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## How do you use the quadratic formula to solve 12sin^2x-13sinx+3=0 in the interval [0,2pi)?

Andy Y.
Featured 6 months ago

$x \approx 0.3398$, $x \approx 0.8481$, $x = 2.294$, and $x = 2.802$ radians.

#### Explanation:

Strategy: Rewrite this equation as a quadratic equation using $u = \sin \left(x\right)$. Solve the quadratic equation for $u$ by factoring. Then replace $u$ with $\sin \left(x\right)$ again and solve using $\arcsin$.

Step 1. Rewrite this equation as a quadratic using $\textcolor{red}{u} = \textcolor{red}{\sin} \left(x\right)$

You are given
$12 {\left(\textcolor{red}{\sin \left(x\right)}\right)}^{2} - 13 \left(\textcolor{red}{\sin \left(x\right)}\right) + 3 = 0$

Replace $\textcolor{red}{\sin \left(x\right)}$ with $\textcolor{red}{u}$
$12 {\textcolor{red}{u}}^{2} - 13 \textcolor{red}{u} + 3 = 0$

Step 2. Factor the quadratic equation.
$\left(3 u - 1\right) \left(4 u - 3\right) = 0$

Solving gives us
$3 u - 1 = 0$ and $4 u - 3 = 0$
$u = \frac{1}{3}$ and $u = \frac{3}{4}$

Step 3. Replace $u$ with $\sin \left(x\right)$ again and solve with $\arcsin$
$\sin \left(x\right) = \frac{1}{3}$ and $\sin \left(x\right) = \frac{3}{4}$

${\sin}^{- 1} \left(\sin \left(x\right)\right) = {\sin}^{- 1} \left(\frac{1}{3}\right)$ and ${\sin}^{- 1} \left(\sin \left(x\right)\right) = {\sin}^{- 1} \left(\frac{3}{4}\right)$

$x \approx 0.3398$ radians and $x \approx 0.8481$ radians

These answer work because we were asked to find the solutions in $\left[0 , 2 \pi\right] \approx \left[0 , 6.2832\right]$

However, the graph of $y = 12 {\left(\sin \left(x\right)\right)}^{2} - 13 \left(\sin \left(x\right)\right) + 3$ is

Which shows four solutions in the interval $\left[0 , 2 \pi\right]$, not two.

We must find the other solutions by recognizing that sine functions are periodic with respect to $\pi$

So, $x = \pi - 0.8481 = 2.294$ and $x = \pi - 0.3398 = 2.802$

The other two solutions are
$x = 2.294$ and $x = 2.802$

## How do you solve \sec ^ { 2} \theta = 7+ \tan \theta?

Ali A.
Featured 4 months ago

theta=tan^-1(3)+npi " or "theta=tan^-1(-2)+npi;

$n \in \mathbb{Z}$

#### Explanation:

${\sec}^{2} \theta = 7 + \tan \theta$

color(red)(uarr sec^2theta=tan^2theta+1

color(red)(tan^2theta+1)=7+tanthetacolor(orange)(larr "subtract"(7+tantheta) " from both sides"

${\tan}^{2} \theta + 1 - 7 - \tan \theta = 0$

${\tan}^{2} \theta - \tan \theta - 6 = 0$

Let $\tan \theta = u$

${u}^{2} - u - 6 = 0$

• now we've got a quadratic equation that can be solved by using the quadratic formula $u = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$.
we can get the values the values of $a , b , \mathmr{and} c$ by comparing our equation to the standard form $a {u}^{2} + b u + c$, and the values are $a = 1 , b = - 1 , c = - 6$

$u = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a} \implies \tan \theta = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$= \frac{- \left(- 1\right) \pm \sqrt{{\left(- 1\right)}^{2} - 4 \left(1\right) \left(- 6\right)}}{2 \left(1\right)}$

$\tan \theta = \frac{1 \pm 5}{2} \implies$

$\tan \theta = \frac{1 + 5}{2} \text{ or } \tan \theta = \frac{1 - 5}{2}$

$\tan \theta = 3 \text{ or } \tan \theta = - 2$

• a period of the function $\tan \theta$ is $\pi$ therefore $\tan \left(\theta + \pi\right) = \tan \theta$

Let $\tan {\theta}_{1} = 3 \text{ and } \tan {\theta}_{2} = - 2$

theta_1=tan^-1(3)+npi; "where "n inZZ

${\theta}_{2} = {\tan}^{-} 1 \left(- 2\right) + n \pi \text{ where } n \in \mathbb{Z}$

## (1-sin(2x)) /(1+sin(2x))=tan^2(45^@-x) How to prove this without integration? Can we use rationalizing here?

Douglas K.
Featured 4 months ago

One does not use integration or rationalizing to prove this. One proves this by changing only 1 side of the equation, until it is identical to the other side.

