Question

if y=asinx+bcosx , prive that y2 + (dy/dx)2 = a2 +b2

Answer 1

Here we will use the following three concepts:


$\boxed{\frac{d}{dx}\sin x = \cos x} \\ \\ \\ \boxed{\frac{d}{dx}\cos x = -\sin x} \\ \\ \\ \boxed{\sin^2x+\cos^2x=1}$

__________________________


Coming to the question:

We have:
$y = a\sin x + b\cos x \quad ---(1)$

Differentiating

$\frac{dy}{dx} = a\cos x + b (-\sin x) \\ \\ \\ \implies \frac{dy}{dx} = a\cos x - b\sin x \quad ---(2)$


Squaring and adding (1) and (2)

$(1)^2+(2)^2\\ \\ \begin{array}{cc}\implies y^2+\left(\frac{dy}{dx}\right)^2=&(a\sin x+b\cos x)^2+(a\cos x-b\sin x)^2\\ \\ \implies y^2+\left(\frac{dy}{dx}\right)^2=&a^2\sin^2x+2ab\sin x\cos x+b\cos^2x+\\ \\&a^2\cos^2x-2ab\sin x\cos x+b\sin^2x\\ \\ \implies y^2+\left(\frac{dy}{dx}\right)^2 = & a^2(\sin^2x+\cos^2x)+b^2(\cos^2x+\sin^2x)\\ \\ \implies y^2 + \left(\frac{dy}{dx}\right)^2=&a^2(1)+b^2(1) \end{array}\\ \\ \\ \implies \boxed{y^2 + \left( \frac{dy}{dx} \right)^2 = a^2+b^2}$

Hence Proved

Answer 2

Step-by-step explanation:

This is the ans required.Hope this helps.