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Write cos(2 tan⁻¹(x)) as an algebraic expression in x.

\(\begin{array}{|rcll|} \hline && \cos(~2\cdot \underbrace{ tan^{-1}(x) }_{=\varphi}~) \qquad & | \qquad tan^{-1}(x) = \varphi \text{ or } x = \tan(\varphi) \\\\ \text{Formula:} \\ \cos(~2\cdot \varphi) &=& \dfrac{1-\tan^2(\varphi)}{1+\tan^2(\varphi)} \qquad & | \qquad \tan(\varphi) = x\\ \cos(~2\cdot \varphi) &=& \dfrac{1-x^2 }{1+x^2} \\ \mathbf{ \cos(~2\cdot \tan^{-1}(x) ) }& \mathbf{=} & \mathbf{\dfrac{1-x^2 }{1+x^2}} \\ \hline \end{array} \)

I drew a right angled triangle to help me answer this question.  sides 1,x and sqrt(1+x^2)

If

\(if\;\;x\ge0\;\;and\;\;\theta=tan^{-1}x\\then\\sin\theta=\ \frac{x}{\sqrt{1+x^2}}\\ cos\theta = \frac{1}{\sqrt{1+x^2}}\\ cos(tan^{-1}x)\\=cos(2\theta)\\=cos^2\theta-sin^2\theta\\ =\frac{1}{1+x^2}-\frac{x^2}{1+x^2}\\ =\frac{1-x^2}{1+x^2}\)

EDIT:

It's actually PERPENDICULAR, not PARALLEL.

... I've been working on this for so much time and was wondering why k= 11/9 wasn't my answer haha

k=11/9 is the answer provided in the solution of the book.

Happy birthday sir!

A huge thanks for all the amazing work and help you bring here!

I wish you only the best!

Hi Kulki   

It is nice to meet you :)

limit( ((((x+2)/(x-2)))^(x+1)), x=infinity )

Mmm  it is good that you use lots of brackets but maybe if some of them were square brackets it might be easier to make sense of.   Not to worry, lots of backets is good.

\(\displaystyle\lim_{x\rightarrow\infty}\;\left[ \frac{x+2}{x-2}       \right]^{x+1}\)

\(\displaystyle\lim_{x\rightarrow\infty}\;\left[ \frac{x+2}{x-2}       \right]^{x+1}\\ =\displaystyle\lim_{x\rightarrow\infty}\;e^{ln\left[ \frac{x+2}{x-2}       \right]^{x+1}}\\ =\displaystyle\lim_{x\rightarrow\infty}\;e^{(x+1)ln\left[ \frac{x+2}{x-2}       \right]}\\ =\displaystyle\lim_{x\rightarrow\infty}\;e^{x(1+\frac{1}{x})ln\left[ \frac{x+2}{x-2}       \right]}\\ =\displaystyle\lim_{x\rightarrow\infty}\;e^{x(\frac{x+1}{x})ln\left[ \frac{x+2}{x-2}       \right]}\\ =\displaystyle\lim_{x\rightarrow\infty}\;e^{xln\left[ \frac{x+2}{x-2}      \right]*(\frac{x+1}{x}) }\\ \text{The limit as x tends to infinity of (x+1)/x = 1 so}\\ =\displaystyle\lim_{x\rightarrow\infty}\;e^{xln\left[ \frac{x+2}{x-2}      \right]}\\ =e^\left[{\displaystyle\lim_{x\rightarrow\infty}\;{xln\left[ \frac{x+2}{x-2}       \right]}}\right]\\ \)

Now  I will look at the nw limit

\(\displaystyle\lim_{x\rightarrow\infty} \frac{ln\left[ \frac{x+2}{x-2} \right]}{\frac{1}{x}}\\ \qquad \text{The numerator and the denominator both tend to 0 so I can use L'Hopital's rule}\\ \qquad \frac{d}{dx}\;ln\left[ \frac{x+2}{x-2}\right]=\frac{(x-2)-(x+2)}{(x-2)^2}=\frac{-4}{(x-2)^2}\\ \qquad \frac{d}{dx}\;x^{-1}=-x^{-2}=\frac{-1}{x^2}\\ \qquad =\displaystyle\lim_{x\rightarrow\infty}\left[ \frac{-4}{(x-2)^2}\div \frac{-1}{x^2}\right]\\ \qquad=\displaystyle\lim_{x\rightarrow\infty}\left[ \frac{4x^2}{(x-2)^2}\right]\\ \qquad=\displaystyle\lim_{x\rightarrow\infty}\;4\left[ \frac{x^2}{(x-2)^2}\right]\\ \qquad=\displaystyle\lim_{x\rightarrow\infty}\;4\left[ \frac{x^2}{x^2-4x+4} \;\;\;\frac{\div x^2}{\div x^2}\right]\\ \qquad=\displaystyle\lim_{x\rightarrow\infty}\;4\left[ \frac{1}{1-\frac{4}{x}+\frac{4}{x^2}} \right]\\ \qquad=4\left(\frac{1}{1-0+0}\right)\\ \qquad=4\\ \text{so the original limit}=e^4 \)

