Question

im a beginner in trigo please see my problem, i will surely give MEDALS

Answer 1

prove that -sinx = (cotx-cscx)(cosx+1)

i know how to prove (cotx-cscx)(cosx+1) = -sinx but i can't do the reverse solution.

Answer 2

\(\normalsize\color{black}{-\sin x = (\cot x-\csc x)(\cos x+1)}\) I'll provide the steps, and you do it.

Answer 3

okay thank you

Answer 4

1) re-write the problem to `cos(x)` and `sin(x)`.

(knowing that:
cot(x) = cos(x) / sin(x)
csc(x) = 1 / sin(x) )

2) add the fractions inside the 1st parenthesis (since they have the common denominator).

3) expand the expression that you are having after step 2.

4) multiply every thing in the equation times `sin(x)`

5) apply `sin²x + cos²x = 1`

Answer 5

I would prefer that you post your work here, don't care whether you use the drawing tool equation editor or just latex.

Answer 6

Doing them in my head :PPP love it

Answer 7

im sorry but i just dont get it :/ could you please show me the solution, then after i learn the solutions, i'll answer the my other problems on my own

Answer 8

@SolomonZelman please help me

Answer 9

\(\normalsize\color{blue}{ -\sin x = (\cot x-\csc x)(\cos x+1) ~~~~~~\rm {write~~in~~terms~~of~sin~~and~~cos,} }\)


\(\normalsize\color{blue}{ -\sin x = ( \Large{ \frac{\cos x}{\sin x} }-\Large{ \frac{1}{\sin x} })(\cos x+1) ~~~~~~\rm \normalsize{ expand~~~the~~~right~~~side} }\)


\(\normalsize\color{blue}{ -\sin x = ( \Large{ \frac{\cos x-1}{\sin x} })(\cos x+1) ~~~~~~\rm \normalsize{ expand~~~the~~~right~~~side} }\)

Answer 10

\(\normalsize\color{blue}{ -\sin x = ( \Large{ \frac{\cos^2x-\cos x}{\sin x} + \Large{ \frac{\cos x-1}{\sin x}} })~~~~~~\rm \normalsize{} }\)

Answer 11

multiply every thing times `sin(x)`

Answer 12

can you finish ?

Answer 13

let me try

Answer 14

so that would cancel the positive denominator sinx ?

Answer 15

yes, if you multiply everything times `sin(x)`

Answer 16

tell me what you get after doing that.

Answer 17

\[\cos ^{2}x - cosx+cosx-1\]

Answer 18

yes, but \(\normalsize\color{blue}{ -\sin^2x=\cos^2x - \cos x + \cos x - 1 \rm \normalsize{} }\)

Answer 19

by the way, why do i need to multiply everything to sin(x) is it in the identities right?

Answer 20

you need to multiply everything times `sin(x)` to cancel the denominators. Doesn't it look more simplified like this ?

Answer 21

ohh i get it then after i get the \[-\sin ^{2}x=\cos ^{2}-cosx+\cos\] i will simplify it?

Answer 22

By the way I just want to show that you can get the magic `\` by

1) Pressing and Holding `ALT`
2) Click 9 2 on the number pad on the bottom right corner of your keyboard (numbers below Esc, F1, F2, F3.... WON'T WORK ! )
3) release ALT

Answer 23

\(\normalsize\color{blue}{ -\sin^2x=\cos^2x - \cos x + \cos x - 1 \rm \normalsize{} }\)

Don't leave the `-1` out :)

Answer 24

ohhh okay :)

Answer 25

\(\normalsize\color{blue}{ -\sin^2x=\cos^2x - \cos x + \cos x - 1 \rm \normalsize{} }\)
\(\normalsize\color{blue}{ -\sin^2x=\cos^2x + \color{red}{- \cos x }\color{red}{+ \cos x } - 1 \rm \normalsize{} }\) they are added, and by adding canceled

Answer 26

but wait, the given is -sinx not\[-\sin ^{2}x\]

Answer 27

What do you get now?
after you get it, play around a little with the rule `sin²x + cos²x =1`

steps for playing around,

subtract `cos²x` from both sides
multiply both sides by `-1`

is it verified ?

Answer 28

Nooo, you multiplied EVERYTHING times sin(x), also the -sin(x) on the left hand side

Answer 29

i got \[-\sin ^{2}x = \cos ^{2}x\]

Answer 30

-1

Answer 31

\[-\sin ^{2}x=\cos ^{2}x\]-1 this is what i got

Answer 32

with the -1 attach on the x.. i just can't attach it

Answer 33

No, you are leaving `-1` out again. Your teacher won't forget and take 10 points off, I do, but telling you, please keep the `-1` inside :D

Answer 34

okay okay i will :)

Answer 35

\(\normalsize\color{blue}{ -\sin^2x =\cos^2x -1~\rm { } }\)

Answer 36

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

Answer 37

No follow the steps to play with the identity (if you don't see that it is verified ). You like playing, don't you ?

Answer 38

Now follow (not "No follow)

Answer 39

i will multiply both sides by \[\cos ^{2}x\] ?

Answer 40

No, look at this rule, `sin²x + cos²x = 1`

1) subtract `cos²x` from both sides
2) multiply everything times `-1`
3) Put `cos²x` in front of `-1` (if necessary)

Answer 41

for number 1:\[\cos ^{2}x-\sin ^{2}x+\cos ^{2}x=1-\cos ^{2}x\]

Answer 42

I am taking about this,

You get \(\normalsize\color{blue}{ -\sin^2x =\cos^2x -1~\rm { } }\) after canceling `-cos(x)` and `+cos(x)`
Now look at the identity (prove from the identity ─ posted below in black)

\(\normalsize\color{black}{ \cos^2x +\sin^2x =1~~~~~~~\rm { subtract~~~\cos^2x ~~~from~~~both~~~sides, } }\)
\(\normalsize\color{black}{ \sin^2x =1-\cos^2x~~~~~~~\rm { multiply~~~times~~~~-1~~~~on~~~both~~~sides, } }\)
\(\normalsize\color{black}{ -\sin^2x =-1+\cos^2x~~~~~~~\rm { and~~~this~~is~~same~~as,} }\)
\(\normalsize\color{black}{ -\sin^2x =\cos^2x-1~~~~~~~\rm { and~~~this~~is~~same~~as,} }\)

Answer 43

and your identity is verified, right ?

Answer 44

And I only touched 1 side of the problem

Answer 45

yes that's right

Answer 46

Any questions regarding what we did ?

Answer 47

wait..
i got -sin^2x-cos^2x-1 right? but how will i get/prove -sinx = (cotx-cscx)(cosx+1)

Answer 48

i got \[\cos ^{2}x-1\] instead of (cotx-cscx)(cosx+1)

Answer 49

I think we went through showing that `sinx = (cotx-cscx)(cosx+1)` is the same as
`-sin²x = cos²x - 1` ?

Answer 50

so we got the right answer? \[-\sin ^{2}x-\cos ^{2}x-1\] ?