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Evaluate : 6(cos^10θ + sin^10θ) - 15(cos^8θ + sin^8θ) + 10(cos^6θ + sin^6θ)

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nasirjam | Student, Grade 9 | Honors

Posted September 13, 2012 at 4:51 PM via web

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Evaluate :

6(cos^10θ + sin^10θ) - 15(cos^8θ + sin^8θ) + 10(cos^6θ + sin^6θ)

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jeew-m | College Teacher | (Level 1) Educator Emeritus

Posted September 14, 2012 at 2:05 AM (Answer #1)

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To do this we need the following.

`(x-y)^2 = x^2-2xy+y^2`

`(x-y)^3 = x^3-3x^2y+3xy-y^3`

`sin^2theta+cos^2theta = 1`

`a^m*a^n = a^(m+n)`

 

`6(cos^10theta + sin^10theta) - 15(cos^8theta + sin^8theta) + 10(cos^6theta +sin^6theta)`

`= cos^6theta(6cos^4theta-15cos^2theta+10)+sin^6theta(6sin^4theta-15sin^2theta+10)`

 

`sin^6theta`

`= (sin^2theta)^3`

`= (1-cos^2theta)^3`

`= 1-3cos^2theta+3cos^4theta-cos^6theta`

 

`sin^4theta`

`= (1-cos^2theta)^2`

`= 1-2cos^2theta+cos^4theta`

 

`sin^6theta(6sin^4theta-15sin^2theta+10)`

`= (1-3cos^2theta+3cos^4theta-cos^6theta)[6(1-2cos^2theta+cos^4theta)-15(1-cos^2theta+10]`

`= (1-3cos^2theta+3cos^4theta-cos^6theta)(1+3cos^2theta+6cos^4theta)`

`= 1+3cos^2theta+6cos^4theta-3cos^2theta-9cos^4theta-` `18cos^6theta+3cos^4theta+9cos^6theta+18cos^8theta-` `cos^6theta-3cos^8theta-6cos^10theta`

 

`=1+cos^2theta(3-3)+cos^4theta(6-9+3)+cos^6theta(-18+9-1)+ cos^8theta(18-3)-6cos^10theta`

 

`= 1-10cos^6theta+15cos^8theta-6cos^10theta`

 

`6(cos^10theta + sin^10theta) - 15(cos^8theta + sin^8theta) + 10(cos^6theta +sin^6theta)`

`= cos^6theta(6cos^4theta-15cos^2theta+10)+sin^6theta(6sin^4theta-15sin^2theta+10)`

`= cos^6theta(6cos^4theta-15cos^2theta+10)+1-10cos^6theta+15cos^8theta-6cos^10theta`

`= 6cos^10theta-15cos^8theta+10cos^6theta+1-10cos^6theta+15cos^8theta-6cos^10theta`

`= 1`

 

Therefore;

`6(cos^10theta + sin^10theta) - 15(cos^8theta + sin^8theta) + 10(cos^6theta +sin^6theta)= 1`


 

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