Stuck on last part of trig identity question.

secA(sin²A-1)²/[cosA(sec²A - tan²A)]+(2sin²A + cos²A) = 2

secA(sin²A-1)2 = cos³A
[cosA(sec²A - tan²A)] = cosA

what do I do with
(2sin²A + cos²A) ???

and I am new to student room and how do I write equations in a form like other people do?

(edited 8 years ago)

Reply 1

nerak99

Original post by Chairm

secA(sin2A-1)² / [cosA(sec2A - tan2A)] + (2sin2A + cos2A) = 2

secA(sin2A-1)2 = cos³A
[cosA(sec2A - tan2A)] = cosA

what do I do with
(2sin2A + cos2A) ???

and I am new to student room and how do I write equations in a form like other people do?

You need to use tex. There is a link but essentially, you surround an equation by [] brackets containing tex to open and /tex to close.

Your opener is (in latex) \frac{\secA(\sin2A-1)^2}{[ \cos A (\sec A - \tan 2A)]+(2 \sin 2A + \cos 2A)}=2
Here it with the tex tags on

\frac{ \sec A (\sin2A-1)^2}{[ \cos A (\sec A - \tan 2A)]+(2 \sin 2A + \cos 2A)}=2

(edited 8 years ago)

Reply 2

Zacken

Original post by Chairm

secA(sin2A-1)² / [cosA(sec2A - tan2A)] + (2sin2A + cos2A) = 2

secA(sin2A-1)2 = cos³A

Where did you get this from? It's completely wrong.

\sec A(\sin 2A - 1)^2 = \sec A(\sin^2 2A - 2\sin 2A + 1)

Do you know how to expand brackets of the form

(a+b)^2

and I am new to student room and how do I write equations in a form like other people do?

You haven't replied to the other thread that I answered your question on, why not? You can use

\LaTeX

if you want to typeset equations, but you really need not.

Reply 3

Chairm

Original post by Zacken

Where did you get this from? It's completely wrong.

\sec A(\sin 2A - 1)^2 = \sec A(\sin^2 2A - 2\sin 2A + 1)

Do you know how to expand brackets of the form

(a+b)^2

?

You haven't replied to the other thread that I answered your question on, why not? You can use

\LaTeX

if you want to typeset equations, but you really need not.

oh sorry. I forgotten to put powers.
It's a question given from lesson as a hand out, it doesn't have answer sheet so I cannot check if I am going right way.
tried using latex but it seems a bit difficult...I'll need to look into it more

Oh and for previous post, I have worked out the question and got the answer. Thank you :smile:

Reply 4

Zacken

Original post by Chairm

Not sure I understand, post your corrected working so we can have a look?

Feel free to post your workings using picture, etc... so we can verify them for you if you want. You can make as many threads as you like.

There's no point using LaTex if you're not comfortable with it. Don't worry about it. :smile:

Oh and for previous post, I have worked out the question and got the answer. Thank you :smile:

It's common courtesy to reply back to the person (SeanFM in this case) who helped you on the thread itself. :smile:

Reply 5

Chairm

Original post by Zacken

It's common courtesy to reply back to the person (SeanFM in this case) who helped you on the thread itself. :smile:

I've double checked that question. I have corrected the powers. Does it still seem to be wrong?

And yes I am sorry about that. I will do that from now on. I am just not familiar with this forum yet :wink:

Reply 6

Zacken

Original post by Chairm

secA(sin²A-1)²/[cosA(sec²A - tan²A)]+(2sin²A + cos²A) = 2

secA(sin²A-1)2 = cos³A
[cosA(sec²A - tan²A)] = cosA

what do I do with
(2sin²A + cos²A) ???

So you now have:

Unparseable latex formula:

\displaystyle \frac{\sec A(\sin^2 A - 1)^2}{\cos A (\csec^2 A - \tan^2 A)} = \frac{\cos^3 A}{\cos A} =\cos^2 A

So:

\cos^2 A + 2 \sin^2 A + \cos^2 A = 2 \cos^2 A + 2\sin^2 A = 2(\cos^2 A + \sin^2 A)

Can you think of a familiar trigonometric identity that connects

\cos^2 A + \sin^2 A

and

1

(edited 8 years ago)

Reply 7

Chairm

Original post by Zacken

So you now have:

Unparseable latex formula:

\displaystyle \frac{\sec A(\sin^2 A - 1)^2}{\cos A (\csec^2 A - \tan^2 A)} = \frac{\cos^3 A}{\cos A} =\cos^2 A

So:

\cos^2 A + 2 \sin^2 A + \cos^2 A = 2 \cos^2 A + 2\sin^2 A = 2(\cos^2 A + \sin^2 A)

Can you think of a familiar trigonometric identity that connects

\cos^2 A + \sin^2 A

and

1

\cos^2 A + 2 \sin^2 A + \cos^2 A = 2 \cos^2 A + 2\sin^2 A = 2(\cos^2 A + \sin^2 A)

For this one, how do you get

\cos^2 A + 2 \sin^2 A + \cos^2 A

wouldn't it be

cos^3 A / cosA + (2sin^2A + cos^2A)

and then

cos²A / 2sin²A + cos²A

Reply 8

Zacken

Original post by Chairm

\cos^2 A + 2 \sin^2 A + \cos^2 A = 2 \cos^2 A + 2\sin^2 A = 2(\cos^2 A + \sin^2 A)

For this one, how do you get

\cos^2 A + 2 \sin^2 A + \cos^2 A

wouldn't it be

cos^3 A / cosA + (2sin^2A + cos^2A)

and then

cos²A / 2sin²A + cos²A

The (2 sin^2 A + cos^2 A) aren't in the denominator.

It's

\frac{\cos^3 A}{\cos A} + 2\sin^2 A + \cos^2 A

Reply 9

Chairm

Original post by Zacken

The (2 sin^2 A + cos^2 A) aren't in the denominator.

It's

\frac{\cos^3 A}{\cos A} + 2\sin^2 A + \cos^2 A

oh wow... thought (2 sin^2 A + cos^2 A) was in the denominator the whole time..

Then yes, you are right

2sin^2A +cos^2A + cos^2A = 2sin^2A+2cos^2A

using sin^2A + cos^2A = 1 identity

you get 2...

Thank you so much for your help...
If you didn't correct me that the second part is not in the denominator, I would have never found that suspicious..