Solving Trigonometry Equations and Problems with Examples and Tips

Determine the smallest positive value of x(in degrees) for which:
$\tan (x + 100^{\circ}) = \tan (x + 50^{\circ}) \tan (x) \tan (x - 50^{\circ})$

Nancy Richmond 2022-03-28

Derivative of $f (x) = \arctan (\sqrt{\frac{1 + x}{1 - x}})$
I am trying to find the derivative of $f (x) = \arctan (\sqrt{\frac{1 + x}{1 - x}})$ by only using the formula ${\arctan (u (x))}^{'} = \frac{u^{'} (x)}{u {(x)}^{2} + 1}$ . I don't honestly understand this formula, here are however my calculations. Is this the right way to find the derivative? And how do I proceed?
$\frac{\frac{1}{2 \sqrt{\frac{1 + x}{1 - x}}} \cdot \frac{(1 - x) - (1 + x)}{x^{2} - 2 x + 1}}{\frac{x + 1}{1 - x} + 1}$
I also appreciate if you can explain what that "formula" exactly is.

talpajocotefnf3 2022-03-28

Computing $\int_{0}^{2 π} {(1 + 2 \cos t)}^{n} \cos n t d t$

serpebm2r 2022-03-28

$18 \sec^{2} θ - 16 \tan θ \sec θ - 15 = 0$ Find all solutions in the interval $0^{\circ} \leq θ < 360^{\circ}$

Sage Fry 2022-03-28

$\tan (40^{\circ}) + 4 \tan (10^{\circ}) = \cot (x^{\circ})$ Find x
My Try:
we have $\tan x - \cot x = - 2 \cot 2 x$
Subtracted both sides with $\cot (40)$ we get
$\tan 40 - \cot 40 + 4 \tan 10 = \cot x - \cot 40$
that is
$- 2 \cot 80 + 4 \tan 10 = \cot x - \cot 40$
$\Rightarrow$
$2 \tan 10 + \cot 40 = \cot x$

dotzis16xd 2022-03-28

$\frac{1}{\sqrt{3}} \sum_{r = 0}^{4} \tan (\frac{π}{15} + \frac{r π}{5}) = ?$
Then evaluation of k
solution i try
$\tan (12^{\circ}) + \tan (48^{\circ}) + \tan (84^{\circ}) + \tan (120^{\circ}) + \tan (156^{\circ})$
$= \tan (12^{\circ}) + \tan (48^{\circ}) + \tan (84^{\circ}) + \tan (156^{\circ}) + \sqrt{3}$
$= \frac{\sin (60^{\circ})}{\cos (12^{\circ}) \cos (48^{\circ})} + \frac{\sin (240^{\circ})}{\cos (84^{\circ}) \cos (156^{\circ})} - \sqrt{3}$

Dexter Odom 2022-03-28

$- \sin x - \cos x = - \sqrt{2} \sin (x + \frac{π}{4})$
How does the cosine disappear and how did $\sin x$ turn into $\sin (x + \frac{π}{4})$

michiiiiiakqm 2022-03-28

proving that $lim_{x \to π} \frac{1 + \cos x}{1 + \cos 3 x} = \frac{1}{9}$ without L'Hôpital

svrstanojpkqx 2022-03-28

Proving $\sum_{k = 1}^{n} | \cos (k) | \geq \frac{n}{4}$ for all n>0
I can do it using the fact that $| \cos (x) | \geq {\cos (x)}^{2}$
$\sum_{k = 1}^{n} | \cos (k) | \geq \sum_{k = 1}^{n} {\cos (k)}^{2} = \frac{n}{2} + \frac{1}{2} R e (e^{2 i} \frac{1 - e^{2 \in}}{1 - e^{2 i}})$
$= \frac{n}{2} + \frac{1}{2} R e (e^{i (n + 1)} \frac{\sin (n)}{\sin 1}) = \frac{n}{2} + \frac{\cos (n + 1) \sin (n)}{2 \sin (1)} \geq \frac{n}{2} - \frac{1}{2 \sin (1)}$

