Solving Trigonometry Equations and Problems with Examples and Tips

Equation $x = \sqrt{2 + \sqrt{2 - \sqrt{2 + x}}}$
Using $x = 2 \cos (θ)$ we get $2 \cos (θ) = \sqrt{2 + \sqrt{2 - \sqrt{2 + 2 \cos θ}}}$
$2 \cos θ = \sqrt{2 + \sqrt{2 - 2 \cos \frac{θ}{2}}}$
$2 \cos θ = \sqrt{2 + 2 \sin \frac{θ}{4}}$
$4 \cos^{2} θ = 2 + 2 \sin \frac{θ}{4}$
After this step I am not able to solve it

crealolobk 2022-01-14

Can you solve $\sin x = \sin (\frac{x}{2})$ without a sum/double/half angle identity?

William Cleghorn 2022-01-14

Prove $\tan (\sin^{- 1} t) = \frac{t}{\sqrt{1 - t^{2}}}$

Kathleen Rausch 2022-01-14

Limit of $\sin (\sqrt{x + 2}) - \sin (\sqrt{x + 4}) as x \to \infty$

Russell Gillen 2022-01-14

How to prove $\sin (x - a) + \sin (x + a) = 2 \sin (x) \cos (a)$ ?

Karen Simpson 2022-01-14

$\cos 5 = \sqrt{1 - a^{2}}$ which statement is true?
1) $\sin 20 = 4 a \sqrt{1 - a^{2}} \sqrt{1 - 4 a^{2}}$
2) $\tan 50 = \frac{1 - a}{1 + a}$
3) $\tan 10 = \frac{2}{a \sqrt{1 - a^{2}}}$
4) $\cos 65 = \frac{12}{\sqrt{1 - a^{2}} - \sqrt{3} a}$

Walter Clyburn 2022-01-14

If $\frac{a + b}{\csc x} = \frac{a - b}{\cot x}$ show that $\csc x \cot x = \frac{a^{2} - b^{2}}{4 a b}$

jamessinatraaa 2022-01-14

Find minimum $f (x) = \sin (x) + \cos (x) + \sin (2 x) + 1$ for $x \in [0, 2 π]$

David Lewis 2022-01-14

Verify $\cos (x) = \frac{1 - t^{2}}{1 + t^{2}}$ with $t = \tan (\frac{x}{2})$

William Curry 2022-01-14

Prove $\cos x + \cos y = 2 \cos (\frac{x + y}{2}) \cos (\frac{x - y}{2})$

Michael Maggard 2022-01-14

How to show f′(0) exists for $f (x) = e^{\frac{- 1}{x^{2}}} \sin (\frac{1}{x})$ for $x \neq 0$ and f(0)=0?

b2sonicxh 2022-01-03

Can you please show me the process that determines $\int_{- 1}^{1} \frac{d x}{\sqrt{- x^{2} + 1}} = π$ ? Unfortunately I only know that the derivative of $\sin^{- 1} (x)$ is equal to the integrand of that, but I don't actually know what to do with it.

Bobbie Comstock 2022-01-03

Prove and compute the limit of this function.
$\lim_{x \to 0^{+}} \frac{\sin (6 x)}{\sqrt{\sin (2 x)}}$
Trying converting it into different functions like $\cos (\frac{π}{2} - 2 x)$ or multiplying by the inverse function and so on, but it keep getting back to $0 / 0$ .

elvishwitchxyp 2022-01-02

Find, in radians the general solution of $\cos 3 x = \sin 5 x$
I have said, $\sin 5 x = \cos (\frac{π}{2} - 5 x)$
so
$\cos 3 x = \sin 5 x \Rightarrow 3 x = 2 n π \pm (\frac{π}{2} - 5 x)$
When I add $(\frac{π}{2} - 5 x) to 2 n π$ I get the answer $x = \frac{π}{16} (4 n + 1)$ , which the book says is correct.
But when I subtract I get a different answer to the book. My working is as follows:
$3 x = 2 n π - \frac{π}{2} + 5 x$
$2 x = \frac{π}{2} - 2 n π$
$x = \frac{π}{4} - n π = \frac{π}{4} (1 - 4 n)$

Donald Johnson 2022-01-02

$\frac{1}{2} = \frac{\sin^{2} x}{\tan x}$
What would be my next step to get to $x = 45 °$ ?

Mabel Breault 2022-01-02

How could I approach
$lim_{n \to \infty} {(\frac{1 + \cos (\frac{1}{2^{n}})}{2})}^{n}$
I tried several things to evaluate it, namely looking at it as ${\cos (\frac{1}{2^{n + 1}})}^{2 n}$ instead or as $\exp (2 n \cdot \ln (\cos (\frac{1}{2^{n + 1}}))$ and then trying to show that the limit of $n \cdot \ln (\cos (\frac{1}{2^{n + 1}}))$ is 0 (for example using LHopitals rule), but I havent

Mary Reyes 2022-01-01

How am I supposed to expand $\sin^{2} A + \sin^{4} A = 1$ into $1 + \sin^{2} A = \tan^{2} A$ ?

Terrie Lang 2022-01-01

The general solution of $| \sin x | = \cos x$ is -
(A) $2 n π + \frac{π}{4}, n \in I$
(B) $2 n π \pm \frac{π}{4}, n \in I$
(C) $n π + \frac{π}{4}, n \in I$
(D) None of these
So what I did was - I made a case for when $\sin x$ is greater than 0 and equated it to $\cos x$ to get $\tan x = 1$ which implies $x = \frac{π}{4}$ . The other case was when $\cos x = - \sin x$ . Here, $x = \frac{3 π}{4}$ . I don't understand how to proceed from here.

PEEWSRIGWETRYqx 2022-01-01

To me the most common identity that comes to mind that results in 1 is the trigonometric sum of squared cosine and sine of an angle:
$\cos^{2} θ + \sin^{θ} = 1$ (1)
and maybe
$- e^{i π} = 1$ (2)
Are there other famous (as in commonly used) identities that yield 1 in particular?

lugreget9 2022-01-01

I was trying to find the least value of $| \sin x | + | \cos x |$ and applied the above inequality as :
$| \sin x + \cos x | \leq | \sin x | + | \cos x |$
$\Rightarrow \sqrt{2} \leq | \sin x | + | \cos x |$
But the range of the given function is $[1, \sqrt{2}]$ .

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