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## Cot2θ+tanθ = ? Write Answer With Explanation...

maganbhai P.
Featured 2 weeks ago

$= \csc 2 \theta$

#### Explanation:

We know that,

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

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

Using $\left(1\right) \mathmr{and} \left(2\right)$

$\cot 2 \theta + \tan \theta = \frac{1}{\tan} \left(2 \theta\right) + \tan \theta$

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

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

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

$= \frac{1 + {\tan}^{2} \theta}{2 \tan \theta}$

$= \frac{1}{\frac{2 \tan \theta}{1 + {\tan}^{2} \theta}}$

$= \frac{1}{\sin 2 \theta}$

$= \csc 2 \theta$

## How do you find the critical points to graph y=3sin(1/3x+ pi/2)-2?

sankarankalyanam
Featured 2 weeks ago

As below.

#### Explanation:

$y = A \sin \left(B x - C\right) + D$

color(brown)(Amplitude = |A|, " Period " = (2pi) / |B|, " Phase Shift " = -C / B, " Vertical Shift " = D

$\text{Given Equation is } y = \sin \left(\left(\frac{1}{3}\right) x + \left(\frac{\pi}{2}\right)\right) - 2$

$A m p l i t u \mathrm{de} = | A | = 3$

$\text{Period } = \frac{2 \pi}{|} B | = \frac{2 \pi}{\frac{1}{3}} = 6 \pi$

$\text{Phase Shift " = -C / B = -(pi/2) / (1/3) = - (3pi)/2, color(crimson)((3pi)/2 " to the LEFT}$

$\text{Vertical Shift " = D = -2, color(crimson)(" 2 down}$

graph{3 sin ((1/3)x + (pi/2)) - 2 [-10, 10, -5, 5]}

## Find the exact value? 3-2sin^2x=3cosx

Abhishek K.
Featured 1 week ago

$\rightarrow x = 2 n \pi \pm \frac{\pi}{3}$ OR $x = 2 n \pi \pm \frac{\pi}{2}$

#### Explanation:

$\rightarrow 3 - 2 {\sin}^{2} x = 3 \cos x$

$\rightarrow 3 - 2 \left(1 - {\cos}^{2} x\right) = 3 \cos x$

$\rightarrow 3 - 2 + 2 {\cos}^{2} x = 3 \cos x$

$\rightarrow 2 {\cos}^{2} x - 3 \cos x + 1 = 0$

$\rightarrow 2 {\cos}^{2} x - 2 \cos x - \cos x + 1 = 0$

$\rightarrow 2 \cos x \left(\cos x - 1\right) - 1 \left(\cos x - 1\right) = 0$

$\rightarrow \left(2 \cos x - 1\right) \left(\cos x - 1\right) = 0$

Either, $2 \cos x - 1 = 0$

$\rightarrow \cos x = \frac{1}{2} = \cos \left(\frac{\pi}{3}\right)$

$\rightarrow x = 2 n \pi \pm \frac{\pi}{3}$ where $n \rightarrow Z$

OR, $\cos x - 1 = 0$

$\rightarrow \cos x = 1 = \cos \left(\frac{\pi}{2}\right)$

$\rightarrow x = 2 n \pi \pm \frac{\pi}{2}$ where $n \rightarrow Z$

## How do you graph y=-2sinx+1?

sankarankalyanam
Featured 1 week ago

As below.

#### Explanation:

$\text{Standard form of equation for a sine function is }$

$y = A \sin \left(B x - C\right) + D$

$\text{Given equation is } y = - 2 \sin x + 1$

$A = - 2 , B = 1 , C = 0 , D = 1$

$A m p l i t u \mathrm{de} = | A | = 2$

$\text{Period } = P = \frac{2 \pi}{|} B | = 2 \pi$

$\text{Phase Shift } = - \frac{C}{B} = 0$

$\text{Vertical Shift } = D = + 1$

graph{-2 sin x + 1 [-10, 10, -5, 5]}

## How do you find exact value of sin (19pi/12)?

Mr. Hsia
Featured 6 days ago

$\frac{- \sqrt{2} - \sqrt{6}}{4}$

#### Explanation:

$\sin \left(\frac{19 \pi}{12}\right) = \sin \left(\frac{15 \pi}{12} + \frac{4 \pi}{12}\right)$
$= \sin \left(\frac{5 \pi}{4} + \frac{\pi}{3}\right)$
$= \sin \left(\frac{5 \pi}{4}\right) \cos \left(\frac{\pi}{3}\right) + \cos \left(\frac{5 \pi}{4}\right) \sin \left(\frac{\pi}{3}\right)$
$= \left(- \frac{\sqrt{2}}{2}\right) \left(\frac{1}{2}\right) + \left(- \frac{\sqrt{2}}{2}\right) \left(\frac{\sqrt{3}}{2}\right)$
$= \left(- \frac{\sqrt{2}}{4}\right) + \left(- \frac{\sqrt{6}}{4}\right)$
$= \frac{- \sqrt{2} - \sqrt{6}}{4}$

