Precalculus Solutions | Expert Assistance for Problems & Examples

Calculation of $\int_{0}^{\frac{π}{4}} {(\tan (x))}^{n} d x$

deiteresfp 2022-01-16

Showing that $\frac{1 - \sin \frac{5 π}{18}}{\sqrt{3} \sin \frac{5 π}{18}} = \tan \frac{π}{18}$

Margie Marx 2022-01-16

$\sin 2 x - \sin^{2} x = 20 \sin x + 10 \cos x - 1$ with $90 \leq x \leq 180$

Vikolers6 2022-01-16

Find a limit $lim_{x \to 0} \int_{0}^{x} \frac{\cos (t^{3})}{t + x} d t$

Kaspaueru2 2022-01-16

Solving ${\frac{58}{\cot 36}}^{\circ} = \cos^{3} x$ without substituting the trig values for $36^{\circ}$

eiraszero11cu 2022-01-16

Simplifying $\frac{\sin x + \sin x \tan^{2} x}{\tan x}$ to $\sec x$

tearstreakdl 2022-01-16

$2^{\sin (x) + \cos (y)} = 1, 16^{\sin^{2} (x) + \cos^{2} (y)} = 4$ (system of equations)

Ben Shaver 2022-01-16

Evaluate $\int_{0}^{\frac{π}{2}} \frac{1}{a \sin^{2} x + b \cos^{2} x} d x$

Danelle Albright 2022-01-16

Simplifying $\sin h (2 \arctan h (e^{H x}))$

Dowqueuestbew1j 2022-01-16

Questions about proving $lim_{x \to 0} \frac{\sin x}{x} = 1$

Gwendolyn Willett 2022-01-16

Solving $\cos (z) = \frac{52}{}$
and I'm trying to solve for z but I keep going in circles. I know $\cos z = \frac{52}{=} \frac{12}{e^{i z} + e^{- i z}}$ so then $e^{i z} + e^{- i z} = 5$ but then I'm stuck

Irrerbthist6n 2022-01-16

I want to solve this equation
$\arccos (x) + \arcsin (x^{2} - x + 1) = \frac{π}{2}$
I write this: for all $x \in [- 1, 1]$
$\arcsin (x^{2} - x + 1) = \frac{π}{2} - \arccos (x)$ then $x^{2} - x + 1 = \sin (\frac{π}{2} - \arccos (x)) = \cos (\arccos (x)) = x$
$x^{2} - x + 1 = x \Rightarrow x^{2} - 2 x + 1 = 0 \Rightarrow {(x - 1)}^{2} = 0 \Rightarrow x = 1$
Is it true ?

Terrie Lang 2022-01-16

If $\cos 17 x = f (\cos x)$ , then show that $\sin 17 x = f (\sin x)$

Harold Kessler 2022-01-16

Analytically solving $\frac{1}{\sin 2 x} + \frac{1}{\sin 3 x} = \frac{1}{\sin x}$

keche0b 2022-01-16

Integrate from 0 to $2 π$ with respect to $θ$ the following ${(\sin θ + \cos θ)}^{n}$

William Boggs 2022-01-16

Showing $\frac{d θ}{d \tan θ} = \frac{1}{1 + \tan^{2} θ}$
I suppose that
$\frac{d θ}{d \tan θ} = \frac{d \arctan x}{d x} = \frac{1}{1 + x^{2}} = \frac{1}{1 + \tan^{2} θ}$
So is
$\frac{d θ}{d \tan θ} = \frac{1}{1 + \tan^{2} θ}$
correct? And
$\frac{d θ}{d \tan \frac{θ}{2}} = \frac{2}{1 + \tan^{2} \frac{θ}{2}}$ ?

pogonofor9z 2022-01-16

Solutions of $e^{x} - 1 - k \cdot \arctan x = 0$ ,
consider $h (x) = e^{x} - 1 - k \cdot \arctan x$ then find on which condition on k there will be two solutions for h(x)=0 (k is real).

David Young 2022-01-16

The derivative of $f (x) = \frac{3 \sin x}{2 + \cos x}$
My solution:
The background
$3 \frac{d}{d x} (\frac{\sin x}{2 + \cos (x)})$
$= 3 \frac{\frac{d}{d x} (\sin (x)) (2 + \cos (x)) - \frac{d}{d x} (2 + \cos (x)) \sin (x)}{{(2 + \cos (x))}^{2}}$
$= 3 \frac{\cos (x) (2 + \cos (x)) - (- \sin (x)) \sin x}{{(2 + \cos (x))}^{2}}$
$= \frac{3 + 6 \cos (x)}{{(2 + \cos (x))}^{2}}$
However, I plugged $f (x) = \frac{3 \sin x}{2 + \cos x}$ into the derivative calculator in wolfram alpha and received the following calculation:
$\frac{3 (\sin^{2} (x) + \cos^{2} (x) + 2 \cos (x))}{{(2 + \cos (x))}^{2}}$
Is my solution incorrect and the one from wolfram alpha correct?

Lennie Davis 2022-01-16

Simplify $\frac{4 \cos^{2} (2 x) - 4 \cos^{2} (x) + 3 \sin^{2} (x)}{4 \sin^{2} (x) - \sin^{2} (2 x)}$

Francisca Rodden 2022-01-16

Prove that $\tan^{- 1} \frac{\sqrt{1 + x^{2}} + \sqrt{1 - x^{2}}}{\sqrt{1 + x^{2}} - \sqrt{1 - x^{2}}} = \frac{π}{4} + \frac{1}{2} \cos^{- 1} x^{2}$ ?
Let the above expression be equal to $ϕ$
$\frac{\tan ϕ + 1}{\tan ϕ - 1} = \sqrt{\frac{1 + x^{2}}{1 - x^{2}}}$
$\frac{1 + \tan^{2} ϕ + 2 \tan ϕ}{1 + \tan^{2} ϕ - 2 \tan ϕ} = \frac{1 + x^{2}}{1 - x^{2}}$
$\frac{1 + \tan^{2} ϕ}{2 \tan ϕ} = \frac{1}{x^{2}}$
$\sin 2 ϕ = x^{2}$
$ϕ = \frac{π}{4} - \frac{1}{2} \cos^{- 1} x^{2}$
Where am I going wrong?

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