#### Explanation:

Prove:

$\frac{1 - \sin \left(2 x\right)}{1 + \sin \left(2 x\right)} = {\tan}^{2} \left({45}^{\circ} - x\right)$

The identity for $\tan \left(A - B\right) = \frac{\tan \left(A\right) - \tan \left(B\right)}{1 + \tan \left(A\right) \tan \left(B\right)}$, therefore, we may substitute ${\left(\frac{\tan \left({45}^{\circ}\right) - \tan \left(x\right)}{1 + \tan \left({45}^{\circ}\right) \tan \left(x\right)}\right)}^{2}$ for ${\tan}^{2} \left({45}^{\circ} - x\right)$:

$\frac{1 - \sin \left(2 x\right)}{1 + \sin \left(2 x\right)} = {\left(\frac{\tan \left({45}^{\circ}\right) - \tan \left(x\right)}{1 + \tan \left({45}^{\circ}\right) \tan \left(x\right)}\right)}^{2}$

Use the fact that $\tan \left({45}^{\circ}\right) = 1$:

$\frac{1 - \sin \left(2 x\right)}{1 + \sin \left(2 x\right)} = {\left(\frac{1 - \tan \left(x\right)}{1 + \tan \left(x\right)}\right)}^{2}$

Write $\tan \left(x\right)$ as $\sin \frac{x}{\cos} \left(x\right)$:

$\frac{1 - \sin \left(2 x\right)}{1 + \sin \left(2 x\right)} = {\left(\frac{1 - \sin \frac{x}{\cos} \left(x\right)}{1 + \sin \frac{x}{\cos} \left(x\right)}\right)}^{2}$

Multiply the fraction inside the square by 1 in the form of $\left(\cos \frac{x}{\cos} \left(x\right)\right)$:

$\frac{1 - \sin \left(2 x\right)}{1 + \sin \left(2 x\right)} = {\left(\left(\cos \frac{x}{\cos} \left(x\right)\right) \frac{1 - \sin \frac{x}{\cos} \left(x\right)}{1 + \sin \frac{x}{\cos} \left(x\right)}\right)}^{2}$

Perform the multiplication:

$\frac{1 - \sin \left(2 x\right)}{1 + \sin \left(2 x\right)} = {\left(\frac{\cos \left(x\right) - \cos \left(x\right) \sin \frac{x}{\cos} \left(x\right)}{\cos \left(x\right) + \cos \left(x\right) \sin \frac{x}{\cos} \left(x\right)}\right)}^{2}$

Please observe the cancellation of the embedded denominators:

$\frac{1 - \sin \left(2 x\right)}{1 + \sin \left(2 x\right)} = {\left(\frac{\cos \left(x\right) - \cancel{\cos \left(x\right)} \sin \frac{x}{\cancel{\cos \left(x\right)}}}{\cos \left(x\right) + \cancel{\cos \left(x\right)} \sin \frac{x}{\cancel{\cos \left(x\right)}}}\right)}^{2}$

Write the equation without the cancelled factors:

$\frac{1 - \sin \left(2 x\right)}{1 + \sin \left(2 x\right)} = {\left(\frac{\cos \left(x\right) - \sin \left(x\right)}{\cos \left(x\right) + \sin \left(x\right)}\right)}^{2}$

Perform the multiplication implied by the square:

$\frac{1 - \sin \left(2 x\right)}{1 + \sin \left(2 x\right)} = \frac{{\cos}^{2} \left(x\right) - 2 \sin \left(x\right) \cos \left(x\right) + {\sin}^{2} \left(x\right)}{{\cos}^{2} \left(x\right) + 2 \sin \left(x\right) \cos \left(x\right) + {\sin}^{2} \left(x\right)}$

Use the identity ${\sin}^{2} \left(x\right) + {\cos}^{2} \left(x\right) = 1$:

$\frac{1 - \sin \left(2 x\right)}{1 + \sin \left(2 x\right)} = \frac{1 - 2 \sin \left(x\right) \cos \left(x\right)}{1 + 2 \sin \left(x\right) \cos \left(x\right)}$

Use the identity $\sin \left(2 x\right) = 2 \sin \left(x\right) \cos \left(x\right)$:

$\frac{1 - \sin \left(2 x\right)}{1 + \sin \left(2 x\right)} = \frac{1 - \sin \left(2 x\right)}{1 + \sin \left(2 x\right)}$ Q.E.D.

## What cosine function represents an amplitude of 3, a period of Ï€, no horizontal shift, and a vertical shift of?

Alan P.
Featured 3 months ago

In order to answer this I have assumed a vertical shift of $+ 7$

$\textcolor{red}{3 \cos \left(2 \theta\right) + 7}$

#### Explanation:

The standard cos function $\textcolor{g r e e n}{\cos \left(\gamma\right)}$ has a period of $2 \pi$

If we want a period of $\pi$ we need to replace $\gamma$ with something that will cover the domain "twice as fast" e.g. $2 \theta$.
That is $\textcolor{m a \ge n t a}{\cos \left(2 \theta\right)}$ will have a period of $\pi$.

