heureka

a. Show, by expanding, that 3¹⁰=9⁵

b. Show, using index laws, that 3¹⁰=9⁵

Don't they just mean the same thing?

I assume

a.

$\begin{array}{rcll} 3^{10} &=& 9^5 \\ 3\cdot 3\cdot 3\cdot 3\cdot 3\cdot 3\cdot 3\cdot 3\cdot 3\cdot 3 &=& 9\cdot 9\cdot 9\cdot 9\cdot 9 \\ 59049 &=& 59049 \end{array}$

b.

$\begin{array}{rcll} 3^{10} &=& 9^5 \\ 3^{2\cdot 5} &=& 9^5 \\ (3^2)^5 &=& 9^5 \\ 9^5 &=& 9^5 \\ \end{array}$

The windows of a downtown office building are arranged so that each floor has six fewer windows than the floor below it. If the ground floor has 52 windows, how many total windows are on the first eight floors?

arithmetic series:

 $a_1 = 52,~d = +6$

formular:

$a_n = a_1+(n-1)\cdot d$

$\begin{array}{rcll} a_n~?\\ a_n &=& 52+(n-1)\cdot 6 \\ a_n &=& 52+6n-6 \\ \mathbf{a_n} & \mathbf{=} & \mathbf{46+6n} \end{array}$

sum:

$\begin{array}{rcll} s_n~? \\ s_n &=& \left( \frac{a_1+a_n}{2} \right) \cdot n \\ s_n &=& \left( \frac{ 52+(46+6n) }{2} \right)\cdot n \\ \mathbf{s_n} & \mathbf{=} & \mathbf{\left( \frac{ 98+6n }{2}\right)\cdot n} \end{array}$

total windows on the first eight floors?

n=8

$\begin{array}{rcll} s_8~? \\ s_8 & = & \left( \frac{ 98+6\cdot8 }{2}\right)\cdot 8 \\ s_8 & = & \left( \frac{ 98+48 }{2}\right)\cdot 8 \\ s_8 & = & \left( \frac{ 146 }{2}\right)\cdot 8 \\ s_8 & = & 146 \cdot 4 \\ \mathbf{s_8} & \mathbf{=} & \mathbf{584} \\ \end{array}$

On the first eight floors are total 584 windows.

If mtan(A-30)=ntan(A+120), then prove that-

2*cos2A=[m+n]/[m-n]

$\begin{array}{rcll} m\cdot \tan(A-30^{\circ} ) &=& n\cdot \tan(A+120^{\circ}) \\ &=& n\cdot \tan(A+90^{\circ}+ 30^{\circ}) \\ &=& n\cdot \tan[90^{\circ}+( A+30^{\circ}) ] \quad | \quad \tan(90^{\circ}+x) = - \cot(x)\\ &=& -n\cdot \cot( A+30^{\circ}) \\ m\cdot \tan(A-30^{\circ} ) &=& -n\cdot \frac{1}{\tan( A+30^{\circ}) } \\ m\cdot\tan( A+30^{\circ}) \cdot \tan(A-30^{\circ} ) &=& -n \\ \tan( A+30^{\circ}) \cdot \tan(A-30^{\circ} ) &=& -\frac{n}{m} \quad | \quad \boxed{~ q = -\frac{n}{m} ~}\\ \tan( A+30^{\circ}) \cdot \tan(A-30^{\circ} ) &=& q \\ \end{array}$

$\begin{array}{rcll} \tan( A+30^{\circ}) \cdot \tan(A-30^{\circ} ) &=& \left( \frac{ \tan(A) +\tan(30^{\circ} ) } { 1-\tan(A)\tan(30^{\circ}) } \right) \cdot \left( \frac{ \tan(A) -\tan(30^{\circ} )} { 1+\tan(A)\tan(30^{\circ}) } \right) \\ &=& \frac{\tan^2(A) -\tan^2(30^{\circ}) } { 1-\tan^2(A)\tan^2(30^{\circ}) } \end{array}$

$\boxed{~ \begin{array}{rcll} \frac{\tan^2(A) -\tan^2(30^{\circ}) } { 1-\tan^2(A)\tan^2(30^{\circ}) } &=& q \end{array} ~}$

