MHB Can anyone simplify any of these questions?

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The discussion revolves around simplifying a series of mathematical equations, particularly focusing on a geometric series and a quadratic equation derived from trigonometric identities. The first equation presented is related to the secant function, leading to a quadratic in terms of tangent. Participants are encouraged to solve the quadratic equation for tangent, which is formulated as tan²θ - 2√3 tanθ + 3 = 0. The thread emphasizes the importance of breaking down complex problems into manageable parts for clarity and understanding. The overall goal is to simplify and solve the given mathematical challenges effectively.
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Can anyone simplify any of these questions?
 

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Hello hmhmherath and welcome to MHB! :D

As per forum rules, in future please post at most two problems per thread.

Let's take these one at a time. For the first one (problem 11), do you know the formula for the sum of a geometric series?
 
hmhmherath said:
Can anyone simplify any of these questions?

$\displaystyle \begin{align*} \frac{1}{\cos^2{(x)}} &= 2\,\sqrt{3}\tan{(x)}- 2 \\ 1 &= 2\,\sqrt{3} \sin{(x)}\cos{(x)} - 2\cos^2{(x)} \\ 1 &= \sqrt{3}\sin{(2\,x)} - 1 - \cos{(2\,x)} \\ 2 + 2\cos{(2\,x)} &= \sqrt{3}\sin{(2\,x)} \\ 4 + 8\cos{(2\,x)} + 4\cos^2{(2\,x)} &= 3 \sin^2{(2\,x)} \\ 4 + 8\cos{(2\,x)} + 4\cos^2{(2\,x)} &= 3 - 3\cos^2{(2\,x)} \\ 7\cos^2{(2\,x)} + 8\cos{(2\,x)} + 1 &= 0 \end{align*}$

Now solve the resulting quadratic.
 
$$\frac{1}{\cos^2\theta}=2\sqrt3\tan\theta-2$$

$$\sec^2\theta=2\sqrt3\tan\theta-2$$

$$\tan^2\theta+1=2\sqrt3\tan\theta-2$$

$$\tan^2\theta-2\sqrt3\tan\theta+3=0$$

Can you solve that quadratic for $\tan\theta$ ?
 

Thread 'Proving that convexity implies second order derivative being positive'

There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...
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