If y( )=(2 + 2 )/( 3 +4 2 +5 +2), then at =( )/(2),y^''+y^'+y is equal to :

Question

If $y(	heta )=\frac{2\cos 	heta +\cos 2	heta }{\cos 3	heta +4\cos 2	heta +5\cos 	heta +2}$, then at $	heta =\frac{\pi }{2},{y}^{''}+{y}^{'}+y$ is equal to :

Accepted Answer

$$\begin{aligned}
& y(\theta)=\frac{2 \cos \theta+\cos 2 \theta}{\cos 3 \theta+4 \cos 2 \theta+5 \cos \theta+2} \
& =\frac{2 \cos ^2 \theta+2 \cos \theta-1}{4 \cos ^3 \theta+8 \cos ^2 \theta+2 \cos \theta-2} \
& =\frac{2 \cos ^2 \theta+2 \cos \theta-1}{\left(2 \cos ^2 \theta+2 \cos \theta-1\right)(2 \cos \theta+2)} \
& =\frac{1}{2(1+\cos \theta)}=\frac{1}{4 \cos ^2 \theta / 2}=\frac{\sec ^2 \theta / 2}{4} \
& y^{\prime}(\theta)=\frac{1}{4}\left(2 \sec \frac{\theta}{2} \cdot \sec \frac{\theta}{2} \cdot \tan \frac{\theta}{2} \cdot \frac{1}{2}\right) \
& =\frac{1}{4} \sec ^2 \frac{\theta}{2} \cdot \tan \frac{\theta}{2}
\end{aligned}$$

$$y^{\prime \prime}(\theta)=\frac{1}{4}\left(\tan \frac{\theta}{2}\right)\left(\sec ^2 \frac{\theta}{2} \cdot \tan \frac{\theta}{2}\right) +\frac{1}{4} \sec ^2 \frac{\theta}{2} \cdot \sec ^2 \frac{\theta}{2} \cdot \frac{1}{2}$$

$$\begin{aligned}
& \text { at } \theta=\frac{\pi}{2}, y(\theta)=\frac{1}{2}, y^{\prime}(\theta)=\frac{1}{2}, y^{\prime \prime}(\theta)=1 \
& \therefore \quad y+y^{\prime}+y^{\prime \prime}=2
\end{aligned}$$

Solutions · 1