costheta+ sintheta= √2sintheta find sinetheta- costheta
Anonymous:
√2 cos theta will be the answer
but u didn't explained how to do it.
I can't explain in comment box
then it's ok
check if my explanation is correct
Yes, it is correct..
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Answers
Answered by
2
Sin A + Cos A = √2 sin A
so, Cos A = sin A (√2-1)
cos A/√2-1 = sin A
= cos A * √2 + 1 / 2-1 = sin A [By rationalising]
= √2cos A + cos A = sinA
so Sin A - Cos A = √2 cos A
so, Cos A = sin A (√2-1)
cos A/√2-1 = sin A
= cos A * √2 + 1 / 2-1 = sin A [By rationalising]
= √2cos A + cos A = sinA
so Sin A - Cos A = √2 cos A
Answered by
0
hey friend !!!
cos¢+sin¢=√2sin¢
{squaring on both side }
(cos¢+sin¢)²=(√2sin¢)²
cos²¢+sin²¢+2sin¢×cos¢=2sin²¢
changing place
cos²¢=2sin²¢-sin²¢-2cos¢×sin¢
cos²¢+cos²¢=sin²¢-2sin¢×cos¢+cos²¢
{adding cos²¢ on both side }
then we got like (a-b)²=a²+b²+2ab
so..
2cos²¢=(sin¢-cos¢)²
(sin¢-cos¢)²=2cos²¢
sin¢-cos¢=√2cos²¢
=>sin¢-cos¢=√2 cos¢ Ans
HOPE IT HELPS YOU !!
@RAJUKUMAR111
cos¢+sin¢=√2sin¢
{squaring on both side }
(cos¢+sin¢)²=(√2sin¢)²
cos²¢+sin²¢+2sin¢×cos¢=2sin²¢
changing place
cos²¢=2sin²¢-sin²¢-2cos¢×sin¢
cos²¢+cos²¢=sin²¢-2sin¢×cos¢+cos²¢
{adding cos²¢ on both side }
then we got like (a-b)²=a²+b²+2ab
so..
2cos²¢=(sin¢-cos¢)²
(sin¢-cos¢)²=2cos²¢
sin¢-cos¢=√2cos²¢
=>sin¢-cos¢=√2 cos¢ Ans
HOPE IT HELPS YOU !!
@RAJUKUMAR111
solved wrong
see the question correctly
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