Simorgh's Maths Questions (2 Viewers)

Simorgh · Apr 2, 2016

I can be really crap at Math, so I will post questions here which I need help with.

First Question

1 + cot^2 alpha could become cosec^2 alpha but how do you simply the top properly with 2cot alpha? Should you just convert the top into cos/sin + cos/sin??

Sien · Apr 2, 2016

Hope i didn't make any silly mistakes :/
Basically you need to know your trig formulas to do this

Sent from my D6503 using Tapatalk

KingOfActing · Apr 2, 2016

$\frac{2\cot{\alpha}}{1+\cot^2{\alpha}} = \frac{2\frac{\cos \alpha}{\sin \alpha}}{1 + \frac{\cos^2{\alpha}}{\sin^2{\alpha}}} = \frac{2\cos{\alpha}\sin{\alpha}}{\sin^2{\alpha} + \cos^2{\alpha}} = 2\cos{\alpha}\sin{\alpha} = \sin{2\alpha}$

Step 1: Write cot(x) as cos(x)/sin(x)

Step 2: Multiply by sin^2(x)/sin^2(x)

Step 3: Recognise sin^2(x) + cos^2(x) = 1 [Pythagorean Identity]

Simorgh · Apr 2, 2016

Sien said:
Hope i didn't make any silly mistakes :/
Basically you need to know your trig formulas to do this

Sent from my D6503 using Tapatalk

Thank you Sien, I tried understanding urs and totally got it. Did couple of similar questions and it works perfectly.

Only learnt this yesterday in class, even though its really easy becoz you just have to know those 3 different identities.

Simorgh · Apr 2, 2016

KingOfActing said:
$\frac{2\cot{\alpha}}{1+\cot^2{\alpha}} = \frac{2\frac{\cos \alpha}{\sin \alpha}}{1 + \frac{\cos^2{\alpha}}{\sin^2{\alpha}}} = \frac{2\cos{\alpha}\sin{\alpha}}{\sin^2{\alpha} + \cos^2{\alpha}} = 2\cos{\alpha}\sin{\alpha} = \sin{2\alpha}$

Step 1: Write cot(x) as cos(x)/sin(x)

Step 2: Multiply by sin^2(x)/sin^2(x)

Step 3: Recognise sin^2(x) + cos^2(x) = 1 [Pythagorean Identity]

Thanks there so many different ways to do these questions!

KingOfActing · Apr 2, 2016

Simorgh said:
Thanks there so many different ways to do these questions!

You're welcome

All trigonometric functions are related to each other in one way or another, so often the easiest way to solve these problems is to transform everything into just sine and cosine. That way, you need need to remember the sin/cos identities, and you can figure out the rest yourself.

Green Yoda · Apr 2, 2016

do we then need to simplify sin(2 alpha)?

InteGrand · Apr 2, 2016

Rathin said:
do we then need to simplify sin(2 alpha)?

No, that's a fully simplified form.

Green Yoda · Apr 2, 2016

InteGrand said:
No, that's a fully simplified form.

I have just learned this in school yesterday so soz if im wrong but then when do we use the double angle formula/method? In simplifying or expanding?

Simorgh · Apr 2, 2016

InteGrand said:
No, that's a fully simplified form.

How did you simply 2cos(alpha)sin(alpha) to 2cos(alpha)? I am using Fitzpatrick and the answer is in the form of just 2cos(alpha)sin(alpha)?

Green Yoda · Apr 2, 2016

Simorgh said:
How did you simply 2cos(alpha)sin(alpha) to 2cos(alpha)? I am using Fitzpatrick and the answer is in the form of just 2cos(alpha)sin(alpha)?

2cos(alpha)sin(alpha) would just turn into sin(2 alpha)?

InteGrand · Apr 2, 2016

Rathin said:
I have just learned this in school yesterday so soz if im wrong but then when do we use the double angle formula/method? In simplifying or expanding?

The double angle formula allows the 2*sin(α)*cos(α) to become sin(2α).

Simorgh · Apr 2, 2016

I was using the 2 Unit book and we haven't quite learnt the double angle identities in 3 Unit yet.

InteGrand · Apr 2, 2016

Simorgh said:
I was using the 2 Unit book and we haven't quite learnt the double angle identities in 3 Unit yet.

If it's from a 2U Question, you won't be expected to simplify it to sin(2α), since the double angle identities aren't part of 2U. So in that case you can leave it as 2*sin(α)*cos(α).

leehuan · Apr 2, 2016

^Precisely the above.

sin(2x)=2sin(x)cos(x) is an assumed 3U result however in 2U the point sin(2x) is your official endpoint. You cannot go beyond that.

Same applies for cosine and tangent. The cotangent one isn't really a part of the syllabus though

Simorgh · Apr 2, 2016

Another question

How would you do this? I know this is quite similar to what you needed to do in the last question, but its just seems a little different. It would be great if someone could show me an easy way to do it?

Nailgun · Apr 2, 2016

InteGrand · Apr 2, 2016

Nailgun said:
$=\frac { { a }^{ 2 }\sec ^{ 2 }{ \theta } }{ \sqrt { { a }^{ 2 }\sec ^{ 2 }{ \theta } -{ a }^{ 2 } } } \\ \\ \\ =\frac { { a }^{ 2 }\sec ^{ 2 }{ \theta } }{ \sqrt { { a }^{ 2 }(\sec ^{ 2 }{ \theta } -1) } } \\ \\ \\ =\frac { { a }^{ 2 }\sec ^{ 2 }{ \theta } }{ \sqrt { { a }^{ 2 }(\tan ^{ 2 }{ \theta } ) } } \\ \\ \\ =\frac { { a }^{ 2 }\sec ^{ 2 }{ \theta } }{ a\tan { \theta } } \\ \\ \\ =a\quad \times \quad \frac { \cos { \theta } }{ \sin { \theta } .\cos ^{ 2 }{ \theta } } \\ \\ \\ =\frac { a }{ \sin { \theta } .\cos { \theta } } \\ \\ \\ =a\csc { \theta } \sec { \theta }$

$\noindent This is assuming that $a\tan \theta$ is positive, of course. The general answer is $|a\csc \theta \sec \theta|$, assuming $\sec{\theta}\neq \pm 1$ (if it is $\pm 1$, then the given expression is undefined, as it has a zero denominator).$

leehuan · Apr 2, 2016

Fortunately for a 2U level they will let him off with Nailgun's answer because they don't care too much about the fact that sqrt(x^2)=|x|

Green Yoda · Apr 2, 2016

Question:
If tan(theta) = t , express in terms of t
cot(180-theta)

Simorgh's Maths Questions (2 Viewers)

Active Member

将来: NEET

lukewarm mess

Active Member

Active Member

lukewarm mess

Hi Φ

Well-Known Member

Hi Φ

Active Member

Hi Φ

Well-Known Member

Active Member

Well-Known Member

Well-Known Member

Active Member

Cole World

Well-Known Member

Well-Known Member

Hi Φ

Users Who Are Viewing This Thread (Users: 0, Guests: 2)