To get an amplitude of $3$ we need to multiply all values in the Range generated by $\textcolor{m a \ge n t a}{\cos \left(2 \theta\right)}$ by $\textcolor{b r o w n}{3}$ giving
$\textcolor{w h i t e}{\text{XXX}} \textcolor{b r o w n}{3 \cos \left(2 \theta\right)}$

There is to be no horizontal shift, so the argument for $\cos$ will not be modified by any further addition/subtraction.

In order to achieve the vertical shift (that I assumed would be $\textcolor{red}{+ 7}$ [substitute your own value]) we will need to add $\textcolor{red}{7}$ to all values in our modified range:
$\textcolor{w h i t e}{\text{XXX}} \textcolor{red}{3 \cos \left(2 \theta\right) + 7}$

## Represent the function? 1)y=sinx

Alan N.
Featured 2 months ago

See below

#### Explanation:

There are several definitions of the function $\sin x$. I will give three of the most common ones below.

(i) Probably the simplest is in relation to any right $\triangle A B C$ where $\angle C$ is a right angle (${90}^{0} \equiv \frac{\pi}{2} r a d$). The sides of $\triangle A B C$ are $a , b , c$ opposite their respective angle. Here side $c$ is called the hypotenuse as it is opposite the right angle.

$\sin \theta$ is defined as the ratio: $\text{opposite / hypotenuse}$.

So, in the case above sin A = a/c; sin B =b/c

(ii) Consider the unit circle centred at the origin $\left(O\right)$. A point $P \left(x , y\right)$ moves around the circumference in a counter clockwise direction from the point $\left(1 , 0\right)$. Let the angle formed between $O P$ and the positive $x -$axis be $\theta$

$\sin \theta$ is then defined as the $y -$component of $P$

NB: This definition is equivalent to (i) above for $\theta \in \left(0 , \frac{\pi}{2}\right)$ and generalises the function for $\theta \in \left[0 , 2 \pi\right]$ and, by extension, to all real numbers.

(iii) $\sin x$ is continuous and differentiable for all real numbers - $\left(\forall x \in \mathbb{R}\right)$

Hence, it can be defined by a Taylor series at $x = 0$. This series expansion is:

sinx = x-x^3/(3!)+x^5/(5!)-x^7/(7!)+....

= sum_"n=0"^oo (-1)^n/((2n+1)!) x^(2n+1)

Finally, to "represent" $y = \sin x$ as requested in the question, the graph of $y = \sin x$ is shown below.

graph{sinx [-6.244, 6.244, -3.12, 3.124]}

## How do you prove sin^-1(tanh(x)) = tan^-1(sinh(x))?

Douglas K.
Featured 2 weeks ago

The way that one proves an identity is to make substitutions to only one side until it is identical to the other side:

#### Explanation:

Given: ${\sin}^{-} 1 \left(\tanh \left(x\right)\right) = {\tan}^{-} 1 \left(\sinh \left(x\right)\right)$

Use the property $u = {\sin}^{-} 1 \left(\sin \left(u\right)\right)$ on the right side and mark as equation [1]:

sin^-1(tanh(x)) = sin^-1(sin(tan^-1(sinh(x)))" [1]"

Digress and prove that sin(tan^-1(sinh(x)) = tanh(x)

An alternate form for ${\tan}^{-} 1 \left(u\right) = \frac{i}{2} \ln \left(1 - i u\right) - \frac{i}{2} \ln \left(1 + i u\right)$

Substitute $\sinh \left(x\right)$ for u:

${\tan}^{-} 1 \left(\sinh \left(x\right)\right) = \frac{i}{2} \ln \left(1 - i \sinh \left(x\right)\right) - \frac{i}{2} \ln \left(1 + i \sinh \left(x\right)\right)$

An alternate form for $\sin \left(v\right) = \frac{i}{2} {e}^{- i v} - \frac{i}{2} {e}^{i v}$

sin(tan^-1(sinh(x))) = i/2e^(-i(i/2ln(1-isinh(x))-i/2ln(1+isinh(x)))-i/2e^(i(i/2ln(1-isinh(x))-i/2ln(1+isinh(x))))

Distribute through $- i$:

$\sin \left({\tan}^{-} 1 \left(\sinh \left(x\right)\right)\right) = \frac{i}{2} {e}^{\left(- {i}^{2} / 2 \ln \left(1 - i \sinh \left(x\right)\right) + {i}^{2} / 2 \ln \left(1 + i \sinh \left(x\right)\right)\right)} - \frac{i}{2} {e}^{i \left(\frac{i}{2} \ln \left(1 - i \sinh \left(x\right)\right) - \frac{i}{2} \ln \left(1 + i \sinh \left(x\right)\right)\right)}$