Dominique Pace 2022-03-28

Proving that if $x > 0$ then $\frac{x - (x^{2} + 1) \arctan (x)}{x^{2} (x^{2} + 1)}$ is less than 0

jisu61hbke 2022-03-28

Proving
$\sin ((n - \frac{1}{2}) ϕ) + \sin (\frac{ϕ}{2}) = \sin (\frac{n + 1}{2} ϕ)$

Pizzadililehz 2022-03-28

Proving $\frac{\cos 2 x}{1 + \sin 2 x} = \sec 2 x - \tan 2 x$
I tried
$\frac{1 - \sin^{2} x}{1 + 2 \sin x \cos x} - \frac{1 - \cos^{2} x}{1 + 2 \sin x \cos x}$
but I couldn't simplify. I start with LHS.

justbethlflql 2022-03-28

Proving $lim_{x \to \frac{π}{2^{-}}} \tan {x} = \infty$ , using $ϵ - δ$
I would like to prove the following:
$lim_{x \to \frac{π}{2^{-}}} \tan {x} = \infty$
By the $ϵ - δ$ definition, for any constant $K \in R$ there is some $δ > 0$ such that if $x \in (\frac{π}{2} - δ, \frac{π}{2})$ then $\tan (x) > K$ . What I did was:
$x > \arctan {K}$
Also, we know that $\forall x \in R - \frac{π}{2} < \arctan {x} < \frac{π}{2}$ , but I can't find out, how to choose the $δ$ .

loraliyeruxi 2022-03-28

Prove:
$\cos (3 θ) = 4 \cos^{3} (θ) - 3 \cos (θ)$

Marzadri9lyy 2022-03-28

Prove:
$\cot x = \sin x \sin (\frac{π}{2} - x) + \cos^{2} x \cot x$

diartrosiwk8o 2022-03-28

Prove the inequality:
$(1 + \frac{1}{\sin a}) (1 + \frac{1}{\cos a}) \geq 3 + 2 \sqrt{2}$

Deegan Chase 2022-03-28

Prove this inequality
$\sum \cos A \geq \frac{1}{4} (3 + \sum \cos (A - B))$

abitinomaq1 2022-03-28

Prove that series with ${(- 1)}^{k} \cos^{3} (k)$ has sequence of partial sums bounded.

ropowiec2gkc 2022-03-28

Prove that $arccot \frac{1 - x \sin ϕ}{x \cos ϕ} - arccot \frac{\cos ϕ}{x - \sin ϕ} = ϕ$
I used the formula $\cot (α - β) = \frac{\cot α \cot β + 1}{\cot α - \cot β}$ with $α = arccot \frac{1 - x \sin ϕ}{x \cos ϕ}$ and $β = arccot \frac{\cos ϕ}{x - \sin ϕ}$ .
What I got is as it follows:
$\frac{\frac{1 - x \sin ϕ}{x \cos ϕ} . \frac{\cos ϕ}{x - \sin ϕ} + 1}{\frac{1 - x \sin ϕ}{x \cos ϕ} - \frac{\cos ϕ}{x - \sin ϕ}}$
And after some calculations I got that it's equal to:
$\frac{(1 - x \sin ϕ + x - \sin ϕ) \cos ϕ}{x - \sin ϕ - x^{2} \sin ϕ + x \sin^{2} ϕ - x \cos^{2} ϕ}$

Pasegeabe85xy 2022-03-28

Prove that $\tan (6^{\circ}) \tan (42^{\circ}) = \tan (12^{\circ}) \tan (24^{\circ})$

The majority of Trigonometry Math problems are quite easy to solve when you have the list of answers that help you see the logic and reasoning. Looking at various examples will help you to see how certain questions are related to provided solutions.

Remember that Trigonometry requires patience and analysis. By following your analysis skills, you will help yourself find the answers. If something remains unclear, start with your exploration of examples again and go step-by-step.

Precalculus

Practice Exercises for Trigonometry