MHB Simplifying a Trigonometric Expression

[karush]

$$\frac{\cot^3\left({y}\right)-\tan^3\left({y}\right)}
{\sec^2\left({y}\right)+\cot^2\left({y}\right)}
=2\cot\left({2y}\right)$$

I tried the LHS but could get it to reduce.

 

[Euge]

Hi Karush,

If you use the difference of cubes formula $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$ with $a = \cot y$ and $b = \tan y$, you get

$$\cot^3 y - \tan^3 y = (\cot y - \tan y)(\cot^2 y + \cot y \tan y + \tan^2 y) = (\cot y - \tan y)(\cot^2 y + 1 + \tan^2 y)$$

$$= (\cot y - \tan y)(\cot^2 y + \sec^2 y)$$

In the last step, the identity $1 + \tan^2 y = \sec^2 y$ was used.

Now

$$\cot y - \tan y = \frac{\cos y}{\sin y} - \frac{\sin y}{\cos y} = \frac{\cos^2 y - \sin^2 y}{\sin y \cos y} = \frac{\cos 2y}{(1/2)\sin 2y} = 2\cot 2y$$

Therefore

$$\frac{\cos^3 y - \tan^3 y}{\sec^2 y + \cot^2 y} = \frac{(2\cot 2y)(\cot^2 y + \sec^2 y)}{\sec^2 y + \cot^2 y} = 2\cot 2y,$$

as desired.

 

[karush]

wow that is awesome
I saw another solution to this but it got way to busy with mega steps
and not in latex

this is much more useful and comprehensive

thanks
K

 

[Greg]

Another approach:

$$\sin^2x+\cos^2x=1\Rightarrow1+\cot^2x=\csc^2x$$

$$\sin^2x+\cos^2x=1\Rightarrow\tan^2x+1=\sec^2x$$

$$\tan2x=\dfrac{\sin2x}{\cos2x}=\dfrac{2\sin x\cos x}{2\cos^2x-1}=\dfrac{2\tan x}{2-\sec^2x}=\dfrac{2\tan x}{2-1-\tan^2x}=\dfrac{2\tan x}{1-\tan^2x}$$

$$\cot2x=\dfrac{1-\tan^2x}{2\tan x}=\dfrac{\cot^2x-1}{2\cot x}\Rightarrow2\cot2x=\dfrac{\cot^2x-1}{\cot x}$$

$$\dfrac{\cot^3y-\tan^3y}{\sec^2y+\cot^2y}\cdot\dfrac{\cot y}{\cot y}=\dfrac{\cot^4y-\tan^2y}{(\sec^2y+\cot^2y)\cot y}$$

$$(\sec^2y+\cot^2y)(\cot^2y-1)=\csc^2y+\cot^4y-\sec^2y-\cot^2y=\csc^2y+\cot^4y-\tan^2y-1-\csc^2y+1$$
$$=\cot^4y-\tan^2y$$

hence

$$\dfrac{\cot^3y-\tan^3y}{\sec^2y+\cot^2y}=\dfrac{(\sec^2y+\cot^2y)(\cot^2y-1)}{(\sec^2y+\cot^2y)\cot y}=\dfrac{\cot^2y-1}{\cot y}=2\cot2y$$

 

[karush]

Wow thank you,

That gave me some more insight hp how to do some other problems.

I'll b be posting some more

I really appreciate the latex really a headache without it

Do you cut and paste l latex?

 

[Greg]

Do you cut and paste latex?

Sometimes. I usually copy and paste. :)