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if 2cosA=x+1/x,2cosB=y+1/y then show that 2cos(A-B)=x/y+y/x
Solve these as quadratic equations in x & y so that you get x = cosA + i sinA & cosA – i sinA; y = cosB + i sinB & cosB – isinBIf x = cosA + i sinA then 1/x = cosA – i sinA ;y = cosB + i sinB then 1/y cosB – isinBx + 1/x = 2cosA => cosA = 1/2(x + 1/x)x – 1/x = 2i sinA => sinA = 1/2i(x – 1/x)y + 1/y = 2cosB => cosB = 1/2(y + 1/y)y – 1/y = 2i sinB => sinB = 1/2i(y – 1/y)2cos(A + B) = 2(cosA cosB – sinA sinB)= 2[1/2(x + 1/x) * 1/2(y + 1/y) – 1/2i(x – 1/x) * 1/2i(y – 1/y)]= 2[1/4(xy + x/y + y/x + 1/xy + xy -x/y – y/x + 1/xy)] = xy + 1/xy
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