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Trigonometric Identities Help!!

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Original post by Alex.trin
Well we all did further maths in June so we're just recapping some of the A* content which is likely to appear in the AS before we forget it completely.


Are you in England?

If so, I'd tend to look at C1 first and finish that so you are clear what topics fit where.
Reply 81
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Zacken
22
Original post by RDKGames
NNOOOOOOOOOOOOOOOOOO.!!!!!

Valerie, same pitfall I made. Look onto the first page.



No it's not, she's doing something completely valid.
Reply 82
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Alex.trin
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Original post by RDKGames
You had 1+sin⁡xcos⁡x\frac{1+\sin{x}}{\cos{x}} and now refer back to my comment on what to do.


I have and I am still as equally confused
Original post by Alex.trin
I have and I am still as equally confused


Once you have cos⁡x(1+sin⁡x)1−sin⁡2x\frac{\cos{x}(1+\sin{x})}{1-\sin^2x} you can express the denominator in difference of two squares, then cancel any common factors.
Reply 84
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Alex.trin
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Original post by Muttley79
Are you in England?

If so, I'd tend to look at C1 first and finish that so you are clear what topics fit where.

Yes I'm in England. I have two maths teachers this year. With one of them were doing core 1 (algebra atm) and with the other we're starting some core 2 I think.
Original post by Zacken
No it's not, she's doing something completely valid.


Oops, read back and I misinterpreted it.
Original post by ValerieKR
Try starting from the RHS and multiplying the top and bottom by 1+sinx


Could someone please explain why everyone is using the word sin, its a sin!!!! THIS THREAD IS FULL OF SINNERSSS
Original post by Alex.trin
Yes I'm in England. I have two maths teachers this year. With one of them were doing core 1 (algebra atm) and with the other we're starting some core 2 I think.


Isn't her geographical location a bit private, please think next time when answering potential pedophillic questions alex deary
Reply 88
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Alex.trin
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Original post by RDKGames
Once you have cos⁡x(1+sin⁡x)1−sin⁡2x\frac{\cos{x}(1+\sin{x})}{1-\sin^2x} you can express the denominator in difference of two squares, then cancel any common factors.


Omg thanks again! You're such a lifesaver
Original post by RDKGames
Once you have cos⁡x(1+sin⁡x)1−sin⁡2x\frac{\cos{x}(1+\sin{x})}{1-\sin^2x} you can express the denominator in difference of two squares, then cancel any common factors.


I swear if i see the word sin one more time I'm going to going cardiac arrest
Original post by ellahick
I swear if i see the word sin one more time I'm going to going cardiac arrest


sin⁡(devil)\sin(de^vi_l)
Original post by Alex.trin
Omg thanks again! You're such a lifesaver


onlly god himself can save ones life
Original post by ellahick
I swear if i see the word sin one more time I'm going to going cardiac arrest

Sin
Original post by RDKGames
sin⁡(devil)\sin(de^vi_l)


is there any chance you are related or are infact simon milner?
Original post by asinghj
Sin


you are a nasty person
(edited 7 years ago)
Reply 95
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Alex.trin
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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?
Reply 96
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Zacken
22
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?


So you have sin⁡xsin⁡xcos⁡x1−cos⁡x=sin⁡2xcos⁡x(1−cos⁡x)=1−cos⁡2xcos⁡x(1−cos⁡x)\displaystyle \frac{\sin x \frac{\sin x}{\cos x}}{1 - \cos x} = \frac{\sin^2 x}{\cos x(1-\cos x)} = \frac{1 - \cos^2 x}{\cos x (1-\cos x)}

Now use different of two squares.
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?


stop SINning, put on your swimming COStume and TAN
Original post by Zacken
So you have sin⁡xsin⁡xcos⁡x1−cos⁡x=sin⁡2xcos⁡x(1−cos⁡x)=1−cos⁡2xcos⁡x(1−cos⁡x)\displaystyle \frac{\sin x \frac{\sin x}{\cos x}}{1 - \cos x} = \frac{\sin^2 x}{\cos x(1-\cos x)} = \frac{1 - \cos^2 x}{\cos x (1-\cos x)}

Now use different of two squares.


Zacken back at it again with the cheeky fast replies
Reply 99
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Alex.trin
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Original post by Zacken
So you have sin⁡xsin⁡xcos⁡x1−cos⁡x=sin⁡2xcos⁡x(1−cos⁡x)=1−cos⁡2xcos⁡x(1−cos⁡x)\displaystyle \frac{\sin x \frac{\sin x}{\cos x}}{1 - \cos x} = \frac{\sin^2 x}{\cos x(1-\cos x)} = \frac{1 - \cos^2 x}{\cos x (1-\cos x)}

Now use different of two squares.

Please can you explain what you've done in between steps 1 and 2 <3

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