How to solve this equation? $a^2 \cdot \arcsin(\frac 4a)+4 \cdot\sqrt {a^2-16}=40$ Please, help me solve this equation:
$$a^2 \cdot \arcsin \left(\frac 4a \right)+4 \cdot\sqrt {a^2-16}=40$$
 A: Say $\frac{4}{a}=u$
Then the given equation becomes
$$\arcsin u + u\sqrt{1-u^2}=\frac{5}{2}u^2$$
Now $u$ must be such that $|u|<1$.
So we can assume that $u=\sin x$ or $\cos x$.
Now see if you can solve this trigonometric equation.
A: There may not exist an explicit solution for $a$ in terms of elementary functions.
However, you can apply the Newton-Raphson Method (A numerical method) using a spreadsheet, or with more sophisticated software such as MATLAB:
$$a_{n+1}=a_n-\frac{f(a_n)}{f'(a_n)} \tag{1}$$
You have your function:
$$f(a)=a^2 \cdot \arcsin \left(\frac 4a \right)+4 \cdot\sqrt {a^2-16}-40$$
Differentiating it:
$$f'(a)=\frac{4a}{\sqrt{a^2-16}}-\frac{4}{\sqrt{1-\frac{16}{a^2}}}+2a\arcsin\left(\frac{4}{a}\right)$$
Substituting into $(1)$:
$$a_{n+1}=a_n-\frac{a_n^2 \cdot \arcsin \left(\frac {4}{a_n} \right)+4 \cdot\sqrt {a_n^2-16}-40}{\frac{4a_n}{\sqrt{a_n^2-16}}-\frac{4}{\sqrt{1-\frac{16}{a_n^2}}}+2a_n\arcsin\left(\frac{4}{a_n}\right)} \tag{2}$$
Let us start with an initial value $a_0=11$. You can choose these values by simply 'guessing' a value close to the actual solution.
Now, substitute the value of $a_0$ for $a_n$, and find the value of $a_1$ using equation $(2)$. Continue this iteration until you get closer and closer to the value of $a$.
I get a value of $a\approx 5.52978022$.