\(\displaystyle\lim_{x\rightarrow\infty}\;\left[ \frac{x+2}{x-2}       \right]^{x+1}=e^4\)

1       2     3     4     6     9

108  54   36   27   18   12  

y= mx + b   = equation for a line

m= slope = (y2-y1)/(x2-x1) =  (sqrt2-9)/(5-3) = (sqrt2-9)/2

y = (sqrt2-9)/2  x    +  b

9 = (sqrt2-9)/2  (3) + b

9 = (3 sqrt2-27)/2 + b

b =  9 - (3sqrt2-27)/2

   =  18/2  - (3sqrt2)/2 + 27/2

b= 45/2 - (3sqrt2)/2

so equation becomes    y = (sqrt2-9)/2  x  +   45/2  - (3sqrt2)/2       Multiply everything by '2'

2y = (sqrt2-9)x + 45 - (3sqrt2)      subtract 2y from both sides and you will see your answer !

3x^2 + 36 = 0      divide both sides by 3

x^2 +12 = 0

x^2 = -12

x = +- 2i sqrt3

The slope, m ,  is given by   tan(58) = 1.600334529041  ~ 1.6

y= 1.6x + b        and   -3,0 is a point     substituting

0 =1.6(-3) +b

b= 4.8

y= 1.6 x + 4.8   is your line

144 duhh!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

r=4    each subsequent term is 4x the previous one

Yet ANOTHER aproach

7 (sqrt(10)(sqrt(14)  / (6 sqrt (35)

=  7 sqrt(140) / 6 sqrt(35)

=7 sqrt(4) / 6 = 7*2 /6 = 14/6 = 7/3

The dental kind of calculus....or the mathematical type of calculus ????     LOL

Hi Tony :)

7 (sqrt (10))  (sqrt (14) ) ) / (6 (sqrt (35) ) )

\(\frac{7 \sqrt {10} \sqrt {14}}{6 \sqrt {35} }\\ =\frac{7\sqrt{2} \sqrt {5} \sqrt {2}\sqrt{7}}{6 \sqrt {5}\sqrt{7} }\\ =\frac{7\sqrt{2} \sqrt {2}}{6 }\\ =\frac{7*2}{6 }\\ =\frac{7}{3 }\\ =2\frac{1}{3}\)

You will need to know the density (mg/ml)  to calculate this....or you can do a displacement test where you put the sample into a graduated beaker full of water (if comatible) and see how much the water irs raised in ml.

x * x = 3x

x^2 = 3x

x^2 -3x=0

x(x-3) = 0      so x = 0  or 3

( ( 3 + sqrt(5) )  ( sqrt(5) -2 )  ) / (5 - sqrt (5) )

I know the answer is    sqrt (5) / 5

\(\frac{(3+\sqrt 5)(\sqrt 5-2)}{5-\sqrt 5}\)

\(=\frac{3\sqrt 5-6+5-2\sqrt5}{5-\sqrt5}\)          expand to the 3rd Binom

\(=\frac{\sqrt5-1}{5-\sqrt5}\times \frac{5+\sqrt5}{5+\sqrt5}\)        \(\frac{multiply}{3rdBinom}\)

\(=\frac{5\sqrt5+5-5-\sqrt5}{25-5}\)           add, subtract

\(=\frac{4\sqrt5}{20}\)                           shortened

 \(\frac{(3+\sqrt 5)(\sqrt 5-2)}{5-\sqrt 5}\)\(=\frac{\sqrt 5}{5}\)     !

3x + xi - 2y = 12-iy-i    add iy  and i to both sides

3x + xi -2y +iy +i = 12    and 'reverse distributive property'

3x + i(x+y+1) -2y = 12    now you cannot do much more unless you have another equation relating x and y 

                                           to reach a discrete solution for x,y  (though you can solve for x in terms of y    or y in terms of x as guest 1  did above....but you cannot find an answer for x,y)

or you could say   x(3+i) - y(2-i)  + i = 12

Solve for x:
(3 + i) x - 2 y = (12 - i) + (0 - i) y

(3 + i) x - 2 y = (3 + 1 i) x - 2 y and (12 - i) + (0 - i) y = (0 - i) y + 12 - i:
(3 + 1 i) x - 2 y = (0 - i) y + 12 - i

Add 2 y to both sides:
(3 + i) x = (2 - i) y + 12 - i

Divide both sides by 3 + i:
Answer: |x = (1/2 - i/2) y + 7/2 - (3 i)/2

Let's work on numerator first    expand

3sqrt5 - 6 +5 -2sqrt5 = sqrt5 -1

so you have  ( sqrt5-1)  / ( 5-sqrt5)   now multiply and expand by  (5+sqrt5)/(5+sqrt5)=

(5sqrt5 +5 -5 -sqrt5 ) /  (25 + 5 sqrt5-5sqrt5 -5)     Simplify by collecting like terms

