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Trigonometric Identities Help!!

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Original post by ellahick
Zacken back at it again with the cheeky fast replies


He's being more helpful than you are Miss Hickey
Original post by Alex.trin
Right guys... LAST QUESTION!!! You'll never have to hear from me again so the sooner you help me, the sooner you can forget about me.
tanxsinx/1-cosx = 1 + 1/cosx
Where do I even start?


tanx = sinx / cosx
So
(sinx / cosx * sinx) / 1-cosx

Spoiler

Original post by Alex.trin
Please can you explain what you've done in between steps 1 and 2 <3


Multiplied top and bottom by cos.
Original post by Alex.trin
Please can you explain what you've done in between steps 1 and 2 <3


abc=ab×1c=abc\displaystyle \frac{\frac{a}{b}}{c} = \frac{a}{b} \times \frac{1}{c} = \frac{a}{bc}.

So here: sin⁡xtan⁡x1−cos⁡x=sin⁡2xcos⁡x1−cos⁡x=sin⁡2xcos⁡x(1−cos⁡x)\displaystyle \frac{\sin x \tan x}{1 - \cos x} = \frac{\frac{\sin^2 x}{\cos x}}{1 - \cos x} = \frac{\sin^2 x}{\cos x(1-\cos x)}.

Alternatively:

sin⁡2xcos⁡x1−cos⁡x×cos⁡xcos⁡x=sin⁡2xcos⁡x(1−cos⁡x)\displaystyle \frac{\frac{\sin^2 x}{\cos x}}{1 - \cos x} \times \frac{\cos x}{\cos x} = \frac{\sin^2 x}{\cos x(1 - \cos x)}
(edited 7 years ago)
Original post by Zacken
abc=ab×1c=1bc\displaystyle \frac{\frac{a}{b}}{c} = \frac{a}{b} \times \frac{1}{c} = \frac{1}{bc}.

So here: sin⁡xtan⁡x1−cos⁡x=sin⁡2xcos⁡x1−cos⁡x=sin⁡2xcos⁡x(1−cos⁡x)\displaystyle \frac{\sin x \tan x}{1 - \cos x} = \frac{\frac{\sin^2 x}{\cos x}}{1 - \cos x} = \frac{\sin^2 x}{\cos x(1-\cos x)}.


Wow is it bad that I didn't know you could do that?
Original post by Alex.trin
Wow is it bad that I didn't know you could do that?


A little, yes - I'm afraid. This is where Muttely's advice of reviewing basic algebra and C1 stuff before trying to do identities like this is useful.
Original post by Zacken
A little, yes - I'm afraid. This is where Muttely's advice of reviewing basic algebra and C1 stuff before trying to do identities like this is useful.

Yeah I need to work on my algebra, especially rearranging formulas etc.
Original post by Alex.trin
He's being more helpful than you are Miss Hickey


i reply to every message
Original post by Alex.trin
Wow is it bad that I didn't know you could do that?


its worse than christ being crucified for a second time
Original post by Alex.trin
Yeah I need to work on my algebra, especially rearranging formulas etc.


use christ as inspiration to resurrect yourself
Original post by Zacken
abc=ab×1c=1bc\displaystyle \frac{\frac{a}{b}}{c} = \frac{a}{b} \times \frac{1}{c} = \frac{1}{bc}.


abc=ab×1c=abc\displaystyle\frac{\frac{a}{b}}{c} = \frac{a}{b} \times \frac {1}{c} = \frac{a}{bc}

Not 1bc\displaystyle\frac{1}{bc}
Original post by Alex.trin
Wow is it bad that I didn't know you could do that?


It's not bad, given that you Lear this now and don't forget it for your exam
Original post by asinghj
[...]


Edited, thanks.

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