Question 1089267
.
if tan a and tan b are the roots of equation 4x^2-7x+1=0 then evaluate 4sin^2(a+b)-7sin(a+b)cos(A+B)+cos^2(A+B)
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~


<pre>
if tan(a) and tan(b) are the roots of equation 4x^2-7x+1=0 then

    tan(a) + tan(b) = {{{7/4}}}  and  tan(a)*tan(b) = {{{1/4}}},    (1)

according to Vieta's formulas/theorem.


It implies  tan(a+b) = {{{(tan(a) + tan(b))/(1- tan(a)*tan(b))}}} = {{{((7/4))/(1-1/4)}}} = {{{((7/4))/((3/4))}}} = {{{7/3}}}  and then  {{{cos^2(a+b)}}} = {{{1/(1+tan^2(a+b))}}} = {{{1/(1 + (7/3)^2)}}} = {{{1/(1 + 49/9)}}} = {{{9/(9 + 49)}}} = {{{9/58}}}.       (2)


Then  

  {{{4sin^2(a+b)-7sin(a+b)cos(a+b)+cos^2(a+b)}}} = {{{cos^2(a+b)*(4*tan^2(a+b) -7*tan(a+b) +1)}}} = {{{cos^2(a+b)*(4*(7/3)^2 - 7*(7/3) + 1)}}} = {{{cos^2(a+b)*(4*49/9 - 7*(7/3) + 1)}}} = 

= {{{cos^2(a+b)*((4*49)/9 - (7*7*3)/9 + 9/9)}}} = {{{cos^2(a+b)*((4*49-7*7*3+9)/9)}}} = {{{cos^2(a+b)*(58/9)}}}.


Substitute here  {{{cos^2(a+b)}}} = {{{9/58}}}  from (2),  and you will get

{{{4sin^2(a+b)-7sin(a+b)cos(a+b)+cos^2(a+b)}}} = . . . = {{{(9/58)*(58/9)}}} = 1.
</pre>

<U>Answer</U>.  The value under the question is 1.



Solved.