Solve the Trigonometry Equation: sin^-1(x) + cos^-1(2x) = π/6

[Petrus]

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]

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]

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]

Hello, Petrus!

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

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

The equation becomes: .$$\alpha + \beta \:=\:\tfrac{\pi}{6}$$

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

. . . $$\sin\alpha\cos\beta + \cos\alpha\sin\beta \:=\:\tfrac{1}{2}$$

. . . $$(x)(2x) + \left(\sqrt{1-x^2}\right)\left(\sqrt{1-4x^2}\right) \:=\:\tfrac{1}{2}$$

. . . $$2x^2 + \sqrt{(1-x^2)(1-4x^2)} \:=\:\tfrac{1}{2}$$

. . . $$\sqrt{(1-x^2)(1-4x^2)} \:=\;\tfrac{1}{2} - 2x^2$$Square both sides:

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

. . . $$(1-x^2)(1-4x^2) \;=\;\tfrac{1}{4} -2x^2 + 4x^4$$

. . . $$1-5x^2 + 4x^4 \;=\;\tfrac{1}{4} - 2x^2 + 4x^4$$

. . . $$\tfrac{3}{4} \:=\:3x^2 \quad\Rightarrow\quad x^2 \:=\:\tfrac{1}{4}$$

Therefore: .$$x \:=\:\pm\frac{1}{2}$$

 

[Petrus]

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

 

[I like Serena]

Congratulations!

You have become a Journeyman.

- $\text I\ \lambda\ \Sigma$

 

[chisigma]

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

$\chi$ $\sigma$

 

[I like Serena]

The number of posts.
Starting at 500 posts you're classified as a Journeyman.

 

[MarkFL]

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:

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[kaliprasad]

Hello, Petrus!


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

The equation becomes: .$$\alpha + \beta \:=\:\tfrac{\pi}{6}$$

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

. . . $$\sin\alpha\cos\beta + \cos\alpha\sin\beta \:=\:\tfrac{1}{2}$$

. . . $$(x)(2x) + \left(\sqrt{1-x^2}\right)\left(\sqrt{1-4x^2}\right) \:=\:\tfrac{1}{2}$$

. . . $$2x^2 + \sqrt{(1-x^2)(1-4x^2)} \:=\:\tfrac{1}{2}$$

. . . $$\sqrt{(1-x^2)(1-4x^2)} \:=\;\tfrac{1}{2} - 2x^2$$Square both sides:

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

. . . $$(1-x^2)(1-4x^2) \;=\;\tfrac{1}{4} -2x^2 + 4x^4$$

. . . $$1-5x^2 + 4x^4 \;=\;\tfrac{1}{4} - 2x^2 + 4x^4$$

. . . $$\tfrac{3}{4} \:=\:3x^2 \quad\Rightarrow\quad x^2 \:=\:\tfrac{1}{4}$$

Therefore: .$$x \:=\:\pm\frac{1}{2}$$

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]

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