:=\:x \end{array}$

We have: .$\displaystyle \begin{array}{ccccc}\cos\alpha &=&\dfrac{x}{1} &=& \dfrac{adj}{hyp} \\ \\[-3mm] \sin\beta &=& \dfrac{x}{1} &=& \dfrac{opp}{hyp} \end{array}$

Hence, the two angles are in this triangle.
Code:
                        *
* β*
1   *     *
*        *
*           *
* α            *
*  *  *  *  *  *  *
x

And we see that the two angles are complementary: .$\displaystyle \alpha + \beta \:=\:\frac{\pi}{2}$

So their sum is a constant, but not their difference.