problem solvingtrigonometry
Find $\theta$ on $[0, 2\pi)$ such that
$$\cos{\theta}^{\sin{\theta}^{\cos{\theta}^{\dots}}} = 2 + 2\sec^2{\theta}\tan^2{\theta} – \sec^4{\theta} – \tan^4{\theta}$$
I'm not sure on how to tackle this problem. I've never really dealt with exponentiation of trig functions. Any help is helpful. Thank you very much
Yes, there are nice geometric explanations of the derivative formulas for all six basic trig functions, which ought to be much more widely known. (I hesitate to use the word "proof" for an argument that uses infinitesimals.) They are all based on the following fact about isosceles triangles:
For an isosceles triangle with small vertex angle $d\theta$, the base length $ds$ satisfies $$ ds \;\approx\; r\,d\theta $$ where $r$ is the length of the legs.
As you already pointed out, there is a nice geometric explanation of the derivative formulas for $\sin \theta$ and $\cos \theta$ that uses this fact. The following picture shows a right triangle version of the explanation, as opposed to the unit circle version you gave above:
The following picture shows a geometric explanation of the derivative formulas for $\sec \theta$ and $\tan \theta$, again using right triangles.
Finally, the following picture shows a geometric explanation of the derivative formulas for $\csc \theta$ and $\cot \theta$.
Use,
$$\sin^2(\theta)+\cos^2(\theta)=1$$
Dividing both sides by $\cos^2 (\theta)$ gives:
$$\tan ^2 (\theta)+1=\sec^2 (\theta)$$
This should be enough to conclude.
$$x+1=y^2$$
Best Answer
$$2 + 2\sec^2{\theta}\tan^2{\theta} - \sec^4{\theta} - \tan^4{\theta}=2+\frac{2*\sin^2\theta}{\cos^4\theta}-\frac{1}{\cos^4\theta}-\frac{\sin^4\theta}{\cos^4\theta}=\frac{2\cos^4\theta-(\sin^4\theta-2\sin^2\theta+1)}{\cos^4\theta}=\frac{2\cos^4\theta-(\sin^2\theta-1)^2}{\cos^4\theta}=\frac{2\cos^4\theta-(\cos^2\theta)^2}{\cos^4\theta}=1$$ So the whole exponent has to be 0 which is easy to tackle with(I guess you don't need help with that)