CPhill

Note that    

cos2θ =  1 -2sin^2θ

So

(1/2) ( Ln l 1 - cos2θ l - ln (2) )   =

(1/2) ( Ln l 1 - (1 - 2sin^2θ) l - ln( 2) )  =

(1/2) ( Ln l 2sin^2 θl  - Ln (2) )  =

{By a log property  Ln (a *b)  = Ln a + Ln b )    ( and Ln l 2l    = Ln(2) )

(1/2) ( Ln (2) + Ln l sin^2θ l - Ln (2) ) =

[ Note that   l sin^2 θ l  = sin^2 θ  ]    so we have

(1/2) Ln (sin^2 θ)  =

Ln (sin^2 θ)^(1/2)  =

Ln √ [ sin^2 θ ]

And note, AT   that....... Ln  l sin θ l  is defined  as   Ln √ [ sin^2 θ ]

So....the right side = the left side

HAHAHA!!!....yeah....that can happen   !!!!

LOL!!!!!

Lightning solved it by NOT solving it......!!!!!

Which brings to mind the old paradox...."If every man in the village who is not the barber shaves himself.....who shaves the barber ? "

Note that   sin^3  θ + cos^3  θ  factors as a sum of cubes   =  ( sin  θ  + cos  θ) ( sin^2 θ - sin θ cos  θ  + cos^2  θ)

So....the first  factor  cancels with the factor in the denominator  on the right and we are left with

(sin^2  θ -  sin θ cos  θ + cos^2  θ)  =

( sin^2  θ  + cos ^2  θ)  -  sin θcos θ  =

1     -  (1/2)sin (2 θ)

tan (v/2)  =

sin(v/2)             √  [ (1 - cosv) / 2 ]             √ [ 1 - cos v]   

_______   =       _______________  =    _____________   multiply   top/bottom by √ [1 - cos v ]  and we have

cos(v/2)            √ [ (1 + cos v)/ 2 ]              √ [ 1 + cos v ]

√[1 - cosv ] √ [ 1 - cos v ]            1  -  cos v                 1  - cos v               1  - cos v

____________________ =       ____________  =     ___________  =   _________  =

√ [ 1 + cos v]√ [ 1 - cos v]         √ [ 1 - cos^2v ]            √ sin^2v                   sin v

 1         -      cos v

____           _____    =

sin v            sin v

csc v  -   cot v

6. A spinner is divided into 5 sectors as shown. Each of the central angles of the sectors 1 through 3 measures 60° while each of the central angles of sectors 4 and 5 measures 90°. If the spinner is spun twice, what is the probability that at least one spin lands on an even number? Express your answer as a common fraction.

P(even on one at least of the spins) =   1 - P(odd on both spins)

P(odd on 1st spin)  =  [60 + 60 + 90] / 360  =  210/360  = 7/12

And we have the same probability of an odd on the second spin

So

P(odd on both spins)  =  7/12 * 7/12  =   49/144

So....P (even on at least one of the spins)  =  1 - 49/144  =  95 / 144

Nice, hectictar   !!!!

THX, Rom  !!!

Here's the c part

Using implicit differentiation....the slope of a tangent line at any point on the ellipse is given by

(1/4)x + (1/2)y y'  = 0

(1/2)yy' = -(1/4)x

y' =  -(1/4)x / (1/2)y  =  -x / [ 2y]

And the slope of a tangent line to the hyperbola at any point is

2x - (2/3)yy' = 0

-(2/3)yy' = -2x

y' = 2x /[ (2/3)y ] =   3x / y

To find the itersection points.....we have that

(1/8)x^2 + (1/4)y^2  = 1    [multiply through by - 8 ] ⇒ - x^2 - 2y^2  = -8      (1)

x^2 - (1/3)y^2   = 1     (2)         add (1)  and (2) and we have that

-(7/3)y^2 = -7    [divide through by -7 ]

(1/3)y^2  = 1

y^2  = ±3

y = ±sqrt(3)

And using  (1)

-x^2 - 2(3) = -8

-x^2 - 6 = -8

x^2 + 6 = 8

x^2 = 2

x =±sqrt (2)

So.....the intersection points are

(sqrt(2), sqrt(3) )

(- sqrt(2),sqrt (3) )

(-sqrt(2), -sqrt(3) )

(sqrt (2), - sqrt(3))      

Testing the first point 

The slope of the tagent line to the ellipse at (sqrt(2), sqrt(3)  )   we have     -sqrt(2) / [2 sqrt(3)] = -1 / sqrt(6)

The slope of the tangent line to the hyperbola at this point is  3sqrt(2) / sqrt(3) = sqrt(6)

Since these are negative reciprocals.....the tangent lines are perpendicular

[I'll let you test the other three points ]

THX, hectictar  !!!!

Here's the "b"  part

Since the hperbola has the same foci its center is an equal distance from both foci = (0,0)....and one branch cuts the x axis at x  = 1 , then  a  =  1  and c = 2

Therefore....b  =   sqrt(c^2 - a^2)  =   sqrt (2^2 - 1^2)  = sqrt (4 - 1)  = sqrt (3)

So a^2  = 1  and b^2  = 3

So the equation of the hyperbola is

x^2          y^2

__    -     _____  =     1

 1              3

Here is the graph of the ellipse and hyperbola :  https://www.desmos.com/calculator/ur21asuwwk