Question

If x=asinA+bcosA and y=acosA-bsinA,prove that x^2+y^2=a^2+b^2

Answer 1

Answer:  The proof is done below.

Step-by-step explanation:  We are given that

$x=a\sin A+b\cos A,~~~~~y=a\cos A-b\sin A.$

We are to prove that

$x^2+y^2=a^2+b^2.$

We will be using the following trigonometric identity :

$\cos^2A+\sin^2A=1.$

We have

$L.H.S.\\\\=x^2+y^2\\\\=(a\sin A+b\cos A)^2+(a\cos A-b\sin A)^2\\\\=a^2\sin^2A+2ab\sin A\cos A+b^2\cos^2A+a^2\cos^2A-2ab\cos A\sin A+b^2\sin^2A\\\\=a^2(\sin^2A+\cos^2A)+b^2(\cos^2A+\sin^2A)\\\\=a^2\times1+b^2\times1\\\\=a^2+b^2\\\\=R.H.S.$

Thus, we get

$x^2+y^2=a^2+b^2.$

Hence proved.

Answer 2

Answer:

Given x=asinA+bcosA ---(1)

and

y=acosA-bsinA ------(2)

LHS = x²+y²

=(asinA+bcosA)²+(acosA-bsinA)²

= a²sin²A+2absinAcosA+b²cos²A

+a²cos²A-2absinAcosA+b²sin²A

/* By algebraic identities:

i) (m+n)² = m²+2mn+n²

ii)(m-n)² = m²-2mn+n² */

= a²(sin²A+cos²A)+b²(cos²A+sin²A)

= a²+b²

/* By Trigonometric identity:

cos²A+sin²A = 1 */

= RHS

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