how is that if 1/cos(x) +1/sin(x)=√2 it must be that cos^3(x)+sin^3(x)= -(1/√2)

Question

This is a problem from a book published in 1930, by authors Durell&Robson

Advanced Trigonometry, it is a problem from the last chapter

again, show if

1/cos(x) +1/sin(x)=√2

then cos³(x)+sin³(x)= -(1/√2)

It is not hard numerically to say cos³(x)+sin³(x) equals about -.707

hint sin(2x)=1-√3

so with some discernment can conlude x is a certain radian angle x

and that will compute to about -.707= cos³(x)+sin³(x)

but it takes advanced trigonometry and advanced algebra to prove

when 1/cos(x) +1/sin(x)=√2

then exactly cos³(x)+sin³(x)= -(1/√2)

Answer 1

have from 1/sinx+1/cosx=√2

cosx+sinx=√2sinxcosx=(√2sin2x)/2

squaring,both sides

1+sin2x=(1/2)sin²2x

solving quadradic

sin2x=1±√3 , cannot use 1+√3 as sin(angle)<1

sin2x=1-√3

meaing angle 2x is either in quadrant III or IV

sin^-1(1-√3)=-.82 or -π-(-.82)=-2.32 radians

so x=-.41 or -.1.16

checking when 1/sinx+1/cosx=√2, x=1.16 (we need this to determine ± in following)

sinx=(1/2)[±√(1+sin2x)±√(1-sin2x)]

cosx=(1/2)[±√(1+sin2x)±√(1+sin2x)]

these two half angle formulas come from adding (for sinx) or subtracting (for cosx)

(sinx+cosx)²=1+sin2x

(sinx-cosx)²=1-sin2x

furthermore

sinx+cosx=±√2sin(x+π/4) sin(-1.16+π/4) is negative, use negative sign

sinx-cosx=±√2sin(x-π/4) sin(-1.16-π/4) is negative, use negative sign

sinx+cosx=±√2cos(x-π/4) cos(-1.16-π/4) is negative, use negative sign

sinx-cosx=±√2cos(x+π/4) cos(-1.16+π/4) is positive, use positive sign

sinx=(1/2)[-√(1+1-√3-)-√(1-(1-√3)]=(1/2)[-√(2-√3)-√√3]

cosx=(1/2)[-√(1+1-√3-)+√(1-(1-√3)]=(1/2)[-√(2-√3)+√√3]

[(1/2³)[-√(2-√3)-√√3]]³+ [(1/2³)[-√(2-√3)+√√3]]³=-(1/√2)

there is algebra involved

but indeed [(1/2³)[-√(2-√3)-√√3]]³+ [(1/2³)[-√(2-√3)+√√3]]³=-(1/√2)

that is, when 1/sinx+1/cosx=√2, then sin³x+cos³x=-1/√2

Answer 2

desmos.com/calculator/2bkh095lbf

The above graph also has rows that confirm solutions shown in the video.