Question 1040397
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If cos(alpha)/cos(beta) = a, sin(alpha)/sin(beta) = b, then sinē(beta) is equal to
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From   {{{sin(alpha)/sin(beta)}}} = {{{b}}}  you have  {{{sin(beta)}}} = {{{sin(alpha)/b}}}  and then  {{{sin^2(beta)}}} = {{{sin^2(alpha)/b^2}}}.     (1)

Having this, I would say the problem is just solved, because I just expressed  {{{sin^2(beta)}}}. 

But I understand that you want me to express  {{{sin^2(beta)}}}  via  "a"  and "b"  and exclude  {{{alpha}}},  although 
it is not stated directly and implicitly in the problem formulation (which is the author's fault).

So I will continue.


From the condition, we have this line of identities 

{{{b^2}}} = {{{sin^2(alpha)/sin^2(beta)}}}  --->  {{{sin^2(alpha)}}} = {{{b^2*sin^2(beta)}}}                                        (2)

and this line 

{{{a^2}}} = {{{cos^2(alpha)/cos^2(beta)}}} = {{{(1-sin^2(alpha))/(1-sin^2(beta))}}}  --->  {{{1-sin^2(alpha)}}} = {{{a^2*(1-sin^2(beta))}}}  --->  

{{{1-sin^2(alpha)}}} = {{{a^2 -a^2*sin^2(beta)}}}  --->  {{{sin^2(beta)}}} = {{{(a^2 - 1 + sin^2(alpha))/a^2}}}.                  (3)

Next substitute the last identity of (3) into the last identity of (2). You will get

{{{sin^2(alpha)}}} = {{{(b^2/a^2)*(a^2-1+sin^2(alpha))}}}  --->  {{{(1-(b^2/a^2))*sin^2(alpha)}}} = {{{(b^2/a^2)*(a^2-1)}}}  --->  {{{sin^2(alpha)}}} = {{{(b^2*(a^2-1))/(a^2-b^2)}}}.     (4)

Now we can complete (1) by substituting the found value of {{{sin^2(alpha)}}} from (4) into (1). You will get

{{{sin^2(beta)}}} = {{{(a^2-1)/(a^2-b^2)}}}.

<U>Answer</U>.  {{{sin^2(beta)}}} = {{{(a^2-1)/(a^2-b^2)}}}.
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