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Anestetic [448]
4 years ago
13

PageProve thatTan ( 2 Tan^-l x)=2 Tan ( Tan^-1 x + Tan^-1 x^3)​

Mathematics
1 answer:
Kay [80]4 years ago
5 0

Answer:

The answers are 0, 1 and −2.

Step-by-step explanation:

Let α=arctan(2tan2x) and β=arcsin(3sin2x5+4cos2x).

sinβ2tanβ21+tan2β2tanβ21+tan2β23tanxtan2β2−(9+tan2x)tanβ2+3tanx(3tanβ2−tanx)(tanβ2tanx−3)tanβ2=3sin2x5+4cos2x=3(2tanx1+tan2x)5+4(1−tan2x1+tan2x)=3tanx9+tan2x=0=0=13tanxor3tanx

Note that x=α−12β.

tanx=tan(α−12β)=tanα−tanβ21+tanαtanβ2=2tan2x−13tanx1+2tan2x(13tanx)or2tan2x−3tanx1+2tan2x(3tanx)=tanx(6tanx−1)3+2tan3xor2tan3x−3tanx(1+6tanx)

So we have tanx=0, tan3x−3tanx+2=0 or 4tan3x+tan2x+3=0.

Solving, we have tanx=0, 1, −1 or −2.

Note that −1 should be rejected.

tanx=−1 is corresponding to tanβ2=3tanx. So tanβ2=−3, which is impossible as β∈[−π2,π2].

The answers are 0, 1 and −2.

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Suppose that \nabla f(x,y,z) = 2xyze^{x^2}\mathbf{i} + ze^{x^2}\mathbf{j} + ye^{x^2}\mathbf{k}. if f(0,0,0) = 2, find f(1,1,1).
lesya [120]

The simplest path from (0, 0, 0) to (1, 1, 1) is a straight line, denoted C, which we can parameterize by the vector-valued function,

\mathbf r(t)=(1-t)(\mathbf i+\mathbf j+\mathbf k)

for 0\le t\le1, which has differential

\mathrm d\mathbf r=-(\mathbf i+\mathbf j+\mathbf k)\,\mathrm dt

Then with x(t)=y(t)=z(t)=1-t, we have

\displaystyle\int_{\mathcal C}\nabla f(x,y,z)\cdot\mathrm d\mathbf r=\int_{t=0}^{t=1}\nabla f(x(t),y(t),z(t))\cdot\mathrm d\mathbf r

=\displaystyle\int_{t=0}^{t=1}\left(2(1-t)^3e^{(1-t)^2}\,\mathbf i+(1-t)e^{(1-t)^2}\,\mathbf j+(1-t)e^{(1-t)^2}\,\mathbf k\right)\cdot-(\mathbf i+\mathbf j+\mathbf k)\,\mathrm dt

\displaystyle=-2\int_{t=0}^{t=1}e^{(1-t)^2}(1-t)(t^2-2t+2)\,\mathrm dt

Complete the square in the quadratic term of the integrand: t^2-2t+2=(t-1)^2+1=(1-t)^2+1, then in the integral we substitute u=1-t:

\displaystyle=-2\int_{t=0}^{t=1}e^{(1-t)^2}(1-t)((1-t)^2+1)\,\mathrm dt

\displaystyle=-2\int_{u=0}^{u=1}e^{u^2}u(u^2+1)\,\mathrm du

Make another substitution of v=u^2:

\displaystyle=-\int_{v=0}^{v=1}e^v(v+1)\,\mathrm dv

Integrate by parts, taking

r=v+1\implies\mathrm dr=\mathrm dv

\mathrm ds=e^v\,\mathrm dv\implies s=e^v

\displaystyle=-e^v(v+1)\bigg|_{v=0}^{v=1}+\int_{v=0}^{v=1}e^v\,\mathrm dv

\displaystyle=-(2e-1)+(e-1)=-e

So, we have by the fundamental theorem of calculus that

\displaystyle\int_C\nabla f(x,y,z)\cdot\mathrm d\mathbf r=f(1,1,1)-f(0,0,0)

\implies-e=f(1,1,1)-2

\implies f(1,1,1)=2-e

3 0
3 years ago
An online furniture store sells chairs for $200 each and tables for $450 each. Every day, the store can ship a maximum of 26 pie
Sophie [7]

Answer:

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Step-by-step explanation:

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3 years ago
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serious [3.7K]

Answer:

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Step-by-step explanation:

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2 years ago
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qaws [65]

Answer:

2/3x+5=y

Step-by-step explanation:

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mx+5=y.

Now all we need to do is plug in some points.

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(Subtract 5 from both sides)

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Vinil7 [7]
When you multiply 35 x 40 = 1400

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