Question

Prove the following trigonometric identity using the double angle formula.
[sin(2x)/sinx] - [cos(2x)/cosx] = secx

Answer 1

Try to write left hand side as term
do that by first combining the fractions

Answer 2

as one term*

Answer 3

Make it into [sin(2x)cosx - cos(2x)sinx/sinxcosx]...

Answer 4

Right...

Answer 5

right use sin difference identity on top

Answer 6

\[\sin(a-b)=\sin(a)\cos(b)-\sin(b)\cos(a)\]

Answer 7

?
I'm not getting where the difference identity comes into play.

Answer 8

There isn't any sin(a+b) or sin(a-b)

Answer 9

\[\frac{\sin(2x)\cos(x)-\sin(x)\cos(2x)}{\sin(x)\cos(x)}\]

Answer 10

I see sin(a)cos(b)-sin(b)cos(a) on top...

Answer 11

Oh

Answer 12

Oh!

Answer 13

Holy pellet! Yeah! Didn't catch that...

Answer 14

So sin (2x - x)/sinxcosx...

Answer 15

2x-x=x
so we have sin(x)/(sin(x)cos(x))

Answer 16

something cancels

Answer 17

And I hope you see that we have sec(x)

Answer 18

Wow yeah. Then just the sec defintion

Answer 19

This wasn't the only approach you could have taken

Answer 20

*definition

Answer 21

\[\frac{\sin(2x)}{\sin(x)}-\frac{\cos(2x)}{\cos(x)} \\ \frac{2 \sin(x) \cos(x)}{\sin(x)}-\frac{\cos^2(x)-\sin^2(x)}{\cos(x)} \\ 2\cos(x)-(\cos(x)-\frac{\sin^2(x)}{\cos(x)} ) \\ 2\cos(x)-\cos(x)+\frac{\sin^2(x)}{\cos(x)} \\ \cos(x)+\frac{\sin^2(x)}{\cos(x) } \\ \frac{\cos^2(x)+\sin^2(x)}{\cos(x)} \\ \frac{1}{\cos(x)} \\ \sec(x)\]

Answer 22

But I like the first approach better

Answer 23

1st) Use double angle identities on both sin(2x) and cos(2x)
2nd) Canceled the sin's in the first fraction and wrote the second fraction as two fractions in my mind while canceling a cos in the first fraction of the rewriting of the second fractions
3rd) Distribute the negative in front of the second part there
4th) Combined like-terms
5th) wrote it one term by forcing a common denominator
6th) Used a Pythagorean identity

Answer 24

First way much prettier

Answer 25

My thanks