SOLUTION: 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)

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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)
Answer by ikleyn(52903) About Me  (Show Source):
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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)
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if tan(a) and tan(b) are the roots of equation 4x^2-7x+1=0 then

    tan(a) + tan(b) = 7%2F4  and  tan(a)*tan(b) = 1%2F4,    (1)

according to Vieta's formulas/theorem.


It implies  tan(a+b) = %28tan%28a%29+%2B+tan%28b%29%29%2F%281-+tan%28a%29%2Atan%28b%29%29 = %28%287%2F4%29%29%2F%281-1%2F4%29 = %28%287%2F4%29%29%2F%28%283%2F4%29%29 = 7%2F3  and then  cos%5E2%28a%2Bb%29 = 1%2F%281%2Btan%5E2%28a%2Bb%29%29 = 1%2F%281+%2B+%287%2F3%29%5E2%29 = 1%2F%281+%2B+49%2F9%29 = 9%2F%289+%2B+49%29 = 9%2F58.       (2)


Then  

  4sin%5E2%28a%2Bb%29-7sin%28a%2Bb%29cos%28a%2Bb%29%2Bcos%5E2%28a%2Bb%29 = cos%5E2%28a%2Bb%29%2A%284%2Atan%5E2%28a%2Bb%29+-7%2Atan%28a%2Bb%29+%2B1%29 = cos%5E2%28a%2Bb%29%2A%284%2A%287%2F3%29%5E2+-+7%2A%287%2F3%29+%2B+1%29 = cos%5E2%28a%2Bb%29%2A%284%2A49%2F9+-+7%2A%287%2F3%29+%2B+1%29 = 

= cos%5E2%28a%2Bb%29%2A%28%284%2A49%29%2F9+-+%287%2A7%2A3%29%2F9+%2B+9%2F9%29 = cos%5E2%28a%2Bb%29%2A%28%284%2A49-7%2A7%2A3%2B9%29%2F9%29 = cos%5E2%28a%2Bb%29%2A%2858%2F9%29.


Substitute here  cos%5E2%28a%2Bb%29 = 9%2F58  from (2),  and you will get

4sin%5E2%28a%2Bb%29-7sin%28a%2Bb%29cos%28a%2Bb%29%2Bcos%5E2%28a%2Bb%29 = . . . = %289%2F58%29%2A%2858%2F9%29 = 1.

Answer. The value under the question is 1.


Solved.