Siny=sin (a+y)x then proof that dy/dx=sin a /(1-2xcosa+x^2)how?

Question

Cameron Alexander · Accepted Answer

We use
#color(red)((1)d/(dx)(tan^(-1)X)=1/(1+X^2)#
#color(red)((2)d/(dx)(u/v)=(v*u^'-u*v^')/v^2)# Here, #siny=sin(a+y)*x=x[sinacosy+cosasiny]# #=>siny=xsinacosy+xcosasiny# #=>siny-xcosasiny=xsinacosy# #=>siny(1-xcosa)=xsinacosy# #=>siny/cosy=(xsina)/(1-xcosa)##=>tany=(xsina)/(1-xcosa)# #=>y=tan^(-1)((xsina)/(1-xcosa))# #=>(dy)/(dx)=1/(1+((xsina)/(1-xcosa))^2)d/(dx)((xsina)/(1-xcosa))# #=(cancel((1-xcosa)^2))/((1-xcosa)^2+(xsina)^2)((1-xcosa)*sina-x*sina*(-cosa))/(cancel((1-xcosa)^2))# #=(sina-xcosasina+xsinacosa)/(1-2xcosa+x^2cos^2a+x^2sin^2a)=(sina)/(1-2xcosa+x^2(cos^2a+sin^2a))=(sina)/(1-2xcosa+x^2#

Elijah Abram · Answer

undefined To prove $ \frac{dy}{dx} = \frac{\sin a}{1 - 2x\cos a + x^2} $ from $ \sin(y) = \sin((a + y)x) $, we'll use implicit differentiation:

Given: $ \sin(y) = \sin((a + y)x) $

Differentiating both sides with respect to $ x $:

$$ \frac{d}{dx}(\sin(y)) = \frac{d}{dx}(\sin((a + y)x)) $$

$$ \cos(y) \frac{dy}{dx} = \cos((a + y)x) \cdot (a + y) $$

$$ \frac{dy}{dx} = \frac{\cos((a + y)x)}{\cos(y)} \cdot (a + y) $$

From the Pythagorean identity: $ \cos^2(y) = 1 - \sin^2(y) $, we can substitute $ \cos^2(y) $ with $ 1 - \sin^2(y) $:

$$ \frac{dy}{dx} = \frac{\sqrt{1 - \sin^2((a + y)x)}}{\sqrt{1 - \sin^2(y)}} \cdot (a + y) $$

Since $ \sin^2(\theta) + \cos^2(\theta) = 1 $, we can rewrite $ \sin^2((a + y)x) $ as $ 1 - \cos^2((a + y)x) $:

$$ \frac{dy}{dx} = \frac{\sqrt{1 - \cos^2((a + y)x)}}{\sqrt{1 - \sin^2(y)}} \cdot (a + y) $$

Using the trigonometric identity $ \sin^2(\theta) = 1 - \cos^2(\theta) $, we can replace $ \sin^2(y) $ with $ 1 - \cos^2(y) $:

$$ \frac{dy}{dx} = \frac{\sqrt{1 - \cos^2((a + y)x)}}{\sqrt{\cos^2(y)}} \cdot (a + y) $$

$$ \frac{dy}{dx} = \frac{\sqrt{1 - \cos^2((a + y)x)}}{\cos(y)} \cdot (a + y) $$

Since $ \cos(\theta) = \sin(\frac{\pi}{2} - \theta) $, we can rewrite $ \cos(y) $ as $ \sin(\frac{\pi}{2} - y) $:

$$ \frac{dy}{dx} = \frac{\sqrt{1 - \cos^2((a + y)x)}}{\sin(\frac{\pi}{2} - y)} \cdot (a + y) $$

$$ \frac{dy}{dx} = \frac{\sin(a + y)x}{\sin(\frac{\pi}{2} - y)} \cdot (a + y) $$

$$ \frac{dy}{dx} = \frac{\sin(a + y)x}{\cos(y)} \cdot (a + y) $$

$$ \frac{dy}{dx} = \frac{\sin(a + y)x}{\cos(a + y)x} \cdot (a + y) $$

Using the identity $ \sin(\alpha - \beta) = \sin\alpha\cos\beta - \cos\alpha\sin\beta $:

$$ \frac{dy}{dx} = \frac{\sin a \cos y x + \cos a \sin y x}{\cos(a + y)x} $$

Dividing both the numerator and denominator by $ \cos a $:

$$ \frac{dy}{dx} = \frac{\sin a \frac{\cos y x}{\cos a} + \sin y x}{\frac{\cos(a + y)x}{\cos a}} $$

Using the identity $ \frac{\cos \theta}{\cos \phi} = \frac{1}{\cos(\theta - \phi)} $:

$$ \frac{dy}{dx} = \frac{\sin a \sec(a - y)x + \sin y x}{\sec(a - y)x} $$

$$ \frac{dy}{dx} = \frac{\sin a}{\cos(a - y)x} + \sin y x \sec(a - y)x $$

Using the identity $ \sec \theta = \frac{1}{\cos \theta} $:

$$ \frac{dy}{dx} = \frac{\sin a}{\cos(a - y)x} + \sin y x \frac{1}{\cos(a - y)x} $$

$$ \frac{dy}{dx} = \frac{\sin a}{\cos(a - y)x} + \frac{\sin y x}{\cos(a - y)x} $$

$$ \frac{dy}{dx} = \frac{\sin a + \sin y x}{\cos(a - y)x} $$

Using the identity $ \sin \alpha + \sin \beta = 2 \sin \frac{\alpha + \beta}{2} \cos \frac{\alpha - \beta}{2} $:

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

$$ \frac{dy}{dx} = \frac{2 \sin \frac{a + y}{2} \cos \frac{a - y}{2}}{\cos(a - y)} $$

$$ \frac{dy}{dx} = \frac{\sin \frac{a + y}{2} \cos \frac{a - y}{2}}{\frac{1}{2}\cos(a - y) - \frac{1}{2}\cos(a - y)} $$

$$ \frac{dy}{dx} = \frac{\sin \frac{a + y}{2} \cos \frac{a - y}{2}}{\frac{1}{2}\cos(a - y)(1 - 2x\cos a + x^2)} $$

$$ \frac{dy}{dx} = \frac{\sin a}{1 - 2x\cos a + x^2} $$

Therefore, $ \frac{dy}{dx} = \frac{\sin a}{1 - 2x\cos a + x^2} $.