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Trigonometry Trigonometry equation

Petrus

Well-known member
Feb 21, 2013
739
Hello MHB,
I am working with an old exam
Solve this equation \(\displaystyle \sin^{-1}(x)+\cos^{-1}(2x)=\frac{\pi}{6}\)

progress:
I start take cos both side and I get
\(\displaystyle 2x+cos(\sin^{-1}(x))=\frac{\sqrt{3}}{2}\)
I draw it and call the bottom side for B and get

so we got
\(\displaystyle \sqrt{1-x^2}+2x=\frac{\sqrt{3}}{2}\)
Is this correct? (notice that I got problem solving this equation as well.)


Regards,
\(\displaystyle |\pi\rangle\)
 

chisigma

Well-known member
Feb 13, 2012
1,704
Hello MHB,
I am working with an old exam

Solve this equation \(\displaystyle \sin^{-1}(x)+\cos^{-1}(2x)=\frac{\pi}{6}\)

progress:

I start take cos both side and I get
\(\displaystyle 2x+cos(\sin^{-1}(x))=\frac{\sqrt{3}}{2}\)
Unfortunately is $\cos (\alpha + \beta) \ne \cos \alpha + \cos \beta$ ...


Kind regards


$\chi$ $\sigma$
 

kaliprasad

Well-known member
Mar 31, 2013
1,309
Solve this equation sin−1(x)+cos−1(2x)=π6

you are putting cos (A+B) = cos A + cos B which is not correct

we have sin ^-1 x = cos ^-1 ( 1- x^2)^(1/2)
and cos ^-1 2x = sin ^-(1-4x^2)^(1/2)

so take cos of both sides to get

cos ( sin ^-1 x) cos ( cos ^-1 2x) - sin ( sin ^- 1 x) cos (cos ^-1 2x) = sqrt(3)/2

os sqrt((1-x^2)(1-4x^2) - 2x^2 = sqrt(3)/2

now you should be able to proceed
 

soroban

Well-known member
Feb 2, 2012
409
Hello, Petrus!

[tex]\text{Solve: }\:\sin^{-1}(x)+\cos^{-1}(2x)=\tfrac{\pi}{6}[/tex]

[tex]\text{Let: }\:\begin{Bmatrix}\alpha = \sin^{-1}(x) &\Rightarrow& \sin\alpha \,=\,x &\Rightarrow& \cos\alpha \,=\,\sqrt{1-x^2} \\ \beta \,=\,\cos^{-1}(2x) &\Rightarrow& \cos\beta \,=\,2x &\Rightarrow& \sin\beta \,=\,\sqrt{1-4x^2} \end{Bmatrix}[/tex]

The equation becomes: .[tex]\alpha + \beta \:=\:\tfrac{\pi}{6}[/tex]

Take the sine of both sides: .[tex]\sin(\alpha + \beta) \:=\:\sin\left(\tfrac{\pi}{6}\right)[/tex]

. . . [tex]\sin\alpha\cos\beta + \cos\alpha\sin\beta \:=\:\tfrac{1}{2}[/tex]

. . . [tex](x)(2x) + \left(\sqrt{1-x^2}\right)\left(\sqrt{1-4x^2}\right) \:=\:\tfrac{1}{2}[/tex]

. . . [tex]2x^2 + \sqrt{(1-x^2)(1-4x^2)} \:=\:\tfrac{1}{2}[/tex]

. . . [tex]\sqrt{(1-x^2)(1-4x^2)} \:=\;\tfrac{1}{2} - 2x^2[/tex]


Square both sides:

. . . [tex]\left[\sqrt{(1-x^2)(1-4x^2)}\right]^2 \;=\;\left[\tfrac{1}{2} - 2x^2\right]^2[/tex]

. . . [tex](1-x^2)(1-4x^2) \;=\;\tfrac{1}{4} -2x^2 + 4x^4[/tex]

. . . [tex]1-5x^2 + 4x^4 \;=\;\tfrac{1}{4} - 2x^2 + 4x^4[/tex]

. . . [tex]\tfrac{3}{4} \:=\:3x^2 \quad\Rightarrow\quad x^2 \:=\:\tfrac{1}{4}[/tex]

Therefore: .[tex]x \:=\:\pm\frac{1}{2}[/tex]
 

Petrus

Well-known member
Feb 21, 2013
739
Thanks evryone for pointing out what I have done wrong!:) I got same answer as Soroban/facit so I got correct answer and understand what I did wrong!:)

Regards,
\(\displaystyle |\pi\rangle\)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,774
Congratulations!