$\begin{array}{rcll} \frac{m+n}{m-n} &=& \frac{2m}{m-n}-1 \\ &=& \frac{ 2 } { \frac{m}{m} - \frac{n}{m} } - 1 \\ &=& \frac{ 2 } { 1 - \frac{n}{m} } - 1 \quad | \quad q = -\frac{n}{m}\\ &=& \frac{ 2 } { 1 +q } - 1 \\ &=& \frac{ 2 - (1+q)} { 1 +q } \\ &=& \frac{ 2 - 1-q} { 1 +q } \\ &=& \frac{ 1-q} { 1 +q } \\ \boxed{~ \frac{m+n}{m-n} = \frac{ 1-q} { 1 +q } ~} \end{array}$

$\begin{array}{rcll} \dfrac{m+n}{m-n} &=& \dfrac{ 1-q} { 1 +q } \\\\ &=& \dfrac{ 1-\frac{\tan^2(A) -\tan^2(30^{\circ}) } { 1-\tan^2(A)\tan^2(30^{\circ}) }} { 1 +\frac{\tan^2(A) -\tan^2(30^{\circ}) } { 1-\tan^2(A)\tan^2(30^{\circ}) } } \\\\ &=& \dfrac{ 1-\tan^2(A)\tan^2(30^{\circ}) -\tan^2(A)+ \tan^2(30^{\circ}) } { 1-\tan^2(A)\tan^2(30^{\circ}) +\tan^2(A)- \tan^2(30^{\circ}) } \\\\ \dfrac{m+n}{m-n} &=& \dfrac{ [ 1-\tan^2(A)] [ 1+\tan^2(30^{\circ})] } { [ 1+\tan^2(A)] [ 1-\tan^2(30^{\circ})] } \end{array}$

$\begin{array}{rcll} \sin^2(A)+\cos(A)^2 &=& 1 \quad | \quad :\cos^2(A) \\ \frac{\sin^2(A)}{\cos^2(A)} + \frac{\cos^2(A)}{\cos^2(A)} &=& \frac{1}{\cos^2(A)} \\ \boxed{~ \tan^2(A) + 1 = \frac{1}{\cos^2(A)} ~} \end{array}$

$\begin{array}{rcll} \cos(A)^2 - \sin^2(A) &=& \cos(2A) \quad | \quad :\cos^2(A) \\ \frac{\cos^2(A)}{\cos^2(A)} - \frac{\sin^2(A)}{\cos^2(A)} &=& \frac{\cos(2A)}{\cos^2(A)} \\ \boxed{~ 1-\tan^2(A) = \frac{\cos(2A)}{\cos^2(A)} ~} \end{array}$

$\begin{array}{rcll} \dfrac{m+n}{m-n} &=& \dfrac{ [ 1-\tan^2(A)] [ 1+\tan^2(30^{\circ})] } { [ 1+\tan^2(A)] [ 1-\tan^2(30^{\circ})] } \\\\ \dfrac{m+n}{m-n} &=& \dfrac{ \frac{\cos(2A)}{\cos^2(A)} [ 1+\tan^2(30^{\circ})] } { \frac{1}{\cos^2(A)} [ 1-\tan^2(30^{\circ})] } \\\\ \dfrac{m+n}{m-n} &=& \cos(2A)\cdot \dfrac{ [ 1+\tan^2(30^{\circ})] } { [ 1-\tan^2(30^{\circ})] } \quad |\quad \tan(30^{\circ}) = \frac{1}{ \sqrt{3} } \quad \tan^2(30^{\circ}) = \frac13 \\\\ \dfrac{m+n}{m-n} &=& \cos(2A)\cdot \dfrac{ ( 1+\frac13) } { ( 1-\frac13) } \\\\ \dfrac{m+n}{m-n} &=& \cos(2A)\cdot \dfrac{ \frac43 } { \frac23 } \\\\ \dfrac{m+n}{m-n} &=& \cos(2A)\cdot \frac43 \cdot \frac32 \\\\ \dfrac{m+n}{m-n} &=& \cos(2A)\cdot \frac42 \\\\ \dfrac{m+n}{m-n} &=& \cos(2A)\cdot 2 \\\\ \mathbf{ \dfrac{m+n}{m-n} }& \mathbf{=} & \mathbf{ 2\cdot \cos(2A) } \end{array}$

If tanA+tanB=x & cotA+cotB=y, then prove that-

cot(A+B)=1/x-1/y

$\begin{array}{rcll} \tan(A) + \tan(B) &=& x \\ \cot(A) + \cot(B) &=& y \end{array}$

formulary

$\begin{array}{rcll} \tan(A+B) &=& \dfrac{\tan(A) + \tan(B)}{1-\tan(A) \cdot \tan(B)} \\\\ \cot(A+B) &=& \dfrac{ \cot(A) \cdot \cot(B) - 1 } {\cot(A) + \cot(B)} \end{array}$

1.

