Precalculus Solutions | Expert Assistance for Problems & Examples

$x + y = 120^{\circ}, x \geq 0, y \geq 0$ . max and min of $\sin x + \sin y$

jelentetvq 2022-01-23

How is the derivative of $x \cos x \sin x \frac{12}{\sin 2} x + x \cos 2 x$ ?
I tried using the product rule on $x \cos x$ and got:
$(x \cos x)' = - x \sin x + \cos x$
then I used the product rule to differentiate xcosxsinx and got:
$(x \cos x \sin x)' = (- x \sin x + \cos x) \sin x + (x \cos x \cos x) = - x \sin^{2} x + \sin x \cos x + x \cos^{2} x$

Jayleen Sanders 2022-01-23

Proving $\int_{0}^{\frac{π}{4}} \sin^{n} (x) d x > \frac{1}{2^{\frac{n}{2}} (n + 2)}$

Danitat6 2022-01-23

Evaluating $\sum_{k = 1}^{n - 1} k \cos (k x)$

pozicijombx 2022-01-23

Solve $y = \frac{13}{\sin x + [\sin x + [\sin x]]}$ & $[y + [y]] = 2 \cos x$
$\sin x \in (π, 2 π)$
$y = \frac{13}{\sin x + [\sin x + [\sin x]]}$
y=-1
$[- 1 + [- 1]] = 2 \cos x$
$\cos x = - 1$ which is possible at $x = π$ hence NO SOLUTION $\sin x \in \frac{π}{2}$
$y = \frac{13}{\sin x + [\sin x + [\sin x]]}$
y=1
$[1 + [1]] = 2 \cos x$
$\cos x = 1$ which is possible at x=0 hence NO SOLUTION
Similary if $\sin x \in (0, π) - \frac{π}{2}$
$y = \frac{1}{3} [\sin x + [\sin x + [\sin x]]]$
y=0
$[0 + [0]] = 2 \cos x$
$\cos x = 0$ which is possible at $x = \frac{π}{2}$ hence NO SOLUTION

veceriraby 2022-01-23

Evaluating $\int_{- \frac{π}{4}}^{\frac{π}{4}} {(\sin x + 2 \cos x)}^{3} d x$

Adrien Barron 2022-01-23

Solve $\sec (x) = \tan (x)$
I have been trying to do this problem: Solve $\sec (x) = \tan (x), 0 \leq x < 2 π$ I started by rewriting $\sec (x) as \frac{1}{\cos (x)}$ . I then rewrote $\tan (x) to get \frac{\sin (x)}{\cos (x)}$ Therefore: $\frac{1}{\cos x} = \frac{\sin x}{\cos x}$ Therefore: $1 = \sin x, x = \frac{π}{2}$ However, this is not a solution to the original problem because both $y = \tan (x) and y = \sec (x)$ have asymptotes as $x = \frac{π}{2}$

Danitat6 2022-01-23

General solution of $\tan (2 x) \tan (x) = 1$

2alr8w 2022-01-23

Simplifying $\sqrt{\frac{{(a^{2} \cos^{2} t + b^{2} \sin^{2} t)}^{3}}{{(b^{2} \cos^{2} t + a^{2} \sin^{2} t)}^{3}}}$

Arely Briggs 2022-01-23

Proving $\tan (x + \frac{π}{3}) = \frac{4 \tan x + \sqrt{3} \sec^{2} x}{\sec^{2} x - 4 \tan^{2} x}$

Alisha Pitts 2022-01-23

How to solve $\ln (\csc (z)) = \ln (2) + i z$ ?

spiderifilms6e 2022-01-23

Prove $\sin (a) < a < \tan (a)$ when $0 < a < \frac{π}{2}$

Jacquelyn Sanders 2022-01-23

Does $A \sin b = B \sin b + C \cos b$ imply C=0?

Tori Hines 2022-01-23

How to use Liouville's theorem to prove the formula of $\frac{π}{\sin π z}$

Tori Hines 2022-01-23

Where is the angle for $- \frac{7 π}{8}$ in a unit circle?
I am having difficulty finding the angle for $- \frac{7 π}{8}$ in a unit circle. I have found the coordinates for $\frac{π}{8}$ and have to use the symmetry to locate the coordinates for $- \frac{7 π}{8}$ . Is the angle just below the (1,0) and symmetry to $\frac{π}{8}$ ?

Turnseeuw 2022-01-23

Differentiate $\sqrt{\frac{1 + \sin x}{1 - \sin x}}$

Duncan Reed 2022-01-23

Evaluate and simplify: $lim_{x \to 0} \frac{\sin 2 x}{3 x}$
So, by direct substitution, it is indeterminate. I got the derivative of this functio,n and now Im

Autohelmvt 2022-01-23

If A+B+C=180, then $\frac{\sin 2 A + \sin 2 B + \sin 2 C}{\cos A + \cos B + \cos C - 1} = 8 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}$
Then 2A+2B+2C=360
So
$\sin 2 C = - \sin (2 A + 2 B)$
Putting that in the equation
$2 \sin (A + B) \sin (A - B) - 2 \sin (A + B) \cos (A + B) \cos A + \cos B - \cos (A + B) + 1$
$2 \sin (A + B) [\sin (A - B) - \cos (A + B)] \cos A + \cos B - \cos (A + B) + 1$
I don’t know how to proceed. Please help me continue

Franco Franco 2022-01-23

Proving that $\sec \frac{π}{30} = \sqrt{2 - \sqrt{5} + \sqrt{15 - 6 \sqrt{5}}}$
which I instantly wanted to prove.
I know that I can "reduce" the problem to the evaluation of $\cos \frac{π}{15}$ , as the rest is easy with the use of the half-angle formula.
I know that cos obeys the nice relation
$\cos n x = T_{n} (\cos x)$
where
$T_{n} (x) = \frac{n}{2} \sum_{k = 0}^{| \frac{n}{2} |} \frac{{(- 1)}^{k}}{n - k} (\begin{matrix} n - k \\ k \end{matrix}) {(2 x)}^{n - 2 k}$
Thus, setting $t = \cos \frac{π}{15}$ ,
$T_{15} (t) = - 1$
The only thing left to do is solve for t. We can narrow down our search to the values 0<t<1.
I have never dealt with degree-15 polynomials before, so I was hoping one of you could help me out.

Selena Cowan 2022-01-23

what is x for $\tan^{2} 3 x = 2 \sin^{2} 3 x$

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