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ant101 2 mins ago
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Verify the following identity:
sin((3 x)/2) (sin(x))/(sin(x/2)) = sin(x) + sin(2 x)

Multiply both sides by sin(x/2):
sin(x) sin((3 x)/2) = ^?sin(x/2) (sin(x) + sin(2 x))

sin(x) sin((3 x)/2) = 1/2 (cos(x - (3 x)/2) - cos(x + (3 x)/2)) = 1/2 (cos(-x/2) - cos((5 x)/2)):
(cos(-x/2) - cos((5 x)/2))/(2) = ^?sin(x/2) (sin(x) + sin(2 x))

Use the identity cos(-x/2) = cos(x/2):
(cos(x/2) - cos((5 x)/2))/(2) = ^?sin(x/2) (sin(x) + sin(2 x))

sin(x/2) (sin(x) + sin(2 x)) = sin(x/2) sin(x) + sin(x/2) sin(2 x):
(cos(x/2) - cos((5 x)/2))/(2) = ^?sin(x/2) sin(x) + sin(x/2) sin(2 x)

sin(x/2) sin(x) = 1/2 (cos(x/2 - x) - cos(x/2 + x)) = 1/2 (cos(-x/2) - cos((3 x)/2)):
(cos(x/2) - cos((5 x)/2))/(2) = ^?(cos(-x/2) - cos((3 x)/2))/(2) + sin(x/2) sin(2 x)

Use the identity cos(-x/2) = cos(x/2):
(cos(x/2) - cos((5 x)/2))/(2) = ^?(cos(x/2) - cos((3 x)/2))/(2) + sin(x/2) sin(2 x)

(cos(x/2) - cos((3 x)/2))/(2) = 1/2 cos(x/2) - 1/2 cos((3 x)/2):
(cos(x/2) - cos((5 x)/2))/(2) = ^?(cos(x/2))/(2) - (cos((3 x)/2))/(2) + sin(x/2) sin(2 x)

sin(x/2) sin(2 x) = 1/2 (cos(x/2 - 2 x) - cos(x/2 + 2 x)) = 1/2 (cos(-(3 x)/2) - cos((5 x)/2)):
(cos(x/2) - cos((5 x)/2))/(2) = ^?(cos(x/2))/(2) - (cos((3 x)/2))/(2) + (cos(-(3 x)/2) - cos((5 x)/2))/(2)

Use the identity cos(-(3 x)/2) = cos((3 x)/2):
(cos(x/2) - cos((5 x)/2))/(2) = ^?(cos(x/2))/(2) - (cos((3 x)/2))/(2) + (cos((3 x)/2) - cos((5 x)/2))/(2)

(cos((3 x)/2) - cos((5 x)/2))/(2) = 1/2 cos((3 x)/2) - 1/2 cos((5 x)/2):
(cos(x/2) - cos((5 x)/2))/(2) = ^?(cos(x/2))/(2) - (cos((3 x)/2))/(2) + (cos((3 x)/2))/(2) - (cos((5 x)/2))/(2)

(cos(x/2))/(2) - (cos((3 x)/2))/(2) + (cos((3 x)/2))/(2) - (cos((5 x)/2))/(2) = 1/2 cos(x/2) - 1/2 cos((5 x)/2):
(cos(x/2) - cos((5 x)/2))/(2) = ^?(cos(x/2))/(2) - (cos((5 x)/2))/(2)

1/2 (cos(x/2) - cos((5 x)/2)) = 1/2 cos(x/2) - 1/2 cos((5 x)/2):
1/2 cos(x/2) - 1/2 cos((5 x)/2) = ^?1/2 cos(x/2) - 1/2 cos((5 x)/2)

The left hand side and right hand side are identical:
Answer: | (identity has been verified)

Guest17 mins ago
 #1
avatar+18281 
+1

Right \(\triangle{ABC}\) has AB = 3, BC = 4, and AC = 5.
Square XYZW is inscribed in triangle ABC with X and Y on \(\overline{AC}\),
W on \(\overline{AB}\), and
Z on \(\overline{BC}\).
What is the side length of the square?

 

Let s is the side length oft the square \( = \overline{XY} = \overline{YZ} = \overline{ZW} = \overline{WX}\)

Let h = \(\overline{BT}\)

Let A the area of \(\triangle{ABC}\)

 

 

h = ?

\(\begin{array}{|rcll|} \hline A &=& \frac{\overline{AB} \cdot \overline{BC} }{2} \\ A &=& \frac{3\cdot 4}{2} \\ \mathbf{A} &\mathbf{=}& \mathbf{6} \\\\ A &=& \frac{\overline{AC}\cdot h}{2} \\ A &=& \frac{5\cdot h}{2} \quad & | \quad \mathbf{A=6} \\ 6 &=& \frac{5\cdot h}{2} \\ \mathbf{h} &\mathbf{=}& \mathbf{ \frac{12}{5} } \\ \hline \end{array}\)

 

\(\mathbf{\overline{BW} =\ ?}\)

\(\begin{array}{|rcll|} \hline \frac{ \overline{BW} } {s} &=& \frac{ \overline{AB} } { \overline{AC} } \\ \frac{ \overline{BW} } {s} &=& \frac{ 3 } { 5 } \\ \mathbf{ \overline{BW} } & \mathbf{=} & \mathbf{ \frac{3}{5}s } \\ \hline \end{array}\)

 

\(\mathbf{\overline{BZ} =\ ?} \)

\(\begin{array}{|rcll|} \hline \frac{ \overline{BZ} } {s} &=& \frac{ \overline{BC} } { \overline{AC} } \\ \frac{ \overline{BZ} } {s} &=& \frac{ 4 } { 5 } \\ \mathbf{\overline{BZ}} &\mathbf{=}& \mathbf{\frac{4}{5}s } \\ \hline \end{array}\)

 

s = ?

