Question

If sin X + cos X = 2 , find the value of sinnx + cosnx . n if not equal to zero

Answer 1

Step-by-step explanation:

Certainly this is impossible when x∈R (in which case the maximum possible value of sinx+cosx=2–√sin(x+π/4) is 2–√ ).

However, if x∈C , then we have sinx,cosx as complex numbers, and it is possible for sinx+cosx=2 . Indeed, if a=sinx and b=cosx in this case, then a+b=2 and a2+b2=1 . Moreover, squaring a+b=2 gives a2+2ab+b2=4 ; hence (subtracting a2+b2=1 ), 2ab=3 . Therefore (a−b)2=a2−2ab+b2=1−3=−2 , so that a−b=±i2–√ . This entails that a=1±i2√2 and b=1∓i2√2 .

Now one could certainly solve cosx=b with this value of b ; take x=−iln(b+i1−b2−−−−−√) , using principal values of the square root and logarithm. Then

cosx=eix+e−ix2=eln(b+i1−b2√)+e−ln(b+i1−b2√)2=(b+i1−b2−−−−−√)+1b+i1−b2√2=(b+i1−b2−−−−−√)+(b−i1−b2−−−−−√)2=2b2=b

And moreover, since sin2x=1−cos2x , we conclude that sinx=±a . If sinx=−a , then that’s not too much of a hassle, just change x to −x and that will get the sine to be a and the cosine to be b .

Now, in your question, you ask for sinnx+cosnx . The answer to this is, of course, an+bn=(1±i2√2)n+(1∓i2√2)n . Though this is a real number (it equals its own complex conjugate), I am not sure how much it can be simplified with the exponent n as a variable. However, we can derive a recurrence relation upon setting sn=an+bn :

s1=2

s2=1

sn+2=2sn+1−32sn

These conditions alone determine the sequence sn which answers your question. Basic induction shows that the sn are all dyadic rational numbers (i.e., a/2k with a∈Z,k∈N ).