trig help proof

[shelly89]

prove \(\displaystyle sin3\theta = 3sinh\theta + 4sinh^{3} \theta \)

I have done this part, but I am having troubles with the second and third part of the question.

We want to solve \(\displaystyle y = x^{3} + x \) for x in terms of y. Show that if \(\displaystyle z = \frac{1}{2} \sqrt{3} x \) then the equation

\(\displaystyle 4z^{3} + 3z = \frac{3}{2} \sqrt{3}y \)
.Then let \(\displaystyle z=sinhθ \) and deduce using part(i) that

\(\displaystyle x = \frac{2}{\sqrt{3}} sinh[ \frac{1}{3} sinh^{-1} (\frac{3\sqrt{3}}{2}y)] \)


I have tired to rearrange the equation

\(\displaystyle z = \frac{1}{2} \sqrt{3} x \)

for x and got \(\displaystyle x = \frac{2z}{\sqrt{3}} \)

and than put this into the equation \(\displaystyle y = x^{3} + x \)

but than dont get the same equation as the question wants

 

[HallsofIvy]

prove \(\displaystyle sin3\theta = 3sinh\theta + 4sinh^{3} \theta \)

Typo- you mean \(\displaystyle sinh3\theta\) on the left.

I have done this part, but I am having troubles with the second and third part of the question.

We want to solve \(\displaystyle y = x^{3} + x \) for x in terms of y. Show that if \(\displaystyle z = \frac{1}{2} \sqrt{3} x \) then the equation

\(\displaystyle 4z^{3} + 3z = \frac{3}{2} \sqrt{3}y \)
.Then let \(\displaystyle z=sinhθ \) and deduce using part(i) that

\(\displaystyle x = \frac{2}{\sqrt{3}} sinh[ \frac{1}{3} sinh^{-1} (\frac{3\sqrt{3}}{2}y)] \)


I have tired to rearrange the equation

\(\displaystyle z = \frac{1}{2} \sqrt{3} x \)

for x and got \(\displaystyle x = \frac{2z}{\sqrt{3}} \)

and than put this into the equation \(\displaystyle y = x^{3} + x \)

but than dont get the same equation as the question wants

Yes, \(\displaystyle x= \frac{2z}{\sqrt{3}}\) so that \(\displaystyle x^3= \frac{8z^3}{3\sqrt{3}}\). Then \(\displaystyle x^3+ x= \frac{8z^3}{3\sqrt{3}}+ \frac{2z}{\sqrt{3}}= \frac{8z^3}{3\sqrt{3}}+ \frac{6z}{3\sqrt{3}}= y\).

Now, multiply both sides by \(\displaystyle \frac{3\sqrt{3}}{2}\).

 

[shelly89]

Typo- you mean \(\displaystyle sinh3\theta\) on the left.


Yes, \(\displaystyle x= \frac{2z}{\sqrt{3}}\) so that \(\displaystyle x^3= \frac{8z^3}{3\sqrt{3}}\). Then \(\displaystyle x^3+ x= \frac{8z^3}{3\sqrt{3}}+ \frac{2z}{\sqrt{3}}= \frac{8z^3}{3\sqrt{3}}+ \frac{6z}{3\sqrt{3}}= y\).

Now, multiply both sides by \(\displaystyle \frac{3\sqrt{3}}{2}\).

Hello, thank you, but where did you get \(\displaystyle \frac{3\sqrt{3}}{2}\) ?

 

[JeffM]

Yes, \(\displaystyle x= \frac{2z}{\sqrt{3}}\) so that \(\displaystyle x^3= \frac{8z^3}{3\sqrt{3}}\). Then \(\displaystyle x^3+ x= \frac{8z^3}{3\sqrt{3}}+ \frac{2z}{\sqrt{3}}= \frac{8z^3}{3\sqrt{3}}+ \frac{6z}{3\sqrt{3}}= y\).

Now, multiply both sides by \(\displaystyle \frac{3\sqrt{3}}{2}\).

Hello, thank you, but where did you get \(\displaystyle \frac{3\sqrt{3}}{2}?.\)

Halls noticed that in the expansion of \(\displaystyle x^3 + x\) every denominator contains \(\displaystyle 3\sqrt{3}.\) So he could simplify by multiplying

both sides by \(\displaystyle 3\sqrt{3}.\)

Halls also noticed that 2 is a common factor in every numerator. He could also simplify by dividing both sides by 2.

But being efficient, he condenses both simplifications and simply multiplies both sides by \(\displaystyle \dfrac{3\sqrt{3}}{2}.\)