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 #1
avatar+9460 
+4

To use the sum and difference formulas, we need to find  cos(a),  cos(B),  sin(a),  sin(B),  tan(a)  and  tan(B).

Using the information given, we can find these values with the Pythagorean theorem.

 

 

\(\text{1)}\qquad\sin(a+B)\,=\,\sin( a)\cos (B)+\cos(a)\sin(B)\\~\\ \phantom{\text{1)}\qquad\sin(a+B)\,}=\,(\frac{5}{13})(-\frac12)+(-\frac{12}{13})(\frac{\sqrt3}{2})\\~\\ \phantom{\text{1)}\qquad\sin(a+B)\,}=\,-\frac{5}{26}-\frac{12\sqrt3}{26}\\~\\ \phantom{\text{1)}\qquad\sin(a+B)\,}=\,\frac{-5-12\sqrt3}{26}\\~\\ \text{2)}\qquad\cos(a+B)\,=\,\cos(a)\cos(B)-\sin(a)\sin(B)\\~\\ \phantom{\text{2)}\qquad\cos(a+B)\,}=\,(-\frac{12}{13})(-\frac{1}{2})-(\frac{5}{13})(\frac{\sqrt3}{2})\\~\\ \phantom{\text{2)}\qquad\cos(a+B)\,}=\,\frac{12}{26}-\frac{5\sqrt3}{26}\\~\\ \phantom{\text{2)}\qquad\cos(a+B)\,}=\,\frac{12-5\sqrt3}{26}\\~\\ \text{3)}\qquad\sin(a-B)\,=\,\sin(a)\cos(B)-\cos(a)\sin(B)\\~\\ \phantom{\text{3)}\qquad\sin(a-B)\,}=\,(\frac{5}{13})(-\frac12)-(-\frac{12}{13})(\frac{\sqrt3}{2})\\~\\ \phantom{\text{3)}\qquad\sin(a-B)\,}=\,-\frac{5}{26}+\frac{12\sqrt3}{26}\\~\\ \phantom{\text{3)}\qquad\sin(a-B)\,}=\,\frac{12\sqrt3-5}{26}\\~\\ \text{4)}\qquad\tan(a-B)\,=\,\frac{\tan(a)-\tan(B)}{1+\tan(a)\tan(B)}\\~\\ \phantom{\text{4)}\qquad\tan(a-B)\,}=\,\frac{-\frac{5}{12}--\frac{\sqrt3}{1}}{1+(-\frac{5}{12})(-\frac{\sqrt3}{1})}\\~\\ \phantom{\text{4)}\qquad\tan(a-B)\,}=\,\frac{-\frac{5}{12}+\sqrt3}{1+\frac{5\sqrt3}{12}}\\~\\ \phantom{\text{4)}\qquad\tan(a-B)\,}=\,\frac{-\frac{5}{12}+\sqrt3}{1+\frac{5\sqrt3}{12}}\cdot\frac{12}{12}\\~\\ \phantom{\text{4)}\qquad\tan(a-B)\,}=\,\frac{-5+12\sqrt3}{12+5\sqrt3}\)

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10 may 2019
 #1
avatar+9460 
+4

Since  f(x)  is an odd function defined for all real numbers  x ,

 

f( -x )  =  - f(x)

___

for all real numbers  x .  And  0  is a real number, so....

f( -0 )  =  - f(0)

 

 

 
f(0)  =  - f(0)

 

 

Here we can notice that the only way for  a = -a  to be true is if  a = 0 .

Still we can add  f(0)  to both sides of the equation.

2f(0)  =  0

 

 

Divide both sides of the equation by  2
f(0)  =  0

 

Now let's make this match the form  f(x + 3) - 5 . Rewrite  0  as  -3 + 3
f( -3 + 3 )  =  0

 

 

Subtract  5  from both sides of the equation.
f( -3 + 3 ) - 5  =  -5

 

Substitute  g( - 3 )  in for  f(-3  + 3) - 5
g( -3 )  =  -5

 

 

 

Now we can see that  y  =  g( x )  passes through the point  (-3, -5) .

This makes sense because the graph of  g(x)  is shifted  3 to the left and  5  down from  f(x) .

 

Since  y  =  g(x)  passes through  (2, -2) ,

g( 2 )  =  -2

 

 

Substitute  f(2 + 3) - 5  in for  g( 2 )
f( 2 + 3 ) - 5  =  -2

 

 
f( 5 ) - 5  =  -2

 

 

Add  5  to both sides.
f( 5 )  =  3

 

 

Notice that  (5, 3)  is shifted  3  to the right and  5  up from  (2, -2).

Since  f( -x )  =  - f( x ) ,  f( -5 )  =  - f(5)  =  -3

f( -5 )  =  - 3

 

 

Rewrite  -5  as  -8 + 3
f( -8 + 3 )  =  -3

 

Subtract  5  from both sides of the equation.
f( -8 + 3 ) - 5  =  -8

 

 

Substitute  g( -8 )  in for  f( -8 + 3 ) - 5
g( -8 )  =  -8

 

And notice  (-8, -8)  is shifted  3  to the left and  5  down from  (-5, -3) .

 

Now we can see that  y  =  g( x )  passes through the point  (-8, -8) .

8 may 2019