giving brainliestsuppose cos(x)=-1/3, where pi/2=<x=<pi. what is the value of tan(2x)?

Mathematics

1 answer:

qaws [65]2 years ago

4 0

Answer:

4√2/7

Step-by-step explanation:

Suppose cos(x)=-1/3, where pi/2≤x≤ pi

tan(2x) = sin(2x)/cos(2x)

Since sin2x = 2sinx cosx

cos 2x = 2cos²x-1

Substitute

tan(2x) = 2sinx cosx/ 2cos²x-1

Given cosx = -1/3

adjacent = -1

hyp = 3

opp = √3²-(-1)²

opp = √9-1

opp = √8

sinx = opp/hyp

sinx = √8/3

tan(2x) = 2sinx cosx/ 2cos²x-1

tan(2x) = 2(√8/3)(-1/3)/ 2(-1/3)²-1

tan2x= (-2√8/9 )÷ (2/9 - 1)

tan2x = -4√2/9 * 9/-7

tan 2x = 4√2/7

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F=(2xy +z³)i + x³j + 3xz²k find a scalar potential and work done in moving an object in the field from (1,-2,1) to (3,1,4)

Alex73 [517]

Step-by-step explanation:

Given:

$\textbf{F} = (2xy + z^3)\hat{\textbf{i}} + x^3\hat{\textbf{j}} + 3xz^2\hat{\textbf{k}}$

This field will have a scalar potential $\varphi$ if it satisfies the condition $\nabla \times \textbf{F}=0$ . While the first x- and y- components of $\nabla \times \textbf{F}$ are satisfied, the z-component doesn't.

$(\nabla \times \textbf{F})_z = \left(\dfrac{\partial F_y}{\partial x} - \dfrac{\partial F_x}{\partial y} \right)$

$\:\:\:\:\:\:\:\:\: = 3x^2 - 2x \ne 0$

Therefore the field is nonconservative so it has no scalar potential. We can still calculate the work done by defining the position vector $\vec{\textbf{r}}$ as

$\vec{\textbf{r}} = x \hat{\textbf{i}} + y \hat{\textbf{j}} + z \hat{\textbf{k}}$

and its differential is

$\textbf{d} \vec{\textbf{r}} = dx \hat{\textbf{i}} + dy \hat{\textbf{j}} + dz \hat{\textbf{k}}$

The work done then is given by

$\displaystyle \oint_c \vec{\textbf{F}} • \textbf{d} \vec{\textbf{r}} = \int ((2xy + z^3)\hat{\textbf{i}} + x^3\hat{\textbf{j}} + 3xz^2\hat{\textbf{k}}) • (dx \hat{\textbf{i}} + dy \hat{\textbf{j}} + dz \hat{\textbf{k}})$