Question

sin theta +cosec theta=2 then sin2016+cosec 2016 equals to

Answer 1

NOTE: You have to be patient while going through the answer as it is long.

$\sin(x)+\cosec(x)=2$
$\sin(x)+\frac{1}{\sin(x)}=2$

Let $\sin(x)=y$
Then $\sin(x)+\frac{1}{\sin(x)}}=y+\frac{1}{y}=2$

Here is a formula derivation, if you only want the result you may skip it.

Let
$y + \frac{1}{y} = a$
(here a=2)

Let
$y = {e}^{iz}$
for some complex $z$.

Therefore,
$\frac{1}{y} = {e}^{ - iz}$
Euler's formula:
${e}^{iz} = \cos(z) + i \sin(z)$

Therefore,
${e}^{ - iz} = \cos( - z) + i \sin( - z)$

${e}^{ - iz} = \cos(z) - i \sin(z)$

(since sine is an odd function and cosine is an even function. This means sin(-x) = -sin(x) and cos(-x) = cos(x) )

Thus,
${e}^{iz} + {e}^{ - iz} = 2 \cos(z)$
$y + \frac{1}{y} = 2 \cos(z)$
$a = 2 \cos(z)$
$\frac{a}{2} = \cos(z) \\ arc \cos( \frac{a}{2} ) = z$

Now
${y}^{n} + \frac{1}{ {y}^{n} } = {e}^{izn} + {e}^{ - izn}$
${e}^{izn} + {e}^{ - izn} = 2 \cos(nz)$

(Using de moivre's theorem.
This is easy to understand if you know law of exponents.)

$y^{n} + \frac{1}{ {y}^{n} } = 2 \cos(nz) \\ {y}^{ n} + \frac{1}{ {y}^{n} } = 2 \cos(n \times arc \cos( \frac{a}{2} ) )$

This is the formula which is now going to be used.

Replace n=2016 and a=2 in the formula.

${y}^{2016} + \frac{1}{ {y}^{2016} } = 2 \cos(2016 \times arc \cos( \frac{2}{2} ) ) \\ {y}^{2016} + \frac{1}{ {y}^{2016} } = 2 \cos(2016 \times \: arc \cos(1) )$

As,
$arc \cos(1) = 0$

Therefore,
${y}^{2016} + \frac{1}{y^{2016} } = 2 \cos(0) \\ {y}^{2016} + \frac{1}{y^{2016} } = 2$
(as cos(0)=1)

${ \sin }^{2016} x + \frac{1}{ \sin^{2016}x } = 2 \\ { \sin }^{2016} x + \csc^{2016}x = 2$

That's it.