Question

Any math expert please help! ! !

Answer 1

Explanation:

Hello,

as cosx+sinx=k we can write that

`(cosx+sinx)^2=cos^2x+sin^2x+2\ cosx \ sinx = k^2`

and we know that
`cos^2x+sin^2x=1`

so

`2 \ cosx \ sinx \ = k^2 -1 \\<=> sinx \ cosx = (k^2-1)/(2)`

which is the answer to the question 2

Now, let s estimate

`1=1^2=(cos^2x+sin^2x)^2=cos^4x+sin^4x+2cos^2xsin^2x`

so

`cos^4x+sin^4x=1-2cos^2xsin^2x`

We use the previous result to write

`cos^4x+sin^4x=1-2cos^2xsin^2x = 1-2((k^2-1)/(2))^2 = (2-(k^2-1)^2)/(2)`

and we know that

`(k^2-1)^2=k^4-2k^2+1`

so

`cos^4x+sin^4x=(2-k^4+2k^2-1)/(2)=(-k^4+2k^2+1)/(2)`

this is the answer to the first question

finally, let s estimate

`(sinx-cosx)^2=cos^2x+sin^2x-2cosxsinx= 1 - (k^2-1)=-k^2+2=2-k^2`so
`(sinx-cosx)=√(2-k^2)`

and this is the answer to the last question

do not hesitate if you need further explanation

hope this helps