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its 15

tan²θ + 1/tan²θ=2 then value of θ=?

\(\begin{array}{rcll} \tan^2(\phi) + \frac{ 1 } { \tan^2(\phi) } &=& 2 \qquad & |\qquad \cdot \tan^2(\phi)\\ \tan^4(\phi) + 1 &=& 2 \cdot \tan^2(\phi) \qquad & |\qquad -2\cdot \tan^2(\phi)\\ \tan^4(\phi) -2\cdot \tan^2(\phi) + 1 &=& 0 \qquad & |\qquad \text{binomial}\\ ( \tan^2(\phi) - 1 ) ^2 &=& 0 \qquad & |\qquad \sqrt{} \\ \tan^2(\phi) - 1 &=& 0 \qquad & |\qquad +1 \\ \tan^2(\phi) &=& 1 \qquad & |\qquad \pm\sqrt{} \\ \tan(\phi) &=& \pm\sqrt{1} \\ \tan(\phi) &=& \pm 1 \qquad & |\qquad \arctan{()} \\ \phi &=& \arctan{(\pm 1)} \pm k\cdot \pi \qquad & |\qquad k = 0,1,2,\cdots \\ \phi &=& \pm\arctan{(1)} \pm k\cdot \pi \qquad & |\qquad \arctan{(1)} = \frac{\pi}{4} \\ \phi &=& \pm\frac{\pi}{4} \pm k\cdot \pi \\ \phi_1 &=& \frac{\pi}{4} \pm k\cdot \pi \\ \phi_2 &=& -\frac{\pi}{4} \pm k\cdot \pi \\\\ \phi &=& \cdots -\frac34\pi, -\frac{\pi}{4}, \frac{\pi}{4}, \frac34\pi, \cdots \end{array}\)

x^3+2x=90

x=~4.3327

Hello guest!

8+7-6x5 divided by two cubed? the answer isn't 11.25

Is two cubed 2³ ? Than is

(8+7-6x5) / 2³ = -1.875

(8+7-6x5) / -1.333.. = 11.25

asinus :- )

  !

Hello!

cosine(34)=10/x

cos(34) = 10 / x

x = 10 / cos 34

x = 0 / -0.84857

x = -11.7845278077

cos (10 * pi)  = 10 / x

x = 10 / cos (10 * pi) 

x = 10 / 1

x = 10

cos(34°) = 10 / x

x = 10 / cos 34°

x = 10 / 0.8290376

x = 12.0622

asinus :- )

  !

Is that x to the power of three?

tan^2(x) +1/tan^2(x)=2

use trig identities to solve.since

tan^2(x) =sec^2(x) -1

then write

sec^2(x)-1 +1/(sec^2(x)-1)=2

so     a + 1/a = 2     where a = sec^2(x)-1

solve quadratic in a

a^2-2a+1=0   gives (a-1)^2=0,so a=1  giving sec^2(x)=2

sec(x)=plus or minus root2

cos(x) = 1/(plus or minus root 2)

x= 45 deg, 315 deg, etc

30+7 s > 2   

Subtract 30 from both sides

7s > -28

Divide both sides by 7

s > -4

tan(45°) = 1   so   tan(45°) + 1/tan(45°) = 1 + 1 → 2

θ = 45°   (or θ = pi/4  radians)

.

remainder of 2^2011/13

\(\begin{array}{rcll} 2^{2011} \pmod {13} &=& \ ? \end{array}\)

\(\small{ \begin{array}{lrcll} 1. & gcd(13,2) &=& 1 \qquad | \qquad 13 \text{ and } 2 \text{ are relatively prim } \\ 2. & 13 \text{ is a prim number } \\ 3. & \phi() \text{ is Euler's totient function, Euler's phi function }\\ & \phi(p) &=& p-1 \qquad p \text{ is a prim number} \\ & \phi(13) &=& {\color{red}12} \\ 4. & 2^{\phi(13)} &\equiv& 1 \pmod{13} \\ & 2^{{\color{red}12}} &\equiv& 1 \pmod{13} \\ \hline &\text{ Let } \phi(n) \text{ denote the totient function. } \\ &\text{Then } a^{\phi(n)} \equiv 1 \pmod {n} \text{ for all } a \text{ relatively prime to } n. \\ \end{array} }\)