4sqrt5 / 20        and now, simplify the fraction

sqrt5 / 5

Simplify the following:
(7 sqrt(10) sqrt(14))/(6 sqrt(35))

sqrt(10) sqrt(14) = sqrt(10×14):
(7 sqrt(10×14))/(6 sqrt(35))

10×14 = 140:
(7 sqrt(140 ) )/(6 sqrt(35))

sqrt(140) = sqrt(2^2×35) = 2 sqrt(35):
(7×2 sqrt(35))/(6 sqrt(35))

(7×2 sqrt(35))/(6 sqrt(35)) = (sqrt(35))/(sqrt(35))×(7×2)/6 = (7×2)/6:
(7×2)/6

2/6 = 2/(2×3) = 1/3:
Answer: |7/3

Volume=4/3 x Pi x r^3, r=radius

Use the key labled "x^y", and then press = key to get your answer.

Find the following limit:
lim_(x->∞) ((x + 2)/(x - 2))^(x + 1)

lim_(x->∞) ((x + 2)/(x - 2))^(x + 1) = lim_(x->∞) exp(log(((x + 2)/(x - 2))^(x + 1))) = lim_(x->∞) exp((x + 1) log((x + 2)/(x - 2))):
lim_(x->∞) exp(log((x + 2)/(x - 2)) (x + 1))

lim_(x->∞) exp((x + 1) log((x + 2)/(x - 2))) = exp(lim_(x->∞) (x + 1) log((x + 2)/(x - 2))):
exp(lim_(x->∞) log((x + 2)/(x - 2)) (x + 1))

lim_(x->∞) (x + 1) log((x + 2)/(x - 2)) = lim_(x->∞) x log((x + 2)/(x - 2)) (x + 1)/x:
exp(lim_(x->∞) x log((x + 2)/(x - 2)) (x + 1)/x)

By the product rule, lim_(x->∞) x (x + 1)/x log((x + 2)/(x - 2)) = (lim_(x->∞) (x + 1)/x) (lim_(x->∞) x log((x + 2)/(x - 2))):
exp(lim_(x->∞) x log((x + 2)/(x - 2)) lim_(x->∞) (x + 1)/x)

The leading term in the denominator of (x + 1)/x is x. Divide the numerator and denominator by this:
exp(lim_(x->∞) x log((x + 2)/(x - 2)) lim_(x->∞) (1 + 1/x)/1)

The expression 1/x tends to zero as x approaches ∞:
exp(lim_(x->∞) x log((x + 2)/(x - 2)))

To prepare the product x log((x + 2)/(x - 2)) for solution by l'Hôpital's rule, write it as (log((x + 2)/(x - 2)))/(1/x):
exp(lim_(x->∞) (log((x + 2)/(x - 2)))/(1/x))

Applying l'Hôpital's rule, we get that
lim_(x->∞) (log((x + 2)/(x - 2)))/(1/x) | = | lim_(x->∞) ( d/( dx) log((x + 2)/(x - 2)))/( d/( dx)(1/x))
 | = | lim_(x->∞) (((x - 2) (1/(x - 2) - (x + 2)/(x - 2)^2))/(x + 2))/(-1/x^2)
 | = | lim_(x->∞) (4 x^2)/((x - 2) (x + 2))
exp(lim_(x->∞) (4 x^2)/((x - 2) (x + 2)))

Replace -2 by 0 in 1/(x - 2):
exp(lim_(x->∞) (4 x^2)/(x (x + 2)))

The leading term in the denominator of (4 x)/(x + 2) is x. Divide the numerator and denominator by this:
exp(lim_(x->∞) 4/(1 + 2/x))

The expression 2/x tends to zero as x approaches ∞:
Answer: |e^4

Simplify the following:
((3 + sqrt(5)) (sqrt(5) - 2))/(5 - sqrt(5))

(3 + sqrt(5)) (sqrt(5) - 2) = 3 (-2) + 3 sqrt(5) + sqrt(5) (-2) + sqrt(5) sqrt(5) = -6 + 3 sqrt(5) - 2 sqrt(5) + 5 = sqrt(5) - 1:
(sqrt(5) - 1)/(5 - sqrt(5))

Multiply numerator and denominator of (sqrt(5) - 1)/(5 - sqrt(5)) by 5 + sqrt(5):
((sqrt(5) - 1) (5 + sqrt(5)))/((5 - sqrt(5)) (5 + sqrt(5)))

(5 - sqrt(5)) (5 + sqrt(5)) = 5×5 + 5 sqrt(5) - sqrt(5)×5 - sqrt(5) sqrt(5) = 25 + 5 sqrt(5) - 5 sqrt(5) - 5 = 20:
((sqrt(5) - 1) (5 + sqrt(5)))/(20)

(sqrt(5) - 1) (5 + sqrt(5)) = -5 - sqrt(5) + sqrt(5)×5 + sqrt(5) sqrt(5) = -5 - sqrt(5) + 5 sqrt(5) + 5 = 4 sqrt(5):
(4 sqrt(5))/(20)

4/20 = 4/(4×5) = 1/5:
Answer: |(sqrt(5))/(5)