If sin theta +cos theta =1.2, then what is sin^3 theta + cos^3 theta?
Hmm...I don't understand how to proceed. I know I must apply a trig Identity, but which one?
Thanks in advance
sinØ + 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!
Thanks a lot :)
To solve this problem, we can start by manipulating the equation sin(theta) + cos(theta) = 1.2. Let's square both sides of the equation:
(sin(theta) + cos(theta))^2 = (1.2)^2
Expanding the left side using the binomial formula, we get:
sin^2(theta) + 2sin(theta)cos(theta) + cos^2(theta) = 1.44
Since sin^2(theta) + cos^2(theta) = 1 (using the Pythagorean identity), we can rearrange the equation:
1 + 2sin(theta)cos(theta) = 1.44
Subtracting 1 from both sides gives us:
2sin(theta)cos(theta) = 0.44
Now, let's cubed both sides of the equation:
(2sin(theta)cos(theta))^3 = (0.44)^3
Expanding the left side of the equation:
8sin^3(theta)cos^3(theta) = 0.44^3
Finally, dividing both sides by 8:
sin^3(theta)cos^3(theta) = (0.44^3)/8
Therefore, sin^3(theta) + cos^3(theta) is equal to (0.44^3)/8.