arcsin and arctan: arcsin(1/sqrt[5]) = (1/2) arctan(4/3); arcsin(th x) = arctan(sh x)

YM30

New member
Joined
Oct 12, 2018
Messages
2
Hello,
could anybody give me a hint for 1) and 2) ?

\(\displaystyle \mbox{1) Montrer: }\, \arcsin\left(\dfrac{1}{\sqrt{5\,}}\right)\, =\, \dfrac{1}{2}\, \arctan\left(\dfrac{4}{3}\right)\)

\(\displaystyle \mbox{2) Montrer, pour tout }\, x\, \in\, \mathbb{R}:\, \arcsin(\mbox{th}\, x)\, =\, \arctan(\mbox{sh}\, x)\)

For 1) I have tried using sin^2(X) on the left side and tan^2(X)/1+tan^2(X) on the right side but it didn't work...
 

Attachments

  • Screenshot_20181012-181252_1.jpg
    Screenshot_20181012-181252_1.jpg
    26.9 KB · Views: 4
Last edited by a moderator:
Hello,
could anybody give me a hint for 1) and 2) ?
For 1) I have tried using sin^2(X) on the left side and tan^2(X)/1+tan^2(X) on the right side but it didn't work...

That's an arcsin(), not a sin(). What are you squaring?

Are we to prove the equivalence in 1? Typically, one would:

1) Draw a right triangle.

2) Designate an acute angle, maybe \(\displaystyle \alpha\)

3) Label lengths of two sides, Opposite = 1 and Hypotenuse = \(\displaystyle \sqrt{5}\), in this case.

4) Calculate the length of the third side. Adjacent = \(\displaystyle \sqrt{(\sqrt{5})^{2} - 1^{2}} = \sqrt{5-1} = 2\)

5) Calculate \(\displaystyle \cos(\alpha) = \dfrac{2}{\sqrt{5}}\)

6) Calculate \(\displaystyle \sin(2\alpha)\;and\;\cos(2\alpha)\) using your double angle formulas. -- You tell me why!

7) We're almost done...
 
Last edited:
Hello,
could anybody give me a hint for 1) and 2) ?
For 1) I have tried using sin^2(X) on the left side and tan^2(X)/1+tan^2(X) on the right side but it didn't work...

The problems are:

1) Show that arcsin(1/sqrt(5)) = (1/2) arctan(4/3)
2) Show, for all real numbers x, that arcsin(th x) = arctan(sh x)

I believe you are using th for what I call tanh, and sh for sinh, the hyperbolic functions.

I'd do both of these essentially as tkhunny described. First, though, I'd restate the problems:

1) Show that, if sin(θ) = 1/sqrt(5), then tan(2θ) = 4/3.
2) Show that, for all real numbers x, if sin(θ) = tanh(x), then tan(θ) = sinh(x).

(Some consideration of ranges of functions is needed, but it will be trivial. All angles are in the first quadrant.)

For each of these problems, I draw a triangle and label sides; in the first, I'd use the double-angle formula for the tangent -- which you wrote incorrectly, if that is what you intended. The reality is that tan(2θ) = 2tan(θ)/(1 - tan^2(θ)). Possibly that is your only error, as it is entirely possible to do this using identities rather than drawing triangles. You have sin^2(θ) = 1/5, from which you can find cos(θ), and so on.

The second ends up using a hyperbolic identity.
 
Top