Prove the identity: cos2x/ sinx + 2 sinx = cscx?

Question

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Answer 1

cos (2x)/sin x + 2 sin x
(cos² x - sin² x)/sin x + 2 sin x
(cos² x - sin² x)/sin x + 2 sin² x/sin x
(cos² x + sin² x)/sin x
1/sin x
csc x

And we are done.

Answer 2

Do you see the double identity cos2x?

We need to rewrite this guy in such a way that it would alllow us to break down the left side fraction.

There are three types of cos2x:

(1) cos^2x - sin^2x

(2) 2cos^2x - 1

(3) 1 - 2sin^2x

It does not matter which one you decide to use in place of cos2x.

Is this clear so far?

I will use number (3).

(1 - 2sin^2x)/(sinx + 2sinx) = cscx

Place (1 - 2sin^2x + 2sin^2) over sinx

Notic that -2sin^2 and +2sin^2 cancel out leaving us with

1/sinx, which is a reciprocal of cscx.

Then:

1/sinx = cscx

cscx = cscx

Done!

Answer 3

From the double angle formula

cos2x = 1 - 2*(sinx)^2

So cos2x/sinx + 2sinx = (1-2*(sinx)^2)/sinx + 2sinx

Now write this with a common denominator of sinx and you get

(1-2*(sinx)^2 + 2*(sinx^2))/sinx = 1/sinx = cscx

Answer 4

cos2x/sinx = (1 - 2sin^2 x)/sinx

2sinx= (2sin^2 x)/sinx

add them to get: 1/(sinx)
which is, of course, cscx

Know your trig rules!