$\begin{array}{rcll} \tan(A+B)=\dfrac{1}{\cot(A+B)} &=& \dfrac{x} {1-\tan(A) \cdot \tan(B)} \\\\ 1-\tan(A) \cdot \tan(B) &=& x\cdot \cot(A+B) \\\\ \tan(A) \cdot \tan(B)-1 &=& -x\cdot \cot(A+B) \\\\ \tan(A) \cdot \tan(B) &=& 1 -x\cdot \cot(A+B) \\\\ \dfrac{1}{\cot(A)} \cdot \dfrac{1}{\cot(B)} &=& 1 -x\cdot \cot(A+B) \\\\ \mathbf{ \cot(A) \cdot \cot(B) } &\mathbf{=}& \mathbf{ \dfrac{1}{ 1 -x\cdot \cot(A+B) } } \end{array}$

2.

$\begin{array}{rcll} \cot(A+B) &=& \dfrac{ \cot(A) \cdot \cot(B) - 1 } {y} \\\\ y\cdot \cot(A+B) &=& \cot(A) \cdot \cot(B) - 1 \\\\ 1+y\cdot \cot(A+B) &=& \cot(A) \cdot \cot(B) \\ && \cot(A) \cdot \cot(B) =\dfrac{1}{ 1 -x\cdot \cot(A+B) } \\ 1+y\cdot \cot(A+B) &=& \dfrac{1}{ 1 -x\cdot \cot(A+B) } \\ [ 1+y\cdot \cot(A+B)]\cdot[1 -x\cdot \cot(A+B)] &=& 1 \\ 1 -x\cdot \cot(A+B) + y\cdot \cot(A+B) - yx\cdot \cot^2(A+B) &=& 1 \\ -x\cdot \cot(A+B) + y\cdot \cot(A+B) - yx\cdot \cot^2(A+B) &=& 0 \quad | \quad : \cot(A+B)\\ -x + y - yx\cdot \cot(A+B) &=& 0 \\ yx\cdot \cot(A+B) &=& y -x \quad | \quad : xy\\ \cot(A+B) &=& \dfrac{y -x}{xy} \\ \cot(A+B) &=& \dfrac{y}{xy}-\dfrac{x}{xy} \\ \mathbf{ \cot(A+B)} &\mathbf{ =}&\mathbf{ \dfrac{1}{x}-\dfrac{1}{y} } \end{array}$

Lisa has $45 more than John. Gina has $25 more than John.

If they have $85 altogether, how much does each creature have?

x = Lisa
y = Gina
z = John

$\begin{array}{lrcll} (1) & x &=& $45+z\\ (2) & y &=& $25+z\\\\ (1)+(2) & x+y &=& $45+z+$25+z \\ & x+y &=& $70+2z \\ \hline \\ (3) & x+y+z &=& $85\\\\ & (x+y)+z &=& $85 \quad & | \quad x+y = $70+2z\\ & $70+2z+z &=& $85 \\ & $70+3z &=& $85 \quad & | \quad -$70\\ & 3z &=& $85 -$70\\ & 3z &=& $15 \quad & | \quad :3 \\ & z &=& \frac{ $15 }{ 3 } \\ & \mathbf{ z } & \mathbf{=} & \mathbf{ $5 }\\ \hline \\ (1) & x &=& $45+z \quad & | \quad z=$5 \\ & x &=& $45+$5 \\ & \mathbf{x} &\mathbf{=}& \mathbf{$50} \\ \hline \\ (2) & y &=& $25+z \quad & | \quad z=$5 \\ & y &=& $25+$5 \\ & \mathbf{y} &\mathbf{=}& \mathbf{$30} \end{array}$

Lisa $50

Gina $30

John $5

ich soll überprüfen, ob diese trigonometrische umformung gilt:

(sin(ax))5 = 10/16 sin(ax) - 5/16 sin(3ax) + 1/16 sin(5ax)

Es genügt $\sin^5(x) = \frac{10}{16} \sin(x) - \frac{5}{16} \sin(3x) + \frac{1}{16} \sin(5x)$ zu beweisen.