## How do you evaluate arcsin(2) #?

Dean R.
Featured 10 hours ago

$\arcsin 2 = \left(\frac{\pi}{2} + 2 \pi k\right) \pm i \ln \left(2 - \sqrt{3}\right) \quad$ integer $k$

#### Explanation:

Obviously no regular angle, no real number, has a sine of $2$. We're getting in the realm of complex numbers.

Euler's Formula is valid for complex numbers too:

${e}^{i z} = \cos z + i \sin z$

We want to know about ${e}^{i \left(- z\right)} = {e}^{- i z} .$ If $z$ is real that would be the conjugate, but it's not for complex $z$. But even for complex $z$ we want cosine to be even, $\cos z = \cos \left(- z\right) ,$ and sine to be odd, $\sin z = - \sin \left(- z\right) .$ So

${e}^{- i z} = {e}^{i \left(- z\right)} = \cos \left(- z\right) + i \sin \left(- z\right) = \cos z - i \sin z$

Subtracting,

${e}^{i z} - {e}^{- i z} = 2 i \sin z$

$\sin z = \frac{1}{2 i} \left({e}^{i z} - {e}^{- i z}\right)$

From here out we'll use that essentially as the definition of $\sin z .$

It's actually easier to get a relatively nice closed form for the inverse sine of a real number greater than one than of a regular sine.

We want to solve

$\sin z = a$ for real $a > 1.$

$\frac{1}{2 i} \left({e}^{i z} - {e}^{- i z}\right) = a$

Let $y = {e}^{i z} .$ Then ${e}^{- i z} = \frac{1}{y} .$

$y - \frac{1}{y} = 2 i a$

${y}^{2} - 2 i a y - 1 = 0$

I call it the Shakespeare Quadratic Formula ($2 b$ or $- 2 b$): ${x}^{2} - 2 b x + c$ has zeros $x = b \pm \sqrt{{b}^{2} - c} .$

$y = i a \pm \sqrt{{\left(- i a\right)}^{2} + 1} = i a \pm \sqrt{1 - {a}^{2}}$

Since $a > 1$ the radicand is negative. Let's make that explicit.

$y = i a \pm \sqrt{\left({a}^{2} - 1\right) \left(- 1\right)} = i \left(a \pm \sqrt{{a}^{2} - 1}\right)$

${e}^{i z} = i \left(a \pm \sqrt{{a}^{2} - 1}\right)$

Since $i = {e}^{i \frac{\pi}{2}}$ and ${e}^{2 \pi k i} = 1 \quad$ integer $k ,$

${e}^{i z} = {e}^{i \frac{\pi}{2}} {e}^{2 \pi k i} {e}^{\ln \left(a \pm \sqrt{{a}^{2} - 1}\right)}$

$i z = i \frac{\pi}{2} + 2 \pi k i + \ln \left(a \pm \sqrt{{a}^{2} - 1}\right)$

$z = \left(\frac{\pi}{2} + 2 \pi k\right) - i \ln \left(a \pm \sqrt{{a}^{2} - 1}\right)$

That's the general solution to $\sin z = a$ for real $a > 1.$ There are a few things to note about it. Let's look at the two imaginary parts first.

$\left(a + \sqrt{{a}^{2} - 1}\right) \left(a - \sqrt{{a}^{2} - 1}\right) = {a}^{2} - \left({a}^{2} - 1\right) = 1$

$a - \sqrt{{a}^{2} - 1} = \frac{1}{a + \sqrt{{a}^{2} - 1}}$

$\ln \left(a - \sqrt{{a}^{2} - 1}\right) = - \ln \left(a + \sqrt{{a}^{2} - 1}\right)$

$z = \left(\frac{\pi}{2} + 2 \pi k\right) \pm i \ln \left(a - \sqrt{{a}^{2} - 1}\right)$

The two imaginary parts are negations of each other! That means the $z$s come in complex conjugate pairs.

Considering $k = 0$, when we add the two roots together we'll get $\pi .$ In other words those two roots, even though complex, are supplementary angles who share the same sine, just like real angles. Just like real angles, we can add or subtract $2 \pi$ as many times as we like and get another number with the same sine.

OK, let's plug in $a = 2$ for the big finish.

$\arcsin 2 = \left(\frac{\pi}{2} + 2 \pi k\right) \pm i \ln \left(2 - \sqrt{3}\right) \quad$ integer $k$

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