MHB Simplifying a Trigonometric Expression

AI Thread Summary
The discussion focuses on simplifying the trigonometric expression $$\frac{\cot^3\left({y}\right)-\tan^3\left({y}\right)}{\sec^2\left({y}\right)+\cot^2\left({y}\right)}$$ to show that it equals $$2\cot\left({2y}\right)$$. The first contributor uses the difference of cubes formula and trigonometric identities to derive the result, demonstrating that $$\cot y - \tan y$$ simplifies to $$2\cot 2y$$. Another participant shares an alternative approach involving various trigonometric identities, ultimately reaching the same conclusion. The use of LaTeX is appreciated for clarity in presenting the mathematical steps. Overall, the thread highlights effective methods for simplifying complex trigonometric expressions.
karush
Gold Member
MHB
Messages
3,240
Reaction score
5
$$\frac{\cot^3\left({y}\right)-\tan^3\left({y}\right)}
{\sec^2\left({y}\right)+\cot^2\left({y}\right)}
=2\cot\left({2y}\right)$$

I tried the LHS but could get it to reduce.
 
Last edited:
Mathematics news on Phys.org
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.
 
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
 
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

$$\dfrac{\cot^3y-\tan^3y}{\sec^2y+\cot^2y}=\dfrac{(\sec^2y+\cot^2y)(\cot^2y-1)}{(\sec^2y+\cot^2y)\cot y}=\dfrac{\cot^2y-1}{\cot y}=2\cot2y$$
 
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?
 
karush said:
Do you cut and paste latex?

Sometimes. I usually copy and paste. :)
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
Thread 'Imaginary Pythagorus'
I posted this in the Lame Math thread, but it's got me thinking. Is there any validity to this? Or is it really just a mathematical trick? Naively, I see that i2 + plus 12 does equal zero2. But does this have a meaning? I know one can treat the imaginary number line as just another axis like the reals, but does that mean this does represent a triangle in the complex plane with a hypotenuse of length zero? Ibix offered a rendering of the diagram using what I assume is matrix* notation...

Similar threads

Back
Top