<img src="https://tex.z-dn.net/?f=%5Cfrac%7B%28csc%20x%2B1%29%28csc%20x-1%29%7D%7Bcsc%5E2x%7D%5C%5C" id="TexFormula1" title="\fr

All categories

2 years ago

ac{(csc x+1)(csc x-1)}{csc^2x}\\" alt="\frac{(csc x+1)(csc x-1)}{csc^2x}\\" align="absmiddle" class="latex-formula">

cscθ·cos^2θ+sinθ= cscθ

Mathematics

1 answer:

tigry1 [53]2 years ago

7 0

Step-by-step explanation:

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

b. Since csc theta = 1/sin theta, we can multiply both sides by sin theta and you will end up with

$\cos^{2} \theta + \sin^{2} \theta = 1$

which is an identity.

You might be interested in

Triangle WXY is isosceles. ∠YWX and ∠YXW are the base angles. YZ bisects ∠WYX. m∠XYZ = (15x)°. m∠YXZ = (2x + 5)°. What is the me

ozzi

Answer:

150 degrees

Step-by-step explanation:

I just answered it on edginuty

8 0

2 years ago

Read 2 more answers

What is negative 5 5/6–2 3/5

dangina [55]

Answer:

the answer is -8 13/30

Step-by-step explanation:

4 0

2 years ago

Show all your work to receive credit!<br><br> What is m∠KNLm∠KNL?<br><br> Question 2 options:

Nataly_w [17]

3(x+14) i think thats it

8 0

2 years ago

thomas has 4 blocks colored blue,red,green and yellow. How many different ways can he combine them on a shelf?

puteri [66]

Answer: 24

Step-by-step explanation:

Take the factorial of that number.

4! = 4*3*2*1=24

8 0

2 years ago

Find the Fourier series of f on the given interval. f(x) = 1, ?7 < x < 0 1 + x, 0 ? x < 7

Zolol [24]

$f(x)=\begin{cases}1&\text{for }-7$

The Fourier series expansion of $f(x)$ is given by

$\dfrac{a_0}2+\displaystyle\sum_{n\ge1}a_n\cos\frac{n\pi x}7+\sum_{n\ge1}b_n\sin\frac{n\pi x}7$

where we have

$a_0=\displaystyle\frac17\int_{-7}^7f(x)\,\mathrm dx$
$a_0=\displaystyle\frac17\left(\int_{-7}^0\mathrm dx+\int_0^7(1+x)\,\mathrm dx\right)$
$a_0=\dfrac{7+\frac{63}2}7=\dfrac{11}2$

The coefficients of the cosine series are

$a_n=\displaystyle\frac17\int_{-7}^7f(x)\cos\dfrac{n\pi x}7\,\mathrm dx$
$a_n=\displaystyle\frac17\left(\int_{-7}^0\cos\frac{n\pi x}7\,\mathrm dx+\int_0^7(1+x)\cos\frac{n\pi x}7\,\mathrm dx\right)$
$a_n=\dfrac{9\sin n\pi}{n\pi}+\dfrac{7\cos n\pi-7}{n^2\pi^2}$
$a_n=\dfrac{7(-1)^n-7}{n^2\pi^2}$

When $n$ is even, the numerator vanishes, so we consider odd $n$ , i.e. $n=2k-1$ for $k\in\mathbb N$ , leaving us with

$a_n=a_{2k-1}=\dfrac{7(-1)-7}{(2k-1)^2\pi^2}=-\dfrac{14}{(2k-1)^2\pi^2}$

Meanwhile, the coefficients of the sine series are given by

$b_n=\displaystyle\frac17\int_{-7}^7f(x)\sin\dfrac{n\pi x}7\,\mathrm dx$
$b_n=\displaystyle\frac17\left(\int_{-7}^0\sin\dfrac{n\pi x}7\,\mathrm dx+\int_0^7(1+x)\sin\dfrac{n\pi x}7\,\mathrm dx\right)$
$b_n=-\dfrac{7\cos n\pi}{n\pi}+\dfrac{7\sin n\pi}{n^2\pi^2}$
$b_n=\dfrac{7(-1)^{n+1}}{n\pi}$

So the Fourier series expansion for $f(x)$ is

$f(x)\sim\dfrac{11}4-\dfrac{14}{\pi^2}\displaystyle\sum_{n\ge1}\frac1{(2n-1)^2}\cos\frac{(2n-1)\pi x}7+\frac7\pi\sum_{n\ge1}\frac{(-1)^{n+1}}n\sin\frac{n\pi x}7$

3 0

2 years ago