How to solve for when this trigonometric function intersects the line $y=1$?

Question

How can I solve for $\alpha$ in $$4\sin\left(\frac{\alpha}{2}\right)\cos^3\left(\frac{\alpha}{2}\right)\left(t-r\right)+\sin\left(\frac{\alpha}{2}\right)=1$$ on the domain $0\leq\alpha\leq\pi$? Clearly, one solution is when $\alpha=\pi$, but through plotting, it seems to only hold true when $t-r$ is less than a value around $0.3$. When $t-r$ is greater than this value, it seems to have different solutions.

Answer 1

Solving for $\alpha$ $$4\sin\left(\frac{\alpha}{2}\right)\cos\left(\frac{\alpha}{2}\right)^{3}\left(t-r\right)+\sin\left(\frac{\alpha}{2}\right)=1$$

$$ 4\sin{\left(\frac{\alpha}{2}\right)}\cos{\left(\frac{\alpha}{2}\right)}\left(1-\sin{\left(\frac{\alpha}{2}\right)^2}\right)\left(t-r\right)+\sin{\left(\frac{\alpha}{2}\right)}=1 $$ Let $x = \sin(\frac{\alpha}{2})$ $$ 4x\cos(\frac{\alpha}{2})(1-x^2)(t-r)+x=1 $$ Because $\cos(\frac{\alpha}{2})=\sqrt{1-\sin^2(\frac{\alpha}{2}})=\sqrt{1-x^2}$ $$ \Longrightarrow4x\sqrt{1-x^2}(1-x^2)(t-r)+x=1 $$

$$ \Longrightarrow4x(1-x^2)^{\frac{3}{2}}(t-r)+x=1 $$

$$ \Longrightarrow(1-x^2)^3=\frac{\left(1-x\right)^2}{16(t-r)^2x^2} $$

$$ \Longrightarrow x^2(1+x)^3(1-x)=\frac{1}{16(t-r)^2} $$

$$ \Longrightarrow -x^6-2x^5+2x^3+x^2-\frac{1}{16(t-r)^2} = 0 $$

Solving for $\alpha$ amounts to solving for the root of the equation above with dependencies on the value of $t,r$

Answer 2

Hint: there are a few ways to proceed in factoring the left hand side. Some ideas:

• Factor out $\sin(\alpha/2)$
• Apply double angle formula for $\sin$
• Use Pythagorean identity on $\cos^2(\alpha/2)$. This will allow you to rewrite the left hand side as a polynomial over the variable $\sin(\alpha/2)$.

Answer 3

Excluding the trivial $a=\pi$, reset the problem as $$(t-r)=\frac{1-\sin\left(\frac{\alpha}{2}\right)}{4\sin\left(\frac{\alpha}{2}\right)\cos^3\left(\frac{\alpha}{2}\right) }=\frac{1}{4} \left(1-\sin \left(\frac{a}{2}\right)\right) \csc \left(\frac{a}{2}\right) \sec ^3\left(\frac{a}{2}\right)\tag 1$$

Let $a=2 \csc ^{-1}(x)$ and $(1)$ becomes $$(t-r)=\frac{x-1}{4 \left(1-\frac{1}{x^2}\right)^{3/2}}$$

The derivative of the rhs is $$\frac{x^2 \left(x^2+x-3\right)}{4 (x-1) (x+1)^2 \sqrt{x^2-1}}$$ So, in the given range, the zero of the first derivative corresponds to $$x^2+x-3=0 \implies x=\frac{\sqrt{13}-1}{2}\implies a=2 \csc ^{-1}\left(\frac{\sqrt{13}-1}{2} \right)$$

At this point, the rhs of $(1)$ $$\frac{\left(\sqrt{13}-3\right) \left(\sqrt{13}-1\right)^3}{8 \left(10-2 \sqrt{13}\right)^{3/2}} =0.287482$$ which is a minimum value.

So, if $(t-r)$ is greater than this value, there are two roots.