Sin theta +cos theta =√2(sin 90-theta) then what is the value of cos theta
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let A = theta
sin A + cos A =√2 sin( 90°- A)
{sin ( 90°- A)= cosA}
(sin A + cos A) ^2=2 cos A
{(a+b)^2= a^2 + b^2 + 2ab}
sin^2 A + cos ^2 A +2 sinA cosA =2 cosA
{sin^2 A + cos ^2 A =1 }
1+ 2 sinA cosA =2 cosA
1+2 sinA cosA /2= cosA
sin A + cos A =√2 sin( 90°- A)
{sin ( 90°- A)= cosA}
(sin A + cos A) ^2=2 cos A
{(a+b)^2= a^2 + b^2 + 2ab}
sin^2 A + cos ^2 A +2 sinA cosA =2 cosA
{sin^2 A + cos ^2 A =1 }
1+ 2 sinA cosA =2 cosA
1+2 sinA cosA /2= cosA
akash2376:
while squaring, you forgot to take cos^2 A on R.H.S.
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Using rationalization after rearrangement.....
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