Distribute through $i$:

sin(tan^-1(sinh(x))) = i/2e^((-i^2/2ln(1-isinh(x))+i^2/2ln(1+i(sinh(x))))-i/2e^((i^2/2ln(1-isinh(x))-i^2/2ln(1+isinh(x))))

use the property ${i}^{2} = - 1$:

$\sin \left({\tan}^{-} 1 \left(\sinh \left(x\right)\right)\right) = \frac{i}{2} {e}^{\left(\frac{1}{2} \ln \left(1 - i \sinh \left(x\right)\right) - \frac{1}{2} \ln \left(1 + i \sinh \left(x\right)\right)\right)} - \frac{i}{2} {e}^{\left(- \frac{1}{2} \ln \left(1 - i \sinh \left(x\right)\right) + \frac{1}{2} \ln \left(1 + i \sinh \left(x\right)\right)\right)}$

Write the $\frac{1}{2}$ in the exponent as a square root:

$\sin \left({\tan}^{-} 1 \left(\sinh \left(x\right)\right)\right) = \frac{i}{2} {e}^{\left(\ln \left(\sqrt{1 - i \sinh \left(x\right)}\right) - \ln \left(\sqrt{1 + i \sinh \left(x\right)}\right)\right)} - \frac{i}{2} {e}^{\left(- \ln \left(\sqrt{1 - i \sinh \left(x\right)}\right) + \ln \left(\sqrt{1 + i \sinh \left(x\right)}\right)\right)}$

Factor out $\frac{i}{2}$:

sin(tan^-1(sinh(x))) = i/2{e^((ln(sqrt(1-isinh(x)))-ln(sqrt(1+isinh(x))))-e^((-ln(sqrt(1-isinh(x)))+ln(sqrt(1+isinh(x)))))}

Use the property of logarithms $\ln \left(a\right) - \ln \left(b\right) = \ln \left(\frac{a}{b}\right)$:

$\sin \left({\tan}^{-} 1 \left(\sinh \left(x\right)\right)\right) = \frac{i}{2} \left\{{e}^{\left(\ln \left(\frac{\sqrt{1 - i \sinh \left(x\right)}}{\sqrt{1 + i \sinh \left(x\right)}}\right)\right)} - {e}^{\left(\ln \left(\frac{\sqrt{1 + i \sinh \left(x\right)}}{\sqrt{1 - i \sinh \left(x\right)}}\right)\right)}\right\}$

Use the property ${e}^{\ln} \left(u\right) = u$:

$\sin \left({\tan}^{-} 1 \left(\sinh \left(x\right)\right)\right) = \frac{i}{2} \left\{\frac{\sqrt{1 - i \sinh \left(x\right)}}{\sqrt{1 + i \sinh \left(x\right)}} - \frac{\sqrt{1 + i \sinh \left(x\right)}}{\sqrt{1 - i \sinh \left(x\right)}}\right\}$

When we make a common denominator we obtain:

$\sin \left({\tan}^{-} 1 \left(\sinh \left(x\right)\right)\right) = \frac{i}{2} \left\{\frac{1 - i \sinh \left(x\right)}{\sqrt{1 + {\sinh}^{2} \left(x\right)}} - \frac{1 + i \sinh \left(x\right)}{\sqrt{1 + {\sinh}^{2} \left(x\right)}}\right\}$

Combine over the common denominator:

$\sin \left({\tan}^{-} 1 \left(\sinh \left(x\right)\right)\right) = \frac{i}{2} \left\{\frac{- 2 i \sinh \left(x\right)}{\sqrt{1 + {\sinh}^{2} \left(x\right)}}\right\}$

The leading coefficient multiplied into the numerator becomes 1:

$\sin \left({\tan}^{-} 1 \left(\sinh \left(x\right)\right)\right) = \sinh \frac{x}{\sqrt{1 + {\sinh}^{2} \left(x\right)}}$

Use the identity $1 + {\sinh}^{2} \left(x\right) = {\cosh}^{2} \left(x\right)$:

$\sin \left({\tan}^{-} 1 \left(\sinh \left(x\right)\right)\right) = \sinh \frac{x}{\sqrt{{\cosh}^{2} \left(x\right)}}$

$\sin \left({\tan}^{-} 1 \left(\sinh \left(x\right)\right)\right) = \sinh \frac{x}{\cosh} \left(x\right)$

$\sin \left({\tan}^{-} 1 \left(\sinh \left(x\right)\right)\right) = \tanh \left(x\right)$

Substitute this into equation [1]:

${\sin}^{-} 1 \left(\tanh \left(x\right)\right) = {\sin}^{-} 1 \left(\tanh \left(x\right)\right)$ Q.E.D.

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