If a cos x + b sin x = m and a sin x – b cos x = n, prove that a^2 + b^2 = m^2 + n^2
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Answered by
30
Hi ,
a cosx + b sinx = m ---( 1 )
a sinx - b cos x = n ---( 2 )
do the Square of both the equations and
add them ,
( a²cos² x + b²sin² x + 2abcosxsinx )+ ( a² sin² x
+ b²cos² x - 2abcosxsinx ) = m² + n²
a²(cos² x + sin² x ) + b²(sin² x+cos² x ) = m²+n²
a² + b² = m² + n²
[ since cos² x + sin² x = 1 ]
Hence proved.
I hope this helps you.
: )
a cosx + b sinx = m ---( 1 )
a sinx - b cos x = n ---( 2 )
do the Square of both the equations and
add them ,
( a²cos² x + b²sin² x + 2abcosxsinx )+ ( a² sin² x
+ b²cos² x - 2abcosxsinx ) = m² + n²
a²(cos² x + sin² x ) + b²(sin² x+cos² x ) = m²+n²
a² + b² = m² + n²
[ since cos² x + sin² x = 1 ]
Hence proved.
I hope this helps you.
: )
Answered by
14
Step-by-step explanation:
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