Question

Trigonometric Identities.
Questions in comments
First two questions have been solved sorry

Answer 1

Prove that these identities are true

\[\frac{1-\cos{a}}{\sin{a}}=\frac{\sin{a}}{1+\cos{a}}\]
\[\sin^{4}{\theta}-\cos^{4}{\theta}=\sin^{2}{\theta}-\cos^{2}{\theta}\]
\[\left(\sin{x}+cos{x}\right)^{4}=\left(1+2\sin{x}\cos{x}\right)^{2}\]

Answer 2

Show that each equation is NOT an identity

\[\sin{2x}=2\sin{x}\]
\[\sec^{2}{x}+\csc^{2}{x}=1\]

*side note: I don't want a complete step by step because I want to learn. Just give me hints, or start off a few steps so I know where to go from there ^_^;

Answer 3

The first equation: try cross multiplying

Answer 4

Sorry, I figured out #1 and #2 @anthonyym ^_^;

Answer 5

Oops sorry

Answer 6

It's okay thank you! :)
Could you help me with the others though? o:

Answer 7

I'll try

Answer 8

Alright thanks :D

Answer 9

For disproving
sin(2x)=2sinx
I think we can plug in a number for x?

Answer 10

Oh! Yeah, that could work

Answer 11

So something like x=30 degrees?

Answer 12

How about dividing by sin(2x) then.
1 does not equal 2

Answer 13

Dividing by sin2x?

Answer 14

I guess, sorry I'm really just learning this right now lol

Answer 15

It's okay you're trying your best :D
So you still get a medal :P

Answer 16

I'm not sure what you meant by dividing by sin2x though D:

Answer 17

I mean sin(2x) divided by sin(2x) is 1

Answer 18

Ah, yeah o:

Answer 19

Well, I was trying out the use a number for x idea.
I put x=30 degrees\[\sin{(2\times30)},2\sin{30}\]That gets me:\[\sin{60},2\sin{30}\]I know those aren't equal so yeah :P