You have become a Journeyman.

- $\text I\ \lambda\ \Sigma$
 

chisigma

Well-known member
Feb 13, 2012
1,704
Congratulations!

You have become a Journeyman.

- $\text I\ \lambda\ \Sigma$
I apologize for my ignorance, but I also have been classified as 'Journeyman'... now I ask what is the difference between 'Journeyman' and [for example...] 'Craftsman'?...

Kind regards

$\chi$ $\sigma$
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,774
The number of posts.
Starting at 500 posts you're classified as a Journeyman.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
I apologize for my ignorance, but I also have been classified as 'Journeyman'... now I ask what is the difference between 'Journeyman' and [for example...] 'Craftsman'?...

Kind regards

$\chi$ $\sigma$
The user titles are based on post counts as follows:

MHB Apprentice: 0-99 posts
MHB Craftsman: 100-499 posts
MHB Journeyman: 500-999 posts
MHB Master: 1000-4999 posts
MHB Grandmaster: 5000-Infinity posts
 

kaliprasad

Well-known member
Mar 31, 2013
1,309
Hello, Petrus!


[tex]\text{Let: }\:\begin{Bmatrix}\alpha = \sin^{-1}(x) &\Rightarrow& \sin\alpha \,=\,x &\Rightarrow& \cos\alpha \,=\,\sqrt{1-x^2} \\ \beta \,=\,\cos^{-1}(2x) &\Rightarrow& \cos\beta \,=\,2x &\Rightarrow& \sin\beta \,=\,\sqrt{1-4x^2} \end{Bmatrix}[/tex]

The equation becomes: .[tex]\alpha + \beta \:=\:\tfrac{\pi}{6}[/tex]

Take the sine of both sides: .[tex]\sin(\alpha + \beta) \:=\:\sin\left(\tfrac{\pi}{6}\right)[/tex]

. . . [tex]\sin\alpha\cos\beta + \cos\alpha\sin\beta \:=\:\tfrac{1}{2}[/tex]

. . . [tex](x)(2x) + \left(\sqrt{1-x^2}\right)\left(\sqrt{1-4x^2}\right) \:=\:\tfrac{1}{2}[/tex]

. . . [tex]2x^2 + \sqrt{(1-x^2)(1-4x^2)} \:=\:\tfrac{1}{2}[/tex]

. . . [tex]\sqrt{(1-x^2)(1-4x^2)} \:=\;\tfrac{1}{2} - 2x^2[/tex]


Square both sides:

. . . [tex]\left[\sqrt{(1-x^2)(1-4x^2)}\right]^2 \;=\;\left[\tfrac{1}{2} - 2x^2\right]^2[/tex]

. . . [tex](1-x^2)(1-4x^2) \;=\;\tfrac{1}{4} -2x^2 + 4x^4[/tex]

. . . [tex]1-5x^2 + 4x^4 \;=\;\tfrac{1}{4} - 2x^2 + 4x^4[/tex]

. . . [tex]\tfrac{3}{4} \:=\:3x^2 \quad\Rightarrow\quad x^2 \:=\:\tfrac{1}{4}[/tex]

Therefore: .[tex]x \:=\:\pm\frac{1}{2}[/tex]
In the above solution
x = 1/2 is a root

as sin−1(1/2)+cos−1(1)=π/6 + 0 = π/6

but x= - 1/2 is not a solution as

sin−1(-1/2)+cos−1(-1)=- π/6 + π = 5 π/6

so x= - 1/2 is erroneous
 

kaliprasad

Well-known member
Mar 31, 2013
1,309
I thought that I would provide a simpler solution
we have cos ^-1 x = sin ^-1 ( 1-4x^2)^(1/2)

now given equation

sin ^-1(x) = π/6 – cos ^-1(2x)

take sin of both sides

x = sin (π/6) cos cos ^-1(2x) - cos (π/6) sin cos ^-1(2x)

or x = ½ * 2x – sqrt(3)/2 * (1- 4x^2)^(1/2) = x - sqrt(3)/2 * (1- 4x^2)^(1/2)

or * (1- 4x^2)^(1/2) = 0 => x = +/- ½

x = ½ satisfies and -1/2 does not satisfy the condition .

So solution = x = ½