sin - cos = 3/5 then sin*cos=?
anuj:
IS IT SIN/COS OR SIN-COS
Answers
Answered by
3
Heya user,
sin x - cos x = 2/3
=> [ sin x - cos x ]² = [ 2/3 ]²
=> sin² x + cos² x - 2sinx cosx = 4/9
=> 1 - 2 sinx cosx = 4/9 <--- [ sin²x + cos²x = 1 ]
=> 1 - 4/9 = 2sinxcosx
=> 5/9 = 2 sinxcosx
=> sinx cosx = 5/18 <---- Desired Result
sin x - cos x = 2/3
=> [ sin x - cos x ]² = [ 2/3 ]²
=> sin² x + cos² x - 2sinx cosx = 4/9
=> 1 - 2 sinx cosx = 4/9 <--- [ sin²x + cos²x = 1 ]
=> 1 - 4/9 = 2sinxcosx
=> 5/9 = 2 sinxcosx
=> sinx cosx = 5/18 <---- Desired Result
answer is wrong but method is correct
read the question
Answered by
3
sin-cos=3/5
squaring both the sides
we get
(sin-cos)^2=(3/5)^2
sin^2 + cos ^2 -2 sin cos=9/25 [since sin^2 + cos ^2=1]
1-2cos sin =9/25
1-9/25=2cos sin
16/(25*2)=sin * cos
ANS = 8/25
squaring both the sides
we get
(sin-cos)^2=(3/5)^2
sin^2 + cos ^2 -2 sin cos=9/25 [since sin^2 + cos ^2=1]
1-2cos sin =9/25
1-9/25=2cos sin
16/(25*2)=sin * cos
ANS = 8/25
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