All categories

3 years ago

How do i work this problem sin(2cos^-1 1/11)

1 answer:

5 0

$\sin\left(2\cos^{-1}\dfrac1{11}\right)=2\sin\left(\cos^{-1}\dfrac1{11}\right)\cos\left(\cos^{-1}\dfrac1{11}\right)$

When $0$ , you have $\cos(\cos^{-1}x)=x$ , so you can simplify the above slightly to get

$\sin\left(2\cos^{-1}\dfrac1{11}\right)=\dfrac2{11}\sin\left(\cos^{-1}\dfrac1{11}\right)$

Now picture a right triangle and pick one of the non-right angles. Let this angle's cosine be $\dfrac1{11}$ . This means the leg adjacent to this angle must have length 1, while the hypotenuse must have length 11. By the Pythagorean theorem, the length of the remaining leg must be

$\sqrt{11^2-1^1}=\sqrt{121-1}=\sqrt{120}=2\sqrt{30}$

This means the sine of this angle is $\dfrac{2\sqrt{30}}{11}$ , and so

$\sin\left(2\cos^{-1}\dfrac1{11}\right)=\dfrac{4\sqrt{30}}{121}$

You might be interested in

On the cross country team, 7th graders are required to run x number of miles each morning. 8th graders are required to run 15% m

Bad White [126]

Answer:

Step-by-step explanation:

7g = x miles

7 grade is x miles

8g = 7g + 15%

8 grad is 15% more than 7 grade

put what 7g is equal to into the 8g equation

8g = x + 15%

5 0

2 years ago

Kris and his dad built an ice rink. The area is 74.4 sq meters. The length is 12 meters what is the width

-BARSIC- [3]

Answer:

6.2 meters

Step-by-step explanation:

1) area = l*w

74.4 = 12x

x=6.2

3 0

2 years ago

How to prove tan z is analytic using cauchy-riemann conditions

Basile [38]

A function $f(z)=f(x+iy)=u(x,y)+i v(x,y)$ is analytic if the C-R conditions are satisfied:

$\begin{cases}u_x=v_y\\u_y=-v_x\end{cases}$

We have

$f(z)=\tan z=\tan(x+iy)$

Recall the angle sum formula for tangent:

$\tan(x+iy)=\dfrac{\tan x+\tan iy}{1-\tan x\tan iy}$

Now recall that

$\tan iy=\dfrac{\sin iy}{\cos iy}=\dfrac{-\frac{e^y-e^{-y}}{2i}}{\frac{e^y+e^{-y}}2}=\dfrac{i\sinh y}{\cosh y}=i\tanh y$

So we have

$\dfrac{\tan x+\tan iy}{1-\tan x\tan iy}=\dfrac{\tan x+i\tanh y}{1-i\tan x\tanh y}$
$=\dfrac{(\tan x+i\tanh y)(1+i\tan x\tanh y)}{(1-i\tan x\tanh y)(1+i\tan x\tanh y)}$
$=\dfrac{\tan x+i\tanh y+i\tan^2x-\tan x\tanh^2y}{1+\tan^2x\tanh^2y}$
$=\dfrac{\tan x(1-\tanh^2y)}{1+\tan^2x\tanh^2y}+i\dfrac{\tanh y(1+\tan^2x)}{1+\tan^2x\tanh^2y}$
$=\underbrace{\dfrac{\tan x\sech^2y}{1+\tan^2x\tanh^2y}}_{u(x,y)}+i\underbrace{\dfrac{\tanh y\sec^2x}{1+\tan^2x\tanh^2y}}_{v(x,y)}$

We could stop here, but taking derivatives may be messy and would be easier to do if we can write this in terms of sines and cosines.

$u(x,y)=\dfrac{\tan x\sech^2y}{1+\tan^2x\tanh^2y}=\dfrac{\frac{\sin x}{\cos x\cosh^2x}}{1+\frac{\sin^2x\sinh^2y}{\cos^2x\cosh^2y}}=\dfrac{\sin x\cos x}{\cos^2x\cosh^2y+\sin^2x\sinh^2y}$
$u(x,y)=\dfrac{\sin2x}{2\cos^2x\cosh^2y+2(1-\cos^2x)(\cosh^2y-1)}=\dfrac{\sin2x}{2\cosh^2y-1+2\cos^2x-1}$
$u(x,y)=\dfrac{\sin2x}{\cos2x+\cosh2y}$

With similar usage of identities, we can find that

$v(x,y)=\dfrac{\sinh2y}{\cos2x+\cosh2y}$

Now we check the C-R conditions.

$u_x=\dfrac{2\cos2x(\cos2x+\cosh2y)-(-2\sin2x)\sin2x}{(\cos2x+\cosh2y)^2}=\dfrac{2+2\cos2x\cosh2y}{(\cos2x+\cosh2y)^2}$
$v_y=\dfrac{2\cosh2y(\cos2x+\cosh2y)-(2\sinh2y)\sinh2y}{(\cos2x+\cosh2y)^2}=\dfrac{2+2\cosh2y\cos2x}{(\cos2x+\cosh2y)^2}$
$\implies u_x=v_y$

Similarly, you can check that $u_y=-v_x$ , hence the C-R conditions are satisfied, and so $\tan z$ is analytic.