Question

$x = a \sin( \alpha ) + b \cos( \alpha ) \\ \\ y = a \cos( \alpha ) + b \sin( \alpha )$
prove that:
${x}^{2} + {y}^{2} = {a}^{2} + {b}^{2}$
​

Answer 1

Given :

▪ x = a(sinα) + b(cosα)

▪ y = a(cosα) + b (sinα)

To Prove :

▪ x² + y² = a² + b²

Explanation :

LHS :

= x² + y²

= [a(sinα) + b(cosα)]² + [a(cosα) + b (sinα)]²

= a²sin²α + b²cos²α + a²cos²α + b²sin²α

= a²(sin²α + cos²α) + b²(sin²α + cos²α)

we know that, sin²Φ + cos²Φ = 1

= a²(1) + b²(1)

= a² + b² = RHS

Hence Proved !!

Learn more :

▪ sin²Φ + cos²Φ = 1

▪ sec²Φ - tan² = 1

▪ cosec²Φ - cot²Φ = 1

▪ sin2Φ = 2(sinΦ)(cosΦ)

▪ cos2Φ = cos²Φ - sin²Φ

Answer 2

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$\huge {\purple {\fbox {\bigstar {\mathbf {\red {ur\: question}}}}}}$

$\huge {\mathbf {\purple {Q.}}}}$$x = a \sin( \alpha ) + b \cos( \alpha ) \\ \\ y = a \cos( \alpha ) + b \sin( \alpha )$

prove that:

${x}^{2} + {y}^{2} = {a}^{2} + {b}^{2}$

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$\huge {\purple {\fbox {\bigstar {\mathbf {\red {ur\: answer✓}}}}}}$

$\huge\purple{\text{given}}$

▪ x = a(sinα) + b(cosα)

▪ y = a(cosα) + b (sinα)

$\pink{\text{to\:prove,}}$

▪ x² + y² = a² + b²

$\tt\implies LHS$

= x² + y²

= [a(sinα) + b(cosα)]² + [a(cosα) + b (sinα)]²

= a²sin²α + b²cos²α + a²cos²α + b²sin²α

= a²(sin²α + cos²α) + b²(sin²α + cos²α)

we know that, sin²Φ + cos²Φ = 1

= a²(1) + b²(1)

= a² + b² = RHS

$\huge{\blue{\fbox{\purple{\bigstar{\mathbf{\red{ a^2 + b^2 = RHS}}}}}}}$

Hence Proved

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