\(\begin{array}{rcll} 5. & 2011 &=& {\color{red}12}\cdot 167 + 7 \\ & 2^{2011 } \pmod{13} &=& 2^{ {\color{red}12}\cdot 167 + 7 } \pmod{13} \\ & &=& ( 2^{ {\color{red}12} } )^{167}\cdot 2^{7} \pmod{13} \\ & &\equiv& ( 1 )^{167}\cdot 2^{7} \pmod{13} \\ & &\equiv& 1\cdot 2^{7} \pmod{13} \\ & &\equiv& 2^{7} \pmod{13} \\ & &\equiv& 128 \pmod{13} \\ & &\equiv& 11 \pmod{13} \end{array} \)

The remainder of \(\frac{2^{2011}} {13}\) is \(\mathbf{11}\)

Lol. I apologize, and I was able to understand well. Thank you!

(Re-posted with v= 4 x 10^3   NOT  NOT  4 x 103 m/s)

OK.....the y velocity is .0135/t    where t is the time it took to travel 56 cm at   (4 x 10^3m/s)  = .56/4000

.0135/ (.56/4000) = ( .0135 x 4000) /.56 = 96.429 m/s

This   velocity change from ZERO to this value occured in the time it took to cover the plate length of 26cm (.26m) at 4 x 10^3 m/s.

   .26/4000 = 6.5 x10^5 seconds

Acceleration in the y direction is thus the change in

velocity/time = (96.429)/.6.5 x 10^-5 = 1483523  m/s^2

Since F= ma    Thus the force is   1483523 ( m )

Lorentz force law :  F= q(E+ v x B) = 1483523 m    divide both sides by m

q/m (E + v x B) = 1483523

or rewritten:

q/m = 1483523/ (E + v x B)   We are given E  and v which we can substitute....but we still do not have 'B'   the magnetic flux density......hmmmmm   .... NOW I'M STUCK!    Wonder if I was given all of the information? 

Think I might be on the wrong path???

All well and good....but what is your question?

Example :

sin (25 )  =  13 / x         multiply both sides by x, divide both sides by sin 25

x = 13 / sin (25)    =  about 30.76  

(4w+3)^2 = (4w+3)(4w+3) =  16w^2 + 12w + 12w + 9 =  16w^2 + 24w + 9

OK.....the y velocity is .0135/t    where t is the time it took to travel 56 cm at   (4 x 103m/s)  = .56/412

.0135/ (.56/412) = ( .0135 x 412) /.56 = 9.932 m/s

This   velocity change from ZERO to this value occured in the time it took to cover the plate length of 26cm (.26m) at 4 x 103 m/s.

(is this written properly?  4 x 103 m/s?)        412m/s (pretty slow)    .26/412 = .00063107 seconds

Acceleration in the y direction is thus the change in

velocity/time = (9.932)/.00063107 = 15738.4 m/s^2

Since F= ma    Thus the force is    15738.4 ( m )

Lorentz force law :  F= q(E+ v x B) = 15738.4 m    divide both sides by m

q/m (E + v x B) = 15738.4

or rewritten:

q/m = 15738.4/ (E + v x B)   We are given E  and v which we can substitute....but we still do not have 'B'   the magnetic flux density......hmmmmm   .... NOW I'M STUCK!    Wonder if I was given all of the information?   And was the initial speed really  4 x 103 m/s ?

Think I might be on the wrong path???

Here's a good explanation with some examples :  http://regentsprep.org/Regents/math/algtrig/ATT12/lawofsinesAmbiguous.htm

If x=1 this is true

What's a 'revolution sign' ?

Maybe.....Maybe not.     Though it is 'I'   not 'i'  and it's 'whoops'  not 'whops'

sqrt2/2   or .707

sqrt2/2   or .707

1

Pretty easy to do using unit circle concepts and   sin/cos = tan

you can do these in your head ....

I would guess that some one got a whole bunch of circles and then measured the cucmfrance, after this they measured each ones diameter. with these two they came up with a value that was universal among all circles when multiplied by the diameter. as i said thats my guess probably not right but eh

idk, but I love eating it!,...

and using it so I don't get to fat!

Maybe it was just always there