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Question

If Tn=sinnx+cosnx, then find the value of 2T63T4+10.

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Solution

Tn=sinnx+cosnx

T4=sin4x+cos4x

T6=sin6x+cos6x

2T63T4+10=2(sin6x+cos6x)3(sin4x+cos4x)+10

=2((sin2x)3+(cos2x)3)3((sin2x)2+(cos2x)2)+10

Formula (i),(ii) and (iii) will be used in the further calculations,
(i)a3+b3=(a+b)(a2ab+b2)

(ii)a2+b2=(a+b)22ab

(iii)sin2θ+cos2θ=1

Using these

=2[(sin2x+cos2x)(sin4x+cos4xsin2xcos2x)]3sin4x3cos4x+10

=2sin4x+2cos4x2sin2xcos2x3sin4x3cos4x+10

=sin4xcos4x2sin2xcos2x+10

=10((sin2x)2+(cos2x)2+2sin2xcos2x)

=10(sin2x+cos2x)2

=101

=9

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