B

Aus der Formelsammlung: $\boxed{~ \sin^2(x) + \cos^2(x) = 1 \qquad \cos^2(x)=1-\sin^2(x) ~}$

1.

$\begin{array}{rcll} \sin(2x) &=& \sin(x+x) \\ &=& \sin(x)\cos(x) + \cos(x)\sin(x)\\ &=& 2\sin(x)\cos(x) \\ \boxed{~ \sin(2x) = 2\sin(x)\cos(x) ~} \end{array}$

2.

$\begin{array}{rcll} \cos(2x) &=& \cos(x+x) \\ &=& \cos(x)\cos(x) - \sin(x)\sin(x)\\ &=& \cos^2(x) - \sin^2(x) \\ &=& [1-\sin^2(x)] - \sin^2(x) \\ &=& 1-2\sin^2(x) \\ \boxed{~ \cos(2x) = 1-2\sin^2(x) ~} \end{array}$

3.

$\begin{array}{rcll} \sin(3x) &=& \sin(x+2x) \\ &=& \sin(x)\cdot \underbrace{\cos(2x)}_{=[1-2\sin^2(x)]} + \cos(x)\cdot \underbrace{\sin(2x)}_{=[2\sin(x)\cos(x)]} \\ &=& \sin(x) \cdot [1-2\sin^2(x)] + \cos(x)\cdot [2\sin(x)\cos(x)] \\ &=& \sin(x) -2\sin^3(x) + 2\cdot \sin(x)\cos^2(x) \\ &=& \sin(x) -2\sin^3(x) + 2\cdot \sin(x)[1-\sin^2(x)] \\ &=& \sin(x) -2\sin^3(x) + 2 \sin(x) -2\sin^3(x) \\ &=& 3\sin(x) -4\sin^3(x) \\ && \boxed{~ \begin{array}{rcll} \sin(3x) &=& 3\sin(x) -4\sin^3(x)\\ \text{ oder } \quad \sin^3(x) &=& \frac14[3\sin(x)-\sin(3x)] \end{array} ~} \end{array}$

4.

$\begin{array}{rcll} \cos(3x) &=& \cos(x+2x) \\ &=& \cos(x)\cdot \underbrace{\cos(2x)}_{=[1-2\sin^2(x)]} - \sin(x)\cdot \underbrace{\sin(2x)}_{=[2\sin(x)\cos(x)]}\\ &=& \cos(x) \cdot [1-2\sin^2(x)] - \sin(x)\cdot [2\sin(x)\cos(x)] \\ &=& \cos(x) -2 \cos(x)\sin^2(x) - 2\sin^2(x) \cos(x)\\ &=& \cos(x) -4 \cos(x)\sin^2(x)\\ &=& \cos(x)[1 -4\sin^2(x)]\\ && \boxed{~ \cos(3x) = \cos(x)\cdot [1 -4\sin^2(x)] ~} \end{array}$

5.