\(\begin{array}{|rcll|} \hline A_{\triangle{ZBW}} = \frac{ \overline{BW}\cdot \overline{BZ} } {2} &=& \frac{s\cdot(h-s)} {2} \\ \overline{BW}\cdot \overline{BZ} &=& s\cdot(h-s) \quad & | \quad \mathbf{ \overline{BW} =\frac{3}{5}s } \quad \mathbf{ \overline{BZ} =\frac{4}{5}s } \quad \mathbf{h=\frac{12}{5}} \\ \frac{3}{5}s \cdot \frac{4}{5}s &=& s\cdot(\frac{12}{5}-s) \\ \frac{12}{25}s &=& \frac{12}{5}-s \\ s+\frac{12}{25}s &=& \frac{12}{5} \\ s \cdot \left(1+\frac{12}{25} \right) &=& \frac{12}{5} \\ s \cdot \left(\frac{25+12}{25} \right) &=& \frac{12}{5} \\ s \cdot \left(\frac{37}{25} \right) &=& \frac{12}{5} \\ s &=& \frac{25}{37} \cdot \frac{12}{5} \\ s &=& \frac{5}{37} \cdot 12 \\ s &=& \frac{60}{37} \\ \mathbf{s} &\mathbf{=}& \mathbf{1.\overline{621}} \\ \hline \end{array}\)

 

The side length oft the square is \(\mathbf{1.\overline{621}}\)

 

 

laugh

heureka 7 hours ago
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Guest11 hours ago
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hectictar Jul 28, 2017 12:19 AM
Jul 27, 2017
 #1
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GuestJul 27, 2017 10:04 PM
 #1
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MIRB15 Jul 27, 2017 7:59:48 PM
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GuestJul 27, 2017 4:36:58 PM
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GuestJul 27, 2017 4:15:29 PM
 #1
avatar+18281 
+2

Let triangle ABC be a triangle such that AB=13, BC=14, and AC=15. 

Meanwhile, D is a point on BC such that AD bisects angle A.

Find the area of triangle ADC

 

Let area of triangle \(\text{ADC} = A_\text{ADC} \)

Let area of triangle \( \text{ADB} = A_\text{ADB}\)
Let area of triangle \(\text{ABC} = A\)

 

Let \(A_\text{ADB} = A - A_\text{ADC}\)

 

\(\begin{array}{|rcll|} \hline \frac { A_\text{ADB} } { A_\text{ADC} } &=& \frac{ \overline{AD}\cdot 13 \cdot \sin(\frac{A}{2}) \cdot \frac12 } { \overline{AD}\cdot 15 \cdot \sin(\frac{A}{2}) \cdot \frac12 } \\ \frac { A_\text{ADB} } { A_\text{ADC} } &=& \frac{ 13 } { 15 } \quad & | \quad A_\text{ADB} = A - A_\text{ADC} \\ \frac { A - A_\text{ADC} } { A_\text{ADC} } &=& \frac{ 13 } { 15 } \\ \frac { A } { A_\text{ADC} } - 1 &=& \frac{ 13 } { 15 } \\ \frac { A } { A_\text{ADC} } &=& 1+ \frac{ 13 } { 15 } \\ \frac { A } { A_\text{ADC} } &=& \frac{ 28 } { 15 } \\ \frac { A_\text{ADC} } { A } &=& \frac{ 15 } { 28 } \\\\ \mathbf{ A_\text{ADC } } & \mathbf{=} & \mathbf{ \frac{ 15 } { 28 } A } \\ \hline \end{array} \)

 

Heron's formula states that the area of a triangle whose sides have lengths a, b, and c is

\( A = \sqrt{s(s-a)(s-b)(s-c)}\)

 

where s is the semiperimeter of the triangle; that is,
\(s=\frac{a+b+c}{2}\).

 

\(\begin{array}{|rcll|} \hline s &=& \frac{a+b+c}{2} \quad & | \quad a= 14, \ b= 15, \ c= 13 \\ s &=& \frac{15+15+13}{2} \\ s &=& \frac{42}{2} \\ \mathbf{ s }& \mathbf{=} & \mathbf{21} \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline A &=& \sqrt{s(s-a)(s-b)(s-c)} \quad & | \quad s= 21, \ a= 14, \ b= 15, \ c= 13 \\ A &=& \sqrt{21(21-14)(21-15)(21-13)} \\ A &=& \sqrt{21\cdot 7 \cdot 6 \cdot 8 } \\ A &=& \sqrt{3\cdot 7 \cdot 7 \cdot 2\cdot 3 \cdot 2\cdot 4 } \\ A &=& \sqrt{4^2\cdot 7^2 \cdot 3^2 } \\ A &=& 4\cdot 7 \cdot 3 \\ \mathbf{ A }& \mathbf{=} & \mathbf{84} \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline \mathbf{ A_\text{ADC } } & \mathbf{=} & \mathbf{ \frac{ 15 } { 28 } A } \quad & | \quad \mathbf{ A } \mathbf{=} \mathbf{84} \\ A_\text{ADC } & = & \frac{ 15 } { 28 } \cdot 84 \\ A_\text{ADC } & = & 15 \cdot 3 \\\\ \mathbf{ A_\text{ADC } }& \mathbf{=} & \mathbf{45} \\ \hline \end{array}\)

 

laugh

heureka Jul 27, 2017

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