(I'll use Q instead of @, so my computer won't think it's 
an email address. Igor hasn't put in the Greek letter "theta"
in his program.  Think I'll ask him to put it in.)

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Given that cosQ =  and Q is in Quadrant 2, find: 

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Draw the picture of Q.  We use the numerator and denominator
of  for x and r.  Since , to put 
q in Quadrant 2, we must take x as  and  as ,
and draw this picture, with the curved line indicating angle Q: 



Now we use the Pythagorean theorm to find 









±

±

But since  goes upward from the x-axis,

we take the positive value, so 



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To find

sinQ, we only need to know that sinQ =  or  

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To find

sin2Q, we need the identity sin2A = 2sinAcosA

sin2Q = 2sinQcosQ

sin2Q = 2

sin2Q = 

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To find

cos3Q, we need the identities cos(A+B) = cosAcosB - sinAsinB and cos2A = 2cos²A-1  

First we rewrite 3Q as 2Q+Q

Then by the first identity,
cos3Q = cos(2Q+Q) = cos2QcosQ - sin2QsinQ
= cos2Q() - sin2Q()

and we have already found sin2Q to be ,
so

= cos2Q() - ()


= ()cos2Q - ()

= cos2Q + 

Now we use the second identity to rewrite cos2Q 

= (2cos²Q - 1) + 

Distribute:

= (2cos²Q) +  + 

= cos²Q +  + 

Combine the last two terms and substitute  for cosQ

=  +  = 

=  + 

=  + 

= 

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To find , we use the identity  = 

 = 
             
= 

= 

= 

= 

= 

= 

= 

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To find sin(Q + 60°) we use identity  sin(A + B) = sinAcosB + cosAsinB

sin(Q + 60°) = sinQcos60° + cosQsin60°

Now we use the fact that cos60° = , sin60° = , cosQ =  and sinQ = 

sin(Q + 60°) =  =

 =



Edwin