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Trigonometry solving trig equation cos(x)=sin(x) + 1/√3

sp3

New member
Sep 3, 2018
8
Hello, I'm trying to solve the following equation : cos(x)=sin(x) + 1/\(\displaystyle \sqrt{3}\)
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/\(\displaystyle \sqrt{3}\))2

if anyone could help me i thank you in advance!
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
Okay, I presume you wrote [tex]\sqrt{1- sin^2(x)}= sin(x)+ \frac{1}{\sqrt{3}}[/tex] and then, squaring both sides, [tex]1- sin^2(x)= sin^2(x)+ \frac{2}{\sqrt{3}}sin(x)+ \frac{1}{3}[/tex].

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

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

Yes, there are two solutions (between 0 and [tex]2\pi[/tex]). Numerically, sin(x)= 0.3568 so x= 0.3649 and sin(x)= -0.9342 so x= -1.2059 to four decimal places.
 

sp3

New member
Sep 3, 2018
8
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 \(\displaystyle \pm\) ? the answer is \(\displaystyle \frac{4}{3\sqrt{3}}\). I thought about using the identity
a3-b3 = (a-b)(a2+2ab+b2) as an alternative method with a=sin(x)+\(\displaystyle \frac{1}{\sqrt{3}}\) and b=sin(x) . I find this :

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

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

Btw people here are so quick to answer this website is bomb, thank you really! Appreciate it ;)
 
Last edited:

kaliprasad

Well-known member
Mar 31, 2013
1,283
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
$\cos^3 x - \sin ^3 x = (\cos\,x - \sin\,x)^3 + 3 (\cos\,x - \sin\,x)(\cos\,x \sin\,x)$
$= \frac{1}{3\sqrt{3}} + 3 \frac{1}{\sqrt{3}} \frac{1}{3}$
$=\frac{1+3}{3\sqrt{3}}$
$=\frac{4}{3\sqrt{3}}$