Precalculus Solutions | Expert Assistance for Problems & Examples

Evaluating $lim_{x \to \infty} \frac{1}{x} \int_{0}^{x} | \sin (t) | d t$

Yahir Haas 2022-01-29

If the sides of a triangle are in Arithmetic progression and the greatest and smallest angles are X and Y, then show that
$4 (1 - \cos X) (1 - \cos Y) = \cos X + \cos Y$
I tried using sine rule but can't solve it.

meteraiqn 2022-01-29

Evaluation of $lim_{x \to 0} \frac{\sin (x + a) - \sin (a)}{\sin 2 x}$
I have got as far as below using $\sin a + \sin b$ in numerator and $\sin 2 a$ in the denominator but am not sure how to expand further, or if these are the right rules to apply.
$lim_{x \to 0} \frac{\sin (x) \cos (a) + \sin (a) [\cos (x) - 1]}{2 \sin (x) \cos (x)}$
I need to simplify it down to apply $lim_{x \to 0} \frac{\sin (x)}{x}$

Mlejd5 2022-01-29

Evaluating $lim_{n \to \infty} \frac{1}{n} \int_{0}^{\frac{π}{2}} \frac{\sin^{2} n x}{\sin^{2} x} f (x) d x$ for continuous f

Mlejd5 2022-01-29

Prove that $1 + \frac{2}{3!} + \frac{3}{5!} + \frac{4}{7!} + \dots = \frac{e}{2}$

kuntungw3 2022-01-29

Evaluating $\int \cot x \csc^{2} x d x$ with $u = \cot x$
Differentiating both sides of u, then making the substitution:
$u = \cot x$
$d u = - \cot x \csc x d x,$
$d x = - \frac{d u}{u \csc x}$
$\int - \frac{u \csc^{2} x d u}{u \csc x} = \int - \csc x d u$
Apparently, this was not an adequate approach, because x is still part of the integrand. What should be done instead?

arrebolyt 2022-01-29

How can I get the sum of two arctangents that is not restricted to the range of $(- π, π]$ ?
$\arctan (\frac{y_{1}}{x_{1}}) + \arctan (\frac{y_{2}}{x_{2}})$
I use $\arctan 2 (y, x)$ function for calculating the above arctangent function. I know that there is this
$\arctan u + \arctan v = \arctan (\frac{u + v}{1 - u v})$
but it does not return the correct value for any x and y.

wingtipti 2022-01-29

Compute $\int \sin (x) (\frac{1}{\cos (x) + \sin (x)} + \frac{1}{\cos (x) - \sin (x)}) d x$

Wendy Gutierrez 2022-01-29

Express x in terms of a and b: $\sin^{- 1} {\frac{2 a}{1 + a^{2}}} + \sin^{- 1} {\frac{2 b}{1 + b^{2}}} = 2 \tan^{- 1} x$
I tried to expand the LHS using the formula
$\sin^{- 1} c + \sin^{- 1} d = \sin^{- 1} (c \sqrt{1 + d^{2}} + d \sqrt{1 + c^{2}})$
But it didn't work out. Could anyone help me out?

Alvin Pugh 2022-01-29

If $\sin (β + γ) = \sin ϵ$ and $\cos β \cos γ < \sin β \sin γ$ then $β + γ = π - ϵ$
I have the following system:
$\sin β, \cos γ + \cos β, \sin γ = \sin ϵ$
$\cos β, \cos γ < \sin β, \sin γ$
In other terms:
$\sin (β + γ) = \sin ϵ$
$\cos β, \cos γ < \sin β, \sin γ$
Can you explain me why $β + γ = π - ϵ$ please?

Jayleen Sanders 2022-01-29

Atypical way to find angle between unit vectors: $θ = 2 \sin^{- 1} (\frac{1}{2} | \hat{A} - \hat{B} |)$
At my work, I have come across code with the following way of calculating the angle between two vectors.
$θ = 2 \sin^{- 1} (\frac{1}{2} | \hat{A} - \hat{B} |)$
(Note the physics convention: $\hat{v}$ indicates the normalization of v; ie, $\hat{v} = \frac{v}{| v |}$ ).

