MHB Solving trig equation cos(x)=sin(x) + 1/√3

  • Thread starter Thread starter sp3
  • Start date Start date
  • Tags Tags
    Trig
AI Thread Summary
The discussion revolves around solving the trigonometric equation cos(x) = sin(x) + 1/√3 and finding the value of cos^3(x) - sin^3(x). Participants derive a quadratic equation for sin(x) and find two solutions for x within the range of 0 to 2π. They also explore using identities to simplify the expression for cos^3(x) - sin^3(x). Ultimately, the correct value is determined to be 4/(3√3) by substituting the derived values into the identity for the difference of cubes.
sp3
Messages
8
Reaction score
0
Hello, I'm trying to solve the following equation : cos(x)=sin(x) + 1/$$\sqrt{3}$$
in order to find cos(x)3 - sin3(x) = ?

i tried to slove for sin(x) using sin2 + cos2 =1 replacing cos(x) by the first equation and i end up with a second degree polynomial using x = sin(x) and there are 2 solutions, it seems off...
btw i used the (a+b)2 identity for (sin(x) + 1/$$\sqrt{3}$$)2

if anyone could help me i thank you in advance!
 
Mathematics news on Phys.org
Okay, I presume you wrote \sqrt{1- sin^2(x)}= sin(x)+ \frac{1}{\sqrt{3}} and then, squaring both sides, 1- sin^2(x)= sin^2(x)+ \frac{2}{\sqrt{3}}sin(x)+ \frac{1}{3}.

Then we have sin^2(x)+ \frac{1}{\sqrt{3}}sin(x)- \frac{1}{3}= 0.

By the quadratic formula, sin(x)= \frac{-\frac{1}{\sqrt{3}}\pm\sqrt{\frac{1}{3}+ \frac{4}{3}}}{2}= \frac{-1\pm \sqrt{5}}{2\sqrt{3}}.

Yes, there are two solutions (between 0 and 2\pi). Numerically, sin(x)= 0.3568 so x= 0.3649 and sin(x)= -0.9342 so x= -1.2059 to four decimal places.
 
thank you! i find the same thing but if i have to solve cos(x)3- sin(x)3 how do i get to one answer with this $$\pm$$ ? the answer is $$\frac{4}{3\sqrt{3}}$$. I thought about using the identity
a3-b3 = (a-b)(a2+2ab+b2) as an alternative method with a=sin(x)+$$\frac{1}{\sqrt{3}}$$ and b=sin(x) . I find this :

$$\frac{12sin(x)^2+\sqrt{3}*4sin(x)+1}{3\sqrt{3}}$$

Idk how to further develop. In the identity I factorized $$(sin+\frac{1}{\sqrt{3}})^2$$, found in (a-b) sinx-sinx =0 so there’s $$\frac{1}{\sqrt{3}}$$ left. any help is welcome!

Btw people here are so quick to answer this website is bomb, thank you really! Appreciate it ;)
 
Last edited:
Hello SP3.
you do not need $\sin\,x$ or $\cos\,x$
you are given
$\cos\,x - \sin\,x = \frac{1}{\sqrt{3}}$
$\cos^3 x - \sin ^3 x = (\cos\,x - \sin\,x)^3 + 3 (\cos\,x - \sin\,x)(\cos\,x \sin\,x)\cdots(1)$
so you need to evaluate $\cos\,x \sin\,x$
$\cos\,x - \sin\,x = \frac{1}{\sqrt{3}}$
square both sides to get $ 1 - 2\cos\,x \sin\,x = \frac{1}{3}$
or $ 2\cos\,x \sin\,x = \frac{2}{3}$
or $ \cos\,x \sin\,x = \frac{1}{3}$
you can put the value of $ \cos\,x \sin\,x = \frac{1}{3}$ and $\cos\,x - \sin\,x = \frac{1}{\sqrt{3}}$ in (1) to get
$\cos^3 x - \sin ^3 x = (\cos\,x - \sin\,x)^3 + 3 (\cos\,x - \sin\,x)(\cos\,x \sin\,x)$
$= \frac{1}{3\sqrt{3}} + 3 \frac{1}{\sqrt{3}} \frac{1}{3}$
$=\frac{1+3}{3\sqrt{3}}$
$=\frac{4}{3\sqrt{3}}$
 
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...
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Thread 'Imaginary Pythagoras'
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...
Back
Top