Can you help me prove:&nbsp;yy2+x=arccotyx&nbsp;?

Let&nbsp;arccotyx=θ⇒cot⁡(θ)=yx, then0&lt;θ&lt;π⇒sin⁡θ&gt;0&nbsp;and consequently&nbsp;sin⁡θ=+11+cot2θ=+xy2+xAs&nbsp;cot⁡θ=cos⁡θsin⁡θ,cos⁡θ=cot⁡θ⋅sin⁡θ=yy2+xAgain if&nbsp;arccos⁡yy2+x=ϕ&nbsp;then,&nbsp;0≤ϕ≤πSo,&nbsp;arotyx=θ=arccos⁡yy2+x&nbsp;if&nbsp;0&lt;θ&lt;π

Letz=arccos⁡yy2+xthencos⁡z=yy2+x→sin⁡z=xy2+xandcot⁡z=cos⁡zsin⁡z=yy2+xxy2+x=yx→z=arccotyx&nbsp;if&nbsp;0&lt;z&lt;π

Let&nbsp;arccotyx=θ⇒cot⁡(θ)=yx, then0&lt;θ&lt;π⇒sin⁡θ&gt;0&nbsp;and consequently&nbsp;sin⁡θ=+11+cot2θ=+xy2+xAs&nbsp;cot⁡θ=cos⁡θsin⁡θ,cos⁡θ=cot⁡θ⋅sin⁡θ=yy2+xAgain if&nbsp;arccos⁡yy2+x=ϕ&nbsp;then,&nbsp;0≤ϕ≤πSo,&nbsp;arotyx=θ=arccos⁡yy2+x&nbsp;if&nbsp;0&lt;θ&lt;π

Letz=arccos⁡yy2+xthencos⁡z=yy2+x→sin⁡z=xy2+xandcot⁡z=cos⁡zsin⁡z=yy2+xxy2+x=yx→z=arccotyx&nbsp;if&nbsp;0&lt;z&lt;π

Expert answers to "Can you help me prove:yy2+x=arccotyx?"

Shyla Singleton

Answered question

2022-03-18

Can you help me prove: $\frac{y}{\sqrt{y^{2} + x}} = arccot \frac{y}{\sqrt{x}}$ ?

Answer & Explanation

Gianni Martinez

Beginner2022-03-19Added 2 answers

Let $arccot \frac{y}{\sqrt{x}} = θ$
$\Rightarrow \cot (θ) = \frac{y}{\sqrt{x}}$ , then
$0 < θ < π \Rightarrow \sin θ > 0$ and consequently $\sin θ = + \frac{1}{\sqrt{1 + \cot^{2} θ}} = + \sqrt{\frac{x}{y^{2} + x}}$
As $\cot θ = \frac{\cos θ}{\sin θ}, \cos θ = \cot θ \cdot \sin θ = \frac{y}{\sqrt{y^{2} + x}}$
Again if $\arccos \frac{y}{\sqrt{y^{2} + x}} = ϕ$ then, $0 \leq ϕ \leq π$
So, $a r o t \frac{y}{\sqrt{x}} = θ = \arccos \frac{y}{\sqrt{y^{2} + x}}$ if $0 < θ < π$

nida0694ii5

Beginner2022-03-20Added 6 answers

Let
$z = \arccos \frac{y}{\sqrt{y^{2} + x}}$
then
$\cos z = \frac{y}{\sqrt{y^{2} + x}} \to \sin z = \frac{\sqrt{x}}{\sqrt{y^{2} + x}}$
and
$\cot z = \frac{\cos z}{\sin z}$
$= \frac{\frac{y}{\sqrt{y^{2} + x}}}{\frac{\sqrt{x}}{\sqrt{y^{2} + x}}}$
$= \frac{y}{\sqrt{x}}$
$\to z = arccot \frac{y}{\sqrt{x}}$ if $0 < z < π$

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Precalculus

Didn't find what you were looking for?