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December 18, 2014

December 18, 2014

Posted by **Sam** on Tuesday, November 5, 2013 at 10:26pm.

Hmm...I don't understand how to proceed. I know I must apply a trig Identity, but which one?

Thanks in advance

- Math - Trigonometry -
**Reiny**, Tuesday, November 5, 2013 at 11:33pmsinØ + cosØ = 1.2

square both sides

sin^2 Ø + 2sinØcosØ + cos^2 Ø = 1.44

but sin^2Ø + cos^2Ø = 1

2sinØcosØ = 0.44

sinØcosØ = .22

now using the fact that

(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3

we see that

a^3 + b^3 = (a+b)^3 - 3a^2b - 3ab^2

= (a+b)^3 - 3ab(a+b)

so sin^3 Ø + cos^3 Ø

= (sinØ + cosØ)^3 - 3sinØcosØ (sinØ + cosØ)

= 1.2^3 - 3(.22)(1.2)

=**0.936**

check:

from 2sinØcosØ = .44

sin 2Ø = .44

2Ø = 26.103...

Ø = 13.0519....

then by calculator: sin^3 13.0519° + cos^3 13.0519°

= .93599...

not bad!

- Math - Trigonometry -
**Sam**, Wednesday, November 6, 2013 at 7:15amThanks a lot :)

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