Question

If the value of $(1 + \sqrt {1 + x} \tan y = 1 + \sqrt {1 - x} )$, then $\sin 4y$ is equal to
A.4x
B.2x
C.x
D.None of these

Answer 1

Hint: Use basic trigonometric conversions like $\sin C + \sin D = 2\sin (\dfrac{{C + D}}{2})\cos (\dfrac{{C - D}}{2}),\cos C + \cos D = 2\cos (\dfrac{{C + D}}{2})\cos (\dfrac{{C - D}}{2})$ and $\tan x = \dfrac{{\sin x}}{{\cos x}}$. Use this to find the value of $\sin 4y$.
$\tan y = \dfrac{{1 + \sqrt {1 - x} }}{{1 + \sqrt {1 + x} }}$……………..(i)

Complete step-by-step answer:
Let $x = \cos \theta $
$ \Rightarrow $$1 + x = 1 + \cos \theta $
                 $ = 2{\cos ^2}(\dfrac{\theta }{2})$
$ \Rightarrow \sqrt {1 + x} = \sqrt 2 \cos \dfrac{\theta }{2}$
Similarly ,
$ \Rightarrow 1 - x = 1 - \cos \theta $
              $ = 2{\sin ^2}\dfrac{\theta }{2}$
$ \Rightarrow \sqrt {1 - x} = \sqrt 2 \sin \dfrac{\theta }{2}$, Hence substituting the value of $\sqrt {1 - x} $ in equation(i)
We get,
$\tan y = \dfrac{{1 + \sqrt {1 - x} }}{{1 + \sqrt {1 + x} }}$
            $
   = \dfrac{{1 + \sqrt 2 \sin (\dfrac{\theta }{2})}}{{1 + \sqrt 2 \cos \dfrac{\theta }{2}}} \\
   = \dfrac{{\sqrt 2 (\dfrac{1}{{\sqrt 2 }} + \sin (\dfrac{\theta }{2}))}}{{\sqrt 2 (\dfrac{1}{{\sqrt 2 }} + \cos (\dfrac{\theta }{2}))}} \\
 $
As we know that $\dfrac{1}{{\sqrt 2 }}$can be written as $\dfrac{\pi }{4}$.
Hence, $\tan y = \dfrac{{\sin \dfrac{\pi }{4} + \sin \dfrac{\theta }{2}}}{{\cos \dfrac{\pi }{4} + \cos \dfrac{\theta }{2}}}$
We know that ,
$
  \sin C + \sin D = 2\sin (\dfrac{{C + D}}{2})\cos (\dfrac{{C - D}}{2}) \\
  \cos C + \cos D = 2\cos (\dfrac{{C + D}}{2})\cos (\dfrac{{C - D}}{2}) \\
 $
Hence substituting the value in $\tan y = \dfrac{{\sin \dfrac{\pi }{4} + \sin \dfrac{\theta }{2}}}{{\cos \dfrac{\pi }{4} + \cos \dfrac{\theta }{2}}}$ , we get
$\therefore $$\tan y = \dfrac{{2\sin (\dfrac{\pi }{8} + \dfrac{\theta }{4})\cos (\dfrac{\pi }{8} - \dfrac{\theta }{4})}}{{2\cos (\dfrac{\pi }{8} + \dfrac{\theta }{4})\cos (\dfrac{\pi }{8} - \dfrac{\theta }{4})}}$
By simplifying we get , \[\tan y = \dfrac{{\sin (\dfrac{\pi }{8} + \dfrac{\theta }{4})}}{{\cos (\dfrac{\pi }{8} + \dfrac{\theta }{4})}}\] hence we know that $\tan x = \dfrac{{\sin x}}{{\cos x}}$
Hence by substituting the value we get ,
$ \Rightarrow $ $\tan y = \tan (\dfrac{\pi }{8} + \dfrac{\theta }{4})$
$
   \Rightarrow y = \dfrac{\pi }{8} + \dfrac{\theta }{4} \\
   \Rightarrow 4y = \dfrac{\pi }{2} + \theta \\
 $
Hence according to question we have to find $\sin 4y$, so
$\sin 4y = \sin (\dfrac{\pi }{2} + \theta )$
Hence , $(\dfrac{\pi }{2} + \theta )$ lies in second quadrant , where $\sin $ is positive , hence it is equal to $\cos \theta $ and initially we have assumed $\cos \theta $$ = x$.
$\therefore \sin 4y = x$

Note: It is always advisable to remember such basic conversions while involving trigonometric questions as it helps save a lot of time. Trigonometric identity and sign of all trigonometric functions in all the quadrants is required to solve this problem.