Question

If anyone has time to check my work on a homework packet due tomorrow I would greatly appreciate it (: It's analytic trig. Packet is attached, there are still a few questions I'm having trouble with but just let me know if you can help me on any

Answer 1

problem #4. one sin(x) is a denominator while the other is a numerator. there factors cancel each other.

Answer 2

also, the negative signs should be multiplied.

Answer 3

So on #4 the (-) sin in the denominator and the (-) sin at the end of the problem cancel out? I'm confused can you cancel a negative with a negative?

Answer 4

the signs can be combined. the product/quotient of two negatives is positive. is it not?

Answer 5

So -cot really equals -cos/-sin which is cos/sin? So would the sin's cancel out and you would be left with cos/1 X 1/cos so the final answer is 1?

Answer 6

-cot is -(cos/sin) and the sign distributes only to terms, not to factors, hence
-cos/sin

Answer 7

So everything cancels out?

Answer 8

luckily, your final answer 1 is correct. let me put some details, and some parentheses too:
\[\cot (-\beta) \sec(-\beta) \sin(-\beta)=-\cot(\beta)*\sec(\beta)*(-\sin(\beta))\]

Answer 9

yes, everything cancels out.

Answer 10

Okay I understand it now, thank you for explaining it (:

Answer 11

next is #6. the numerator on the LHS of the identity is
\[\frac{ 1 }{ \sin x } - \sin x\]

Answer 12

but when you combine, you must follow addition rules of dissimilar fractions. to wit, \[\frac{ 1 }{ \sin x }-\frac{ \sin x }{ 1 }\]

Answer 13

so it's just sin(x)/sin(x) X sin(x)/1?

Answer 14

on the simplify part, in the first two lines sin is not squared in the numerator but is squared in the third line.

Answer 15

\[\frac{ 1 }{ \sin x }-\frac{ \sin x }{ 1 }=\frac{ 1-\sin^2 x }{ \sin x }\]

Answer 16

would it be 1-sin^3(x)/sin(x) after dividing 1/sin(x)?

Answer 17

the third line is correct. i have no idea why the first two lines do not have squared sines in the numerator.

Answer 18

I couldn't tell you I've been working on this all day I keep making mistakes :/

Answer 19

\[\frac{ 1 }{ \sin x } -\frac{ \sin x }{ 1 } =\] \[\frac{ 1-\sin^2 x }{ \sin x }\]
the last expression should be your expression in the 1st and 2nd lines of the simplify part. there is no problem with \[\div \frac{ 1 }{ \sin x }\] or \[\times \frac{ \sin x }{ 1 }\] in the left hand side.

Answer 20

mk I understand it now, thank you again. Sorry I'm slow at catching on tonight, its 1:30am here

Answer 21

in the third line, you should write it this way