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Any math expert please help! ! !

Any math expert please help! ! !-example-1

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Answer:

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^2so
(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

User AmGates
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