Question

prove the identity

Answer 1

So these are the identities

Answer 2

first use the sin and cos double angles to re-write the numerators of the expressions

lmk what you get

Answer 3

\[\sin 2x = 2 \sin x \cos x=\frac{ 2 \tan x }{ 1+\tan^2 x }\]

Answer 4

you don't need that tan part haha

Answer 5

i will rewrite it now

Answer 6

just replace sin(2x) with 2sin(x)cos(x) for now

Answer 7

oh duh lol

Answer 8

okay

Answer 9

\[2 \sin (x) \cos(x)=2\sin x \cos x\]

Answer 10

its the same on both sides

Answer 11

good, so take the original problem, replace "sin(2x)" with "2sin(x)cos(x)'

then replace cos(2x) with 2cos^2(x)-1

Answer 12

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

Answer 13

awesome, now do you notice anything you can cancel out in the first term?

Answer 14

sin x

Answer 15

leaving us with 2 cos( x)?

Answer 16

I think

Answer 17

awesome

now multiply both sides of the whole equation by cos(x), let me know what you get

Answer 18

2cos^2(x)-1=sec (x) * cos (x)

Answer 19

the cos x's cancel out right

Answer 20

don't forget about that first term

2cos^2(x) - [2cos^2(x)-1] = cos(x)sec(x)

now what do you get when you simplify the left side?

Answer 21

(distribute the negative sign)

Answer 22

1

Answer 23

awesome

now, in the next step it says to "apply the sec definition"

sec(x) = 1/cos(x)

so the right hand side becomes

(1/cos(x)) * cos(x)) which equals 1

thus the left hand side and right hand side are both 1

thus proved