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  • Prove that y= 4 sin theta/ 2 + cos theta - theta is an increasing function of theta in 0 to pie by 2

9. Prove that   y = \frac{4 \sin \theta }{(2+ \cos \theta )} - \theta is an increasing function of    \theta\: \: in\: \: [ 0 , \pi /2 ]

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best_answer

Given function is,
f(x) = y = \frac{4 \sin \theta }{(2+ \cos \theta )} - \theta

f^{'}(x) = \frac{dy}{d\theta} = \frac{4 \cos \theta(2+\cos \theta) - (-\sin \theta)4\sin \theta) }{(2+ \cos \theta )^2} - 1
                         = \frac{8 \cos \theta+4\cos^2 \theta + 4\sin^2 \theta - (2+ \cos \theta )^2 }{(2+ \cos \theta )^2}
                         = \frac{8 \cos \theta+4(\cos^2 \theta + \sin^2 \theta) - 4- \cos^2 \theta - 4\cos \theta }{(2+ \cos \theta )^2}
                         = \frac{8 \cos \theta+4 - 4- \cos^2 \theta - 4\cos \theta }{(2+ \cos \theta )^2}                                                 
                         = \frac{4 \cos \theta-\cos^2 \theta }{(2+ \cos \theta )^2}
Now,  for  \theta \ \epsilon \ [0,\frac{\pi}{2}]
\\ 4 \cos \theta \geq \cos^2 \theta\\ 4 \cos \theta - \cos^2 \geq 0\\ and \ (2+\cos \theta)^2 > 0
So, f^{'}(x) > 0 \ for \ \theta \ in \ [0,\frac{\pi}{2}]
Hence, f(x) = y = \frac{4 \sin \theta }{(2+ \cos \theta )} - \theta is increasing function in  \theta \ \epsilon \ [0,\frac{\pi}{2}]
    

Posted by

Gautam harsolia

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