Question

If m=acos theta -b sin theta,n=a sin theta +bcos theta find m^2+n^2=a^2+b^2

Answer 1

Theta is written as A,




➡ m = a cos A - b sin A

Square on both sides,

➡ m² = ( a cosA - b sinA )²

$\bold{ m {}^{2} = a {}^{2} cos {}^{2} A + b {}^{2} sin {}^{2} A - 2ab \times cosA \times sinA } \: \: \: \: \: \: - - - - - - -(1)$





➡ n = a sinA + b cosA

Square on both sides,

➡ n² = ( a sinA + b cosA )²

$\bold{n {}^{2} = a {}^{2} sin {}^{2} A + {b}^{2} cos{}^{2} A + 2ab cosA \times sinA } \: \: \: \: \: - - - - - - - - - (2)$





Adding ( 1 ) & ( 2 ),

⏩ m² + n² = a² cos²A + b² sin²A - 2ab ( sinA cosA ) + a² sin²A + b² cos²A + 2ab ( sinA cosA )



⏩ m² + n² = a² cos²A + a² sin²A + b² sin²A + b² cos²A - 2ab( sinA cosA ) - 2ab( sin² cos²A )


⏩ m² + n² = a²( cos²A + sin²A ) + b²( sin²A + cos²A )




$\fbox{ \text{We know that sin} {}^{2} \text{A +} \: \text{cos} {}^{2} \text{A} = 1}$


So,


⏩ m² + n² = a²( 1 ) + b²( 1 )

▶ m² + n² = a² + b²





Hence, proved.