Solving Trigonometry Equations and Problems with Examples and Tips

Find m for which $\sin (x) + \cos (2 x) = m$ has solutions, no derivatives
Given $f : [0, 2 π] \to R, f (x) = \sin (x) + \cos (2 x)$ find the values of m for which the equations f(x)=m has solutions. The problem it is easy solvable with derivatives but it takes some time and careful computations to end up with the solution $[- 2, \frac{9}{8}]$

gabolzm6d 2022-04-12

How do you simplify and evaluate $\frac{{\tan 80}^{\circ} - {\tan 20}^{\circ}}{1 + {\tan 80}^{\circ} {\tan 20}^{\circ}}$

Magdalena Warren 2022-04-12

How do I prove
$\cos (4 A) = \cos^{4} A - 6 \cos^{2} A \sin^{2} A + \sin^{4} A$ ?

Jaylyn Villarreal 2022-04-12

Find the exact value of $\cos (11 \frac{π}{12})$

Julian Peck 2022-04-12

Find the domain of the following function:
$f (x) = \ln (2 \tan (3 x + \frac{π}{3}) - 2) .$
The first step that I done is that f(x) is more than 0.
Then I got to: $1 < \tan (3 x + \frac{π}{3})$

Annie Rice 2022-04-12

Find real part of $\frac{1}{1 - e^{i \frac{π}{7}}}$
How can you find
$R e (\frac{1}{1 - e^{i \frac{π}{7}}})$
I put it into wolframalpha and got $\frac{12}{}$ , but I have no idea where to begin. I though maybe we could use the fact that
$\frac{1}{z} = \frac{\overset{―}{z}}{{| z |}^{2}}$ ,
where $\overset{―}{z}$ is the conjugate of z. Unfortunately, the magnitude doesn't seem to be a nice number. I feel like this might be a trigonometry question in disguise, but converting $e^{i \frac{π}{7}} = \cos (\frac{π}{7}) + i \sin (\frac{π}{7})$ hasn't been very fruitful.

Aiyana Ayers 2022-04-12

Find $\frac{\tan 54 - \tan 2}{\frac{\sin 2}{\cos 6} + \frac{\sin 6}{\cos 18} + \frac{\sin 18}{\cos 54}}$

Bryanna Leon 2022-04-12

Find maximum of $f (x) = 3 \cos {2 x} + 4 \sin {2 x}$

Aliyah Mendez 2022-04-12

Find all values of x between 0 and 4 for which $\sin (2 x + 1) = \cos (2 x + 1)$
$\tan (2 x + 1) = 1$
Which I believe is correct.
Now the problem is , the range .
If the range is 0 to $2 π$ I would've known how to do but what about the 0 to 4 ?

Regina Ewing 2022-04-12

If $\sin (α) + \cos (α) = a,$ , denote $| \sin (α) - \cos (α) |$ in terms of a

Justine Webster 2022-04-12

Can someone explain why I get wrong answer in simplifying this expression?
$\sin \frac{11 π}{12} \sin \frac{29 π}{12} - \cos \frac{13 π}{12} \cos \frac{41 π}{12}$
If we rewrite the expression with new angles,
$\frac{29 π}{12} \to \frac{5 π}{12} and \frac{13 π}{12} \to \frac{11 π}{12}$
we don't change the value of the expression. But, if we now use sine of sum of angles, we get $\frac{1}{2}$ instead of 0.
Why does this happen? Do angles need to be in the same quadrant for the formula to work?

mikior9t3ue 2022-04-12

Verify Trigonometric Identities
$1 / \sin 50^{\circ} + \sqrt 3 / \cos 50^{\circ} = 4$

Elle Weber 2022-04-12

If $z = \cos θ + i \sin θ$ prove $\frac{z^{2} - 1}{z^{2} + 1} = i \tan θ$
Here is my workings, I'm not sure if I've made a mistake or I'm just not spotting what to do next. Any help would be appreciated.
$\frac{(\cos θ + i \sin θ)^{2} - 1}{(\cos θ + i \sin θ)^{2} + 1}$
$\frac{(\cos^{2} θ + 2 i \sin θ \cos θ - \sin^{2} θ) - 1}{(\cos^{2} θ + 2 i \sin θ \cos θ - \sin^{2} θ) + 1}$
$\frac{(\cos^{2} θ - \sin^{2} θ) + (2 i \sin θ \cos θ) - 1}{(\cos^{2} θ - \sin^{2} θ) + (2 i \sin θ \cos θ) + 1}$
$\frac{\cos 2 θ + i \sin 2 θ - 1}{\cos 2 θ + i \sin 2 θ + 1}$
I understand how I can do it with using $z = e^{i θ}$ , however I want to solve it using double angle identities.

junoonib89p4 2022-04-12

$\sin x + \cos x = \frac{1}{2}$
What is the value of $\tan x$ ? I tried using
$\sin 2 x = \frac{2 \tan x}{1 + \tan^{2} x}$
and
$\cos 2 x = \frac{1 - \tan^{2} x}{1 + \tan^{2} x}$
but we get a quadratic for $\tan (\frac{x}{2})$ . So any better approach would be much appreciated.

Carina Valenzuela 2022-04-12

How is $\sqrt{2 + \sqrt{2 + \sqrt{2 +}}} . . . n$ times = $2 \cos (π / 2^{n + 1})$ ?

Merati4tmjn 2022-04-12

If $A + B + C = 180^{\circ}$ , then prove that
$\sin^{3} A + \sin^{3} B + \sin^{3} C = 3 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2} + \cos \frac{3 A}{2} \cos \frac{3 B}{2} \cos \frac{3 C}{2}$

Lillianna Sandoval 2022-04-11

Use de-Moivre's theorem to find the reciprocal of each number below.
$\sqrt{3} - i$
Given $\sqrt{3} - i$ we need to find the reciprocal of it using de-Moivre's theorem.
$\frac{1}{\sqrt{3} - i}$
$= \frac{1 ({\cos 0}^{c} + i {\sin 0}^{c})}{2 (\frac{\sqrt{3}}{2} - \frac{i}{2})}$
$= \frac{1}{2} \cdot \frac{\cos (\frac{π}{6}) - i \sin (\frac{π}{6})}{(\frac{\sqrt{3}}{2} - \frac{i}{2})}$
$= \frac{1}{2} \cdot \cos (\frac{π}{6}) - i \sin (\frac{π}{6})$
$= \frac{1}{2} \cdot (\frac{\sqrt{3}}{2} + i \frac{1}{2})$
$= \frac{\sqrt{3}}{4} + \frac{i}{4}$

Breanna Fisher 2022-04-11

Understanding equality $\frac{1}{2} \int_{0}^{2 π} \frac{\cos (θ)}{2 + \cos (θ)} d θ = π - \int_{0}^{2 π} \frac{d θ}{2 + \cos (θ)}$

Perla Benitez 2022-04-11

Trigonometric equation $2 \sin 2 x + 3 = 2 \sin^{2} x$

daurialebraslmc 2022-04-11

To find the value of $\sqrt[3]{{\cos 40}^{\circ}}$ . Wolfram says it is a root of the cubic $8 x^{3} - 6 x + 1$ but how to prove that. Note that
${(e^{i 2 π})}^{\frac{1}{9}} = {(\cos 2 π + i \sin 2 π)}^{\frac{1}{9}} = {(1)}^{\frac{1}{9}}$
Can anything be concluded from this?

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