HAravindG Superstar
Hax cos thetaL + Hby sin thetaL = a^2 - b^2 and Haxsin theta  cos^2 thetaL - Hbycos theta  sin^2 thetaL = 0
show that HaxL^2 3 + HbyL^2 3 = Ha^2 - b^2L^2 3
Ha xL^2 3 + Hb yL^2 3 = Ha^2 - b^2L^2 3L
a2 x2
In[12]:=
H Ha xL ^ 2  3 + Hb yL ^ 2  3 == Ha ^ 2 - b ^ 2L ^ 2  3 L ®
b2 y2
+
1
2
Ia2 - b2 M ;
Š
3
3
3
Reset the Mathematica line counter.
In[13]:=
$Line = 0;
The trig equations from the problem statement:
In[1]:=
9a2 Š b2 + b y Csc@ΘD + a x Sec@ΘD, b y Cot@ΘD Csc@ΘD Š a x Sec@ΘD Tan@ΘD=
Out[1]=
9a2 Š b2 + b y Csc@ΘD + a x Sec@ΘD, b y Cot@ΘD Csc@ΘD Š a x Sec@ΘD Tan@ΘD=
Solve the equations above for x and y. Flatten gets rid of an extra set of braces.
In[2]:=
Solve@%1, 8x, y<D  Flatten
I- a2 + b2 M Cot@ΘD Csc@ΘD
Out[2]=
I- a2 + b2 M Sec@ΘD Tan@ΘD
,y®-
:x ® -
>
a ICsc@ΘD2 + Sec@ΘD2 M
b ICsc@ΘD2 + Sec@ΘD2 M
x and y above are replacement rules. Square each side of the ® symbol to obtain new replacement rules.
I- a2 + b2 M Cot@ΘD Csc@ΘD
In[3]:=
2
2
I- a2 + b2 M Sec@ΘD Tan@ΘD
,y ® -
:x ® a ICsc@ΘD2 + Sec@ΘD2 M
>
b ICsc@ΘD2 + Sec@ΘD2 M
2
2
I- a2 + b2 M Cot@ΘD2 Csc@ΘD2
Out[3]=
2
2
:x2 ®
I- a2 + b2 M Sec@ΘD2 Tan@ΘD2
, y2 ®
>
2
2
a2 ICsc@ΘD2 + Sec@ΘD2 M
b2 ICsc@ΘD2 + Sec@ΘD2 M
Apply the replacement rules above to the following expression, In[12]:
a2 x2
b2 y2
1
+
In[4]:=
3
Š
3
2
Ia2 - b2 M . %3
3
2
2
I- a2 + b2 M Cot@ΘD2 Csc@ΘD2
I- a2 + b2 M Sec@ΘD2 Tan@ΘD2
+
Out[4]=
2 2
2
1
Š
2 2
2
3 ICsc@ΘD + Sec@ΘD M
2
Ia2 - b2 M
3
3 ICsc@ΘD + Sec@ΘD M
Factor the LHS of Out[4].
2
2
I- a2 + b2 M Cot@ΘD2 Csc@ΘD2
In[5]:=
FactorB
I- a2 + b2 M Sec@ΘD2 Tan@ΘD2
+
2 2
2
3 ICsc@ΘD + Sec@ΘD M
Ha - bL2 Ha + bL2 ICot@ΘD2 Csc@ΘD2 + Sec@ΘD2 Tan@ΘD2 M
1
Š
2 2
2
Ia2 - b2 M
3
3 ICsc@ΘD + Sec@ΘD M
Out[5]=
2
1
FŠ
2 2
2
2
Ia2 - b2 M
3
3 ICsc@ΘD + Sec@ΘD M
Rearrange the LHS terms of Out[5] collecting the trig functions and the other expressions in that order.
ICot@ΘD2 Csc@ΘD2 + Sec@ΘD2 Tan@ΘD2 M
1
In[6]:=
2
2 2
ICsc@ΘD + Sec@ΘD M
Simplify the trig fraction above.
3
Ha - bL2 Ha + bL2 Š
1
3
2
Ia2 - b2 M ;
2
Untitled-3
ICot@ΘD2 Csc@ΘD2 + Sec@ΘD2 Tan@ΘD2 M
In[7]:=
FullSimplifyB
3
ICsc@ΘD + Sec@ΘD M
1
Ha - bL2 Ha + bL2 H5 + 3 Cos@4 ΘDL Š
Out[7]=
24
1
Out[8]=
3
Solve@%7, ΘD
88Θ ® 0<<
Replace Θ in Out[7] with zero.
In[9]:=
%7 . %8
1
Out[9]=
:
Ha - bL2 Ha + bL2 Š
3
1
2
Ia2 - b2 M >
3
Expand then simplify the LHS of Out[9],
1
In[10]:=
Ha - bL2 Ha + bL2  Expand  Simplify
3
1
Out[10]=
2
Ia2 - b2 M
3
The LHS equals the RHS.
1
In[11]:=
Out[11]=
2
Ia2 - b2 M ==
3
True
1
3
2
Ia2 - b2 M
2
Ia2 - b2 M
Solve Out[7] for Θ.
In[8]:=
1
F
2 2
2
Ha - bL2 Ha + bL2 Š
1
3
2
Ia2 - b2 M