$\begin{array}{rcll} \sin(5x) &=& \sin(2x+3x) \\ &=& \underbrace{\sin(2x)}_{=[2\sin(x)\cos(x)]} \cdot \underbrace{\cos(3x)}_{=\cos(x)\cdot [1 -4\sin^2(x)]} + \underbrace{\cos(2x)}_{=[1-2\sin^2(x)]} \cdot \underbrace{\sin(3x)}_{=[3\sin(x) -4\sin^3(x)]} \\\\ &=& [2\sin(x)\cos(x)] \cdot \cos(x)\cdot [1 -4\sin^2(x)]\\ &+& [1-2\sin^2(x)] \cdot [3\sin(x) -4\sin^3(x)] \\\\ &=& 2\sin(x)\cos^2(x) \cdot [1 -4\sin^2(x)]\\ &+& [1-2\sin^2(x)] \cdot [3\sin(x) -4\sin^3(x)] \\\\ &=& 2\sin(x)[1-\sin^2(x)] \cdot [1 -4\sin^2(x)]\\ &+& [1-2\sin^2(x)] \cdot [3\sin(x) -4\sin^3(x)] \\\\ &=& 2\sin(x)\cdot [1 -4\sin^2(x)]-2\sin^3(x)\cdot [1 -4\sin^2(x)]\\ &+& [1-2\sin^2(x)] \cdot [3\sin(x) -4\sin^3(x)] \\\\ &=& 2\sin(x)-8\sin^3(x)-2\sin^3(x)+8\sin^5(x)\\ &+&3\sin(x)-4\sin^3(x)-6\sin^3(x)+8\sin^5(x) \\\\ \sin(5x) &=& 5\sin(x) -20\cdot \underbrace{\sin^3(x)}_{=\frac14[3\sin(x)-\sin(3x)]} +16\sin^5(x) \\ &=& 5\sin(x) -20\cdot \frac14[3\sin(x)-\sin(3x)] +16\sin^5(x) \\ &=& 5\sin(x) -5\cdot [3\sin(x)-\sin(3x)] +16\sin^5(x) \\ &=& 5\sin(x) -15\sin(x)+5\sin(3x) +16\sin^5(x) \\\\ \sin(5x) &=& -10\sin(x)+5\sin(3x) +16\sin^5(x) \\ 16\sin^5(x) &=& 10\sin(x)-5\sin(3x)+\sin(5x)\\ \sin^5(x) &=& \frac{1}{16}\cdot[ 10\sin(x)-5\sin(3x)+\sin(5x) ] \\\\ \mathbf{ \sin^5(x) }& \mathbf{=} & \mathbf{\frac{10}{16}\sin(x)-\frac{5}{16}\sin(3x)+\frac{1}{16}\sin(5x) } \end{array}$

ich soll überprüfen, ob diese trigonometrische umformung gilt:

(sin(ax))5 = 10/16 sin(ax) - 5/16 sin(3ax) + 1/16 sin(5ax)

Es genügt $\sin^5(x) = \frac{10}{16} \sin(x) - \frac{5}{16} \sin(3x) + \frac{1}{16} \sin(5x)$ zu beweisen.

A Methode 1 mit komplexen Zahlen

B Methode 2 ohne komplexen Zahlen

A

Aus der Formelsammlung: $\boxed{~ \sin(x) = \frac{1}{2i}(e^{ix}-e^{-ix}) ~}$

$\begin{array}{rcll} \sin^5(x) &=& \left[~ \frac{1}{2i}(e^{ix}-e^{-ix}) ~\right]^5 \\\\ &=& \left[~ \frac{1}{2^5i^5}(e^{ix}-e^{-ix}) ~\right]^5 & | \quad i^5 = i\\\\ &=& \frac{1}{32i}(e^{ix}-e^{-ix})^5 \\\\ &=& \frac{1}{16\cdot 2i}(e^{ix}-e^{-ix})^5 \\\\ \text{Binom auflösen:}&&\boxed{~ \binom50=\binom55=1 \quad \binom51=\binom54=5 \quad \binom52=\binom53=10 ~}\\\\ \sin^5(x) &=& \frac{1}{16\cdot 2i} [~ \binom50 e^{i5x}-\binom51 ( e^{i4x} \cdot e^{-ix} )\\ &+&\binom52 ( e^{i3x} \cdot e^{-i2x} )-\binom53 ( e^{i2x} \cdot e^{-i3x} )\\ &+&\binom54 ( e^{ix} \cdot e^{-i4x} )-\binom55 e^{-i5x} ~ ] \\\\ &=& \frac{1}{16\cdot 2i} [~ e^{i5x}-5 \cdot( e^{i4x} \cdot e^{-ix} ) + 10 \cdot( e^{i3x} \cdot e^{-i2x} )-10 \cdot( e^{i2x} \cdot e^{-i3x} )\\ &+&5 \cdot( e^{ix} \cdot e^{-i4x} )- e^{-i5x} ~ ] \\\\ &=& \frac{1}{16\cdot 2i} [~ e^{i5x}-5 \cdot( e^{i4x-ix} ) + 10 \cdot( e^{i3x-i2x} )-10 \cdot( e^{i2x-i3x} )\\ &+&5 \cdot( e^{ix-i4x} )- e^{-i5x} ~ ] \\\\ &=& \frac{1}{16\cdot 2i} [~ e^{i5x}-5 \cdot( e^{ix(4-1)} ) + 10 \cdot( e^{ix(3-2)} )-10 \cdot( e^{ix(2-3)} )\\ &+&5 \cdot( e^{ix(1-4)} )- e^{-i5x} ~ ] \\\\ &=& \frac{1}{16\cdot 2i} [~ e^{i5x}-5 \cdot e^{i3x} + 10 \cdot e^{ix} -10 \cdot e^{-ix} \\ &+&5 \cdot e^{-i3x} - e^{-i5x} ~ ] \\\\ &=& \frac{1}{16\cdot 2i} [~ e^{i5x}- e^{-i5x} -5 \cdot e^{i3x} +5 \cdot e^{-i3x} + 10 \cdot e^{ix} -10 \cdot e^{-ix} ~ ] \\\\ &=& \frac{1}{16\cdot 2i} [~ ( e^{i5x}- e^{-i5x} ) - 5\cdot( e^{i3x} - e^{-i3x} ) + 10\cdot(e^{ix} - e^{-ix} ) ~ ] \\\\ &=& \frac{1}{16} [~ \underbrace{\frac{1}{2i}( e^{i5x}- e^{-i5x} )}_{=\sin(5x)} - 5\cdot \underbrace{\frac{1}{2i}( e^{i3x} - e^{-i3x} )}_{=\sin(3x)} + 10\cdot\underbrace{\frac{1}{2i}(e^{ix} - e^{-ix} )}_{=\sin(x)} ~ ] \\\\ &=& \frac{1}{16} [~ \sin(5x) - 5\cdot\sin(3x)+ 10\cdot\sin(x) ~ ] \\\\ \sin^5(x) &=& \frac{1}{16}\sin(5x) - \frac{5}{16}\cdot\sin(3x)+ \frac{10}{16}\cdot\sin(x) \end{array}$

