Solving Trigonometry Equations and Problems with Examples and Tips

Solve the equation:
$\tan (2 x + 1.426) = - 2.156$

Nathanael Hansen 2022-03-20

Solve the equation:
$\int_{0}^{\frac{π}{4}} \sec^{4} θ \tan^{4} θ d θ$

Zack Mora 2022-03-20

Solve the equation with trigonometry
$\frac{2 \cos x}{(1 + \cos 2 x)} = \frac{(1 - \cos 2 x)}{\sin 2 x}$

calcolare45pj 2022-03-20

Solve the equation
$\sin^{3} x + \sin^{3} (\frac{2 π}{3} + x) + \sin^{3} (\frac{4 π}{3} + x) + \frac{3}{4} \cos 2 x = 0$

ashes86047xhz 2022-03-20

Solve $y^{3} - 3 y - \sqrt{2} = 0$ using trigonometry.

sexoagotadorogyr 2022-03-20

Solve $\tan (θ) + \tan (2 θ) = \tan (3 θ)$

Elisha Frost 2022-03-19

Closed form of $\sum_{n = 1}^{N} \cos^{2} (n x)$
From the Lagrange identity we have:
$\sum_{n = 0}^{N} \cos (n x) = \frac{1}{2} + \frac{\sin [(2 N + 1) \frac{x}{2}]}{2 \sin \frac{x}{2}}$
Mathematica tells me that it is:
$\sum_{n = 1}^{N} \cos^{2} (n x) = \frac{N}{2} + \frac{\sin N x \cos (N x + x)}{2 \sin x}$
I actually want to see the process. If you could include the derivation of the original identity, that would be appreciated!
Also, am I correct to say that:
$\sum_{k = 0}^{n} \cos (k x) = \sum_{k = 1}^{n} \cos (k x) + 1$

poplavileqlo 2022-03-19

Can I factorize ${\sin θ}_{1} {\cos θ}_{1} + {\sin θ}_{2} {\cos θ}_{2}$ into $\sin (θ_{1} + θ_{2}) \cos (θ_{1} - θ_{2})$ ?

Adrien Wong 2022-03-19

Calculate x, if
$\tan (x) = \tan 9 \tan 69 \tan 33$

WesDiectstemiwxg 2022-03-19

By using euler formula,one can obtain:
$2 \sin (\frac{π}{180}) = {({(- 1)}^{\frac{1}{180}})}^{89} - {({(- 1)}^{\frac{1}{180}})}^{91}$

Shyla Singleton 2022-03-18

Can you help me prove: $\frac{y}{\sqrt{y^{2} + x}} = arccot \frac{y}{\sqrt{x}}$ ?

Damaris Huffman 2022-03-18

Can I simplify $n \cdot r \cdot \sin (90^{\circ} - \frac{180^{\circ}}{n}) \sqrt{r^{2} - r^{2} \sin^{2} (90^{\circ} - \frac{180^{\circ}}{n})}$ further?

pinka1hf 2022-03-18

Calculate $\cos (\frac{2 π}{5})$ in terms of $\sin (\frac{π}{5})$
I intend to calculate $\cos (\frac{2 π}{5})$ via the formula
$\sin (\frac{π}{5}) = \sqrt{\frac{1 - \cos (\frac{2 π}{5})}{2}}$
This could be expressed as:
$\sqrt{\frac{5 - \sqrt{5}}{8}} = \sqrt{\frac{1 - \cos (\frac{2 π}{5})}{2}}$ . Squaring both sides and manipulate a bit, this will give $\cos (\frac{2 π}{5}) = - \frac{1 - \sqrt{5}}{4}$ . However, the true result is $\frac{\sqrt{5} - 1}{4}$ . What is wrong with my derivation?

Charity Barr 2022-03-17

Alternative proof of
$\cos^{6} θ + \sin^{6} θ = \frac{1}{8} {(5 + 3 \cos 4 θ)$

fellesturduj 2022-03-17

Alternative form of $y = \frac{1}{1 - {(\frac{x}{a})}^{2}}$ ?
I solved the differential equation $y^{'} (x) = \frac{2}{a^{2}} x y^{2} (x)$ (with y(0)=1 and $x \in (- | a |, | a |)$ ) and came to
$y (x) = \frac{1}{1 - {(\frac{x}{a})}^{2}} .$
This looks like some trigonometric thing to me. Do you know a better form for y?

ohmaig00dnessyb3 2022-03-17

A question on range of trigonometric functions.
If $u = {\sqrt{a^{2} \cos^{2} α + b^{2} \sin^{2} α}} + {\sqrt{a^{2} \sin^{2} α + b^{2} \cos^{2} α}}$ , find the difference between the maximum and minimum value of $u^{2}$
I tried squaring the expression on both sides but i am ending up with some complex expression, i.e, $a^{2} + b^{2} + 2 \sin α \cos α \sqrt{{(a^{2} - b^{2})}^{2} + \frac{a^{2} b^{2}}{\sin^{2} α \cos^{2} α}}$

evighetl2b 2022-03-17

What is the maximum value of k so that $x^{2} + 4 (\sin^{2} {x} \tan^{2} {x} + \cos^{2} {x} \cot^{2} {x} + k^{2} - x \sec {x} \csc {x}) = 0$ has real roots.

WesDiectstemiwxg 2022-03-17

Check the proof that $f (x) = \cos ({x}) \cdot \cos ({\sqrt{3} x})$ is not periodic
If the function is periodic then:
$f (x) = f (x + T)$
$\cos (x) \cdot \cos (\sqrt{3} x) = \cos (x + T) \cdot \cos (\sqrt{3} (x + T))$
Consider the function at 0:
$\cos (T) \cdot \cos (\sqrt{3} T) = 1$
But this equation has only one solution at T=0 which contradicts the initial assumption that there exists a positive period.
Or the other way:
$\cos (T) \cdot \cos (\sqrt{3} T) = 1$
Let $\cos (T) = 1 and \cos (\sqrt{3} T) = 1$ , hence
$T = 2 π m$
$\sqrt{3} T = 2 π n$
Substituting T in the second equation gives:
$2 π m \sqrt{3} = 2 π n$
$\sqrt{3} = \frac{n}{m}$
But $m, n \in N$ and $\sqrt{3} \in {R ∖ Q}$ which gives a contradiction.

voordragh2 2022-03-17

Choose $y \in R$ so that $\sin y = \frac{u}{\sqrt{u^{2} + v^{2}}}$ and $\cos y = \frac{v}{\sqrt{u^{2} + v^{2}}}$
I know the identity $\sin^{2} y + \cos^{2} y = 1$ . Also, I notice that
${(\frac{u}{\sqrt{u^{2} + v^{2}}})}^{2} + {(\frac{v}{\sqrt{u^{2} + v^{2}}})}^{2} = 1$
without u,v vanishing simultaneously. But I am not sure whether $\exists y \in R$ s.t. $\sin y = \frac{u}{\sqrt{u^{2} + v^{2}}}$ and $\cos y = \frac{v}{\sqrt{u^{2} + v^{2}}}$ .Does anyone have an idea?

afadimoz4 2022-03-17

Calculate:
$\tan (α + 15^{\circ}) \cdot {\cot 165}^{\circ}$ , if
${\sin 75}^{\circ} \cdot \sin (α + 15^{\circ}) = 2 \sin α$

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