Calculate cos³x + sin³x , if cos x + sin x = ½

Question

Is there an Addition Theorem for (cos x/sin x) with ...³ exponents? Or just cos²x + sin²x=1 ?

 

Your help is appreciated. ;)

Answer 1

Since Alan shows you how to factor that SUM OF CUBES, the answer is (sin x + cos x)(sin²x - sin x cos x + sin²x) = 

(sin x + cos x)(1 - sin x cos x) =½[1 - sin x cos x].

 

On the other hand, squaring sin x + cos x = ½ gives ¼ = sin²x + 2 sinx cos x + cos²x

so -3/4 = 2 sin x cos x   so  -3/8 = sin x cos x.

 

Substitute the above negative constant into the last equation of my first pararaph, we get

this result:  ½[1 - (-3/8)] = ½(1 + 3/8) = ½(11/8) = 11/16.

 

Disregard my separate comment, which was accidentally sent before it was fully typed.

 

Answer 2

There is an angle in the second quadrant that approximate very well the solution to this problem using numerical methods. That angle is in the close vicinity of 114.295 degrees !

 

Indeed, cos 114.295 + sin 114.295 =~ 0.50001 . I did some computations for you, Franz, ...

 

Knowing the existence of that angle and having at hand an excellent approximation of that angle x, we can fairly calculate cos³x + sin³x.

 

cos³(114.295) + sin³(114.295) = [-0.4114348221]³ + [0.9114391846]³ = -0.069647116 + 0.7571520238 = 0.6875049078  or 0.6875 rounded to 4 decimals.

Answer 3

Franz,

 

Do you know anything else on this question, for example the quadrant of x?

 

Here is a hint which may be helpful, in the meantime.

 

a³ + b³ = (a + b)(a² - ab + b²).

 

If you let a = cos x and b = sin x, you can replace the first factor by ½. As for the second factor, it cannot be factored and you may get stuck unless you are given more information about x.

 

It is true that cos² x + sin² x = 1, a Pythagorean identity, but there is no such identity for cos³ x + sin³ x.

 

Please let me know if there is more to this problem than what you have already given.

 

Thank you.