Babenzgs 2022-01-29

Calculate $\sum_{n = 1}^{\infty} \arctan (\frac{2 \sqrt{2}}{n^{2} + 1})$

Laney Spears 2022-01-29

How does one solve $\sin x - \sqrt{3} \cos x = 1$
I thought this one up, but I am not sure how to solve it. Here is my attempt:
$\sin x - \sqrt{3} \cos x = 1$
${(\sin x - \sqrt{3} \cos x)}^{2} = 1$
$\sin^{2} x - 2 \sqrt{3} \sin x \cos x + 3 \cos^{2} x = 1$
$1 - 2 \sqrt{3} \sin x \cos x + 2 \cos^{2} x = 1$
$2 \cos^{2} x - 2 \sqrt{3} \sin x \cos x = 0$
$2 \cos x (\cos x - \sqrt{3} \sin x) = 0$
$2 \cos x = 0 \Rightarrow x \in {\frac{π}{2} (2 n - 1) : n \in Z}$
But how do I solve
$\cos x - \sqrt{3} \sin x = 0$

iocasq4 2022-01-29

Evaluating $\frac{\sin A + \sin B + \sin C}{\cos A + \cos B + \cos C}$ for a triangle with sides 2, 3, 4

Alvin Pugh 2022-01-29

Solving $\sin^{2} x + 1 = 2 x$

Madilyn Fitzgerald 2022-01-29

Showing that $lim_{x \to 0} \frac{x \sin (\frac{1}{x}) - \cos (\frac{1}{x})}{x}$ does not exist without sequence
Show that
$lim_{x \to 0} \frac{x \sin (\frac{1}{x}) - \cos (\frac{1}{x})}{x}$
does not exist.

trefoniu1 2022-01-29

Simplifying $\frac{\cot (34^{\circ}) \sin (44^{\circ})}{\sin (22^{\circ}) \sin (56^{\circ})}$
We know that $\cot α = \frac{\sin α}{\cos α}$ and $\sin (44^{\circ}) = 2 \sin (22^{\circ}) \cos (22^{\circ})$

Cancelling similar terms out
$\frac{\cot (34^{\circ}) 2 \cos (22^{\circ})}{\sin (56^{\circ})} = \frac{\frac{\cos (34^{\circ})}{\sin (34^{\circ})} 2 \cos (22^{\circ})}{\sin (56^{\circ})}$
Using the identity $\cos (90^{\circ} - α) = \sin (α)$
$\frac{\frac{\sin (56^{\circ})}{\sin (34^{\circ})} 2 \cos (22^{\circ})}{\sin (56^{\circ})} = \frac{2 \sin (56^{\circ}) \cos (22^{\circ})}{\sin (34^{\circ}) \sin (56^{\circ})} = \frac{2 \cos (22^{\circ})}{\sin (34^{\circ})}$
Since the degrees aren't the same, I could not proceed further.

plorinosakl 2022-01-29

Prooving an upper bound for $\sum_{k = 0}^{N} \frac{k^{2}}{N^{3}} \cos (\frac{k}{N} x)$
im struggling to proove, that there exsists an constant C, so that for every $N \in N$ and every $x \in [0, N π]$ , the following holds:
$\sum_{k = 0}^{N} \frac{k^{2}}{N^{3}} \cos (\frac{k}{N} x) \leq C \cdot (x^{- 1} + x^{- 2} + x^{- 3})$

Jay Mckay 2022-01-29

How to solve $\cos (2^{n} x) = \cos (x)$
I am given the function $F : R \to R$ , $x \mapsto 4 x (1 - x)$ and I have to find all n-cycles of the function. I have already reduced this to the problem of solving $\cos (2^{n} x) = \cos (x)$ for $x \in [0, π]$ . Using Mathematica and $n \in {1, 2, 3, 4}$ , I concluded that the solutions must be ${\frac{2 k π}{2^{n} - 1}, \frac{2 k π}{2^{n} + 1}} \cap [0, π]$ . Since the equation obviously has $2^{n}$ solutions and the solution set has $2^{n}$ elements, it only remains to verify that they are indeed solutions to the equation. I tried using $\cos (2 x) = 2 \cos^{2} (x) - 1$ and proving this by induction on n, but I was not able to.

Maximus George 2022-01-29

If $\cos^{4} α + 4 \sin^{4} β - 4 \sqrt{2} \cos α \sin β + 2 = 0$ then find $α, β in (0, \frac{π}{2})$
Try: I am trying to convert it into sum of square of quantity like
${(\cos^{2} α)}^{2} + {(2 \sin^{2} β)}^{2} - 2 \cdot \cos^{2} α \cdot 2 \sin^{2} β - 4 \sqrt{2} \cos α \sin β + 2 + 4 \cos^{2} α \cdot \sin^{2} β$
${(\cos^{2} α - 2 \sin^{2} β)}^{2} - 4 \sqrt{2} \cos α \sin β + 4 \cos^{2} α \cdot \sin^{2} β$
Now i did not know how to solve it

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