Samuel is 4 times Victor's age. In 12 years, Samuel will only be 2 times Victor's age. How old is Victor now?

x = samuel

y = victor

$\begin{array}{lrcll} (1) & 4\cdot y &=& x \quad & | \quad :4 \\ & y &=& \frac{x}{4} \\\\ (2) & (x+12) &=& 2(y+12) \quad & | \quad y = \frac{x}{4} \\ & (x+12) &=& 2(\frac{x}{4}+12) \\ & x+12 &=& 2\cdot \frac{x}{4} +2\cdot 12 \\ & x+12 &=& 2\cdot \frac{x}{4} +24 \\ & x+12 &=& \frac{x}{2} +24 \quad & | \quad -\frac{x}{2} \\ & x-\frac{x}{2}+12 &=& 24 \quad & | \quad -12 \\ & x-\frac{x}{2} &=& 24-12 \\ & x-\frac{x}{2} &=& 12 \\ & \frac{x}{2} &=& 12 \quad & | \quad \cdot 2 \\ & x &=& 12 \cdot 2 \\ & \mathbf{ x } & \mathbf{=} & \mathbf{24} \\\\ & y &=& \frac{x}{4} \quad & | \quad x = 24 \\ & y &=& \frac{24}{4} \\ & \mathbf{ y } & \mathbf{=} & \mathbf{6} \\\\ \end{array}$

Samual was 24 and Victor was 6.

Now Samual is 24+12= 36 and Victor is 6+12= 18 years old.

a cookie recipe requires 1\2 cup of sugar. whitch is equivalent to 1\2

2\3?

3\5?

4\8? or

6\10

cosine of .97444

$\begin{array}{rcll} \cos{(\varphi)} &=& 0.97444 \quad & | \quad \pm\arccos{} \\ \varphi &=& \pm\arccos{(0.97444)} \pm k\cdot 360^{\circ} \quad & | \quad k=0,1,2,\dots \\ \varphi &=& \pm 12.9821752590^{\circ} \pm k\cdot 360^{\circ} \\\\ \mathbf{\varphi_1 }&\mathbf{=}& \mathbf{ 12.9821752590^{\circ} \pm k\cdot 360^{\circ} }\\\\ \varphi_2 &=& -12.9821752590^{\circ} \pm k\cdot 360^{\circ} \\ \varphi_2 &=& 360^{\circ}-12.9821752590^{\circ} \pm k\cdot 360^{\circ} \\ \mathbf{\varphi_2 }&\mathbf{=}& \mathbf{347.017824741\pm k\cdot 360^{\circ}} \end{array}$