Question

If Tn =Sin n A +Cos n A Then Prove That - 6T10- 15 T 8+ 10 T 6 -1= 0

Answer 1

There are two possibilities of this question. As the question is not clear, I will do both ways.

1)
$T^n(A)=Sin^n A+cos^n A\\\\T^2(A)=sin^2A+cos^2A=1,\ \ T(A)=+1\ or \ -1\\\\T^6(A)=1=T^8(A)=T^{10}(A)\\\\LHS=6*1-15*1+10*1-1=0=RHS$

2)  For easiness in writing the solution, let C = CosA ,  S= SinA
 
we know S² + C² = 1

$T_n(A)=Sin^n A+Cos^n A\\\\ T_4=S^4+C^4=(S^2+C^2)^2-2S^2C^2 =1-2S^2C^2\\\\T_6=S^6+C^6=(S^4+C^4)(S^2+C^2)-S^2C^2(S^2+C^2)\\\\=S^4+C^4-S^2C^2=(S^2+C^2)^2-3S^2C^2=1-3S^2C^2\\\\T_8=(S^6+C^6)(S^2+C^2)-S^2C^2(S^4+C^4)\\\\=1-3S^2C^2-S^2C^2(1-2S^2C^2)\\\\=1-4S^2C^2+2S^4C^4\\\\T_{10}=(S^6+C^6)(S^4+C^4)-S^4C^4(S^2+C^2)\\\\=(1-3s^2C^2)(1-2S^2C^2)-S^4C^4\\\\=1-5S^2C^2+5S^4C^4\\\\Substituting\ in\ LHS\ we\ get\ 0.$
 
Substituting these values in 6 * T₁₀ - 15 T₈ + 10 T₆ - 1  we get  0