Precalculus Solutions | Expert Assistance for Problems & Examples

Simplify Trig Identity in $\sin θ$ and/or $\cos θ$
I just need to simplify in terms of $\sin θ$ and/or $\cos θ$
I did this one earlier today but forgot how I did it. I just need help on the other two.
$\frac{\cos θ}{\sin θ} + \frac{1}{\cos θ} = \frac{\cos^{2} θ + \sin θ}{\sin θ \cos θ}$
Question 1: $\sin θ + \frac{1}{\cos θ}$
Question 2: $\frac{\sin θ}{\cos θ} + \frac{1}{\sin θ}$

dooporpplauttssg 2022-04-26

Simplify the equation $16 \cos^{6} (t) + 16 \sin^{6} (t) + 48 \sin^{2} (t) \cos^{2} (t)$

Eileen Garza 2022-04-26

Simplify and find the value of
$\tan θ (1 - \sec \frac{θ}{2}) (1 - \sec θ) (1 - \sec 2 θ) \dots (1 - - {\sec 2}^{n - 1} θ)$
at $n = 1, 2, 3$

Dashawn Robbins 2022-04-25

What is $\tan (f (1) + f (2) + \dots + f (21))$ where $f (x) = \arctan (\frac{1}{x^{2} + x + 1})$

p3sa1bynn 2022-04-25

Period of $\sin^{2} x + \cos^{4} x$ ?

broggesy9 2022-04-25

Since
$D^{- 1} {(1 + x^{2})}^{- \frac{1}{2}} = {sinh}^{- 1} (x) + C$ ,
is it ture that
${sinh}^{- 1} x + C ∣_{0}^{1} = \log (1 + \sqrt{2})$ ?

pacificak8m 2022-04-25

Simplify: $\frac{1 + \cos 2 x}{1 - \cos x} - \frac{4 \cos^{2} (\frac{x}{2})}{\tan^{2} x}$

Essence Byrd 2022-04-25

Simplify this expression with $\cos$ and $\sin$
$\frac{\cos 12 x}{\cos 4 x} - \frac{\sin 12 x}{\sin 4 x} = \frac{\cos^{2} 6 x - \sin^{2} 6 x}{\cos^{2} 2 x - \sin^{2} 2 x} - \frac{2 \sin 6 x \cos 6 x}{2 \sin 2 x \cos 2 x}$

Michael Rivers 2022-04-25

Simplify $\prod_{k = 1}^{5} \tan \frac{k π}{11}$ and $\sum_{k = 1}^{5} \tan^{2} \frac{k π}{11}$

pettingyg0 2022-04-25

Simplify the expression
$(1 - \frac{{\cos 61}^{\circ}}{{\cos 1}^{\circ}}) (1 - \frac{{\cos 62}^{\circ}}{{\cos 2}^{\circ}}) \cdot \dots \cdot (1 - \frac{{\cos 119}^{\circ}}{{\cos 59}^{\circ}})$

kabutjv7 2022-04-24

Number of values that satisfy
$2 \sin^{2} (x) - 3 = 3 \cos (x), 90^{\circ} < x < 270^{\circ}$

et3atissb 2022-04-24

Need help proving this expression.
$\sin^{2} (\frac{x}{2}) = \frac{1}{2} (1 - c o s (x))$

Deven Livingston 2022-04-24

Is a function periodic $f (x) = \cos (x) + \cos (x^{2})$
This one I gave my students today, nobody solve it.
Is a following function periodic $f : R \to R$
$f (x) = \cos (x) + \cos (x^{2})$
If someone is interested I can show a solution later.

Halle Marsh 2022-04-24

Inverse Trigonometry System of Equations
$2 \tan^{- 1} (\sqrt{x - x^{2}}) = \tan^{- 1} (x) + \tan^{- 1} (1 - x)$
I have a feeling solution includes drawing triangles but cannot make the leap to get the solution

topoit8d 2022-04-24

Integrate the function
$\frac{2 x^{3} + 5 x^{2} + 8 x + 4}{{(x^{2} + 2 x + 2)}^{2}}$

dooporpplauttssg 2022-04-24

Integrate $\tan^{2} (\frac{π}{12} \cdot y)$

luminarc24lry 2022-04-24

In any triangle, if $\frac{\cos A + 2 \cos C}{\cos A + 2 \cos B} = \frac{\sin B}{\sin C}$ , prove that the triangle is either isosceles or right angled.
My Attempt:
Given:
$\frac{\cos A + 2 \cos C}{\cos A + 2 \cos B} = \frac{\sin B}{\sin C}$
$\frac{\frac{b^{2} + c^{2} - a^{2}}{2 b c} + 2 \frac{a^{2} + b^{2} - c^{2}}{2 a b}}{\frac{b^{2} + c^{2} - a^{2}}{2 b c} + 2 \frac{a^{2} + c^{2} - b^{2}}{2 a c}} = \frac{b}{c}$
On simplification,
$\frac{a b^{2} + a c^{2} - a^{3} + 2 a^{2} c + 2 b^{2} c - 2 c^{3}}{a b^{2} + a c^{2} - a^{3} + 2 a^{2} b + 2 b c^{2} - 2 b^{3}} = \frac{b}{c}$

Esther Hoffman 2022-04-24

In a $△ A B C, \frac{a}{b} = 2 + \sqrt{3}$ and $∠ C = 60^{\circ}$ Then find the ordered pairs $(∠ A, ∠ B) =$

gaitaprepeted05u 2022-04-24

Im trying to show that
$z = \tan [\frac{1}{i} \log (\sqrt{\frac{1 + i z}{1 - i z}})]$

predorsomwy 2022-04-24

If $y = - e^{x} \cos 2 x$ , show that $\frac{d^{2} y}{d x^{2}} = 5 e^{x} \sin (2 x + α)$ where $α = \tan^{- 1} (\frac{3}{4})$

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