Question

If ${[1-\sin (\pi +\alpha )+\cos (\pi +\alpha )]}^2+{[1-\sin (\frac{3\pi }{2}+\alpha )+\cos (\frac{3\pi }{2}-\alpha )]}^2=a+b\sin 2\alpha$ then the value of $a$ & $b$ are:

• A. $a=4$ & $b=-2$
• B. $a=2$ & $b=-4$
• C. $a=2$ & $b=-2$
• D. None of these

Answer 1

Answer: (A)

$a=4$ & $b=-2$

As we know that, $sin(\pi +\alpha )=-sin\alpha cos(\pi +\alpha )=-cos\alpha sin(\frac{3\pi }{2}+\alpha )=-cos\alpha cos(\frac{3\pi }{2}-\alpha )=-sin\alpha$

So, we can write

${[1-\sin (\pi +\alpha )+\cos (\pi +\alpha )]}^2+{[1-\sin (\frac{3\pi }{2}+\alpha )+\cos (\frac{3\pi }{2}-\alpha )]}^2=[1+\sin \alpha -\cos \alpha {]}^2+[1+\cos \alpha -\sin \alpha {]}^2=1+{\sin }^2\alpha +2\sin \alpha +{\cos }^2\alpha -2\cos \alpha -2\cos \alpha \sin \alpha +1+{\cos }^2\alpha +2\cos \alpha -2\sin \alpha -2\sin \alpha \cos \alpha +{\sin }^2\alpha =4-4\sin \alpha \cos \alpha =4-2\sin 2\alpha$
Comparing $4-2\sin \alpha$ with $a+b\sin 2\alpha$
we get, $a=4,b=-2$