Question

any math expert please help! ! !

Answer 1

Answer:

Step-by-step 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

    $2 \ cosx \ sinx \ = k^2 -1 \\<=> sinx \ cosx = \dfrac{k^2-1}{2}$

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

$cos^4x+sin^4x=1-2cos^2xsin^2x = 1-2(\dfrac{k^2-1}{2})^2 = \dfrac{2-(k^2-1)^2}{2}$

and we know that

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

so

$cos^4x+sin^4x=\dfrac{2-k^4+2k^2-1}{2}=\dfrac{-k^4+2k^2+1}{2}$

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 $(sinx-cosx)=\sqrt{2-k^2}$

and this is the answer to the last question

do not hesitate if you need further explanation

hope this helps