Solving an equation ...

alesha98

if sina = 4/5
cosa = 3/5
tana = 4/c

how do i solve cos((90-a)/2)?

Reply 1

Zacken

Original post by alesha98

if sina = 4/5
cosa = 3/5
tana = 4/c

how do i solve cos((90-a)/2)?

Could you post a picture of the question?

Reply 2

alesha98

Original post by Zacken

Could you post a picture of the question?

Here is the question, the one i am having difficulty with is from question 8(b):

m1 problem.png102.2KB

Reply 3

Notnek

Original post by alesha98

Here is the question, the one i am having difficulty with is from question 8(b):

That's a lot different to your original question! Please always post the full question.

I recommend you watch this video : it should help you with force by a string on a pulley questions:

https://www.youtube.com/watch?v=rC2IQ3R5im4

If you want us to help you with your attempt, please post your working in full.

Reply 4

Zacken

Original post by alesha98

Here is the question, the one i am having difficulty with is from question 8(b):

It's a mechanics question (this is why you always need to post the full question!!) and the intended method (I think) is instead of evaluating it exactly is to simply find

\alpha

by doing

\alpha = \tan^{-1} \frac{4}{3}

and then plugging this in into your cosine using the 'ANS' button on your calculator. (I'm assuming you already know how to find the force on a pulley since you seem to have gotten

2T\cos \left(\frac{90 - \alpha}{2}\right)

Reply 5

Notnek

Original post by Zacken

It's a mechanics question (this is why you always need to post the full question!!) and the intended method (I think) is instead of evaluating it exactly is to simply find

\alpha

by doing

\alpha = \tan^{-1} \frac{4}{3}

and then plugging this in into your cosine using the 'ANS' button on your calculator. (I'm assuming you already know how to find the force on a pulley since you seem to have gotten

2T\cos \left(\frac{90 - \alpha}{2}\right)

I really should have read the question first - I thought this was a standard question with a given angle.

But finding the angle using arctan is a simple way to do this, as you say.

(edited 8 years ago)

Reply 6

alesha98

Original post by notnek

i got the right equation, but somehow i solved the equation wrong , the only part i got wrong is what i posted.
i have set an equation :2(T)(cos((90-a)/2))
=2(2g/3)(cos((90-a)/2))
then here is where i got wrong: =2(2g/3)(cos(45-a/2))
=2(2g/3)((cos45)-((3/5)/2))

Reply 7

Notnek

Original post by alesha98

\tan \alpha = \frac{4}{3} \Rightarrow \alpha = \tan^{-1}\frac{4}{3}

You can use your calculator and then plug

\alpha

into

\displaystyle \cos \left(\frac{90-\alpha}{2}\right)

Reply 8

alesha98

Original post by Zacken

It's a mechanics question (this is why you always need to post the full question!!) and the intended method (I think) is instead of evaluating it exactly is to simply find

\alpha

by doing

\alpha = \tan^{-1} \frac{4}{3}

and then plugging this in into your cosine using the 'ANS' button on your calculator. (I'm assuming you already know how to find the force on a pulley since you seem to have gotten

2T\cos \left(\frac{90 - \alpha}{2}\right)

Thankyou so much, now i got the right answer . I was trying to solve this for an hour and got nothing ...
Re picture:sorry about that because i have a really bad laptop...

Reply 9

alesha98

Original post by notnek

\tan \alpha = \frac{4}{3} \Rightarrow \alpha = \tan^{-1}\frac{4}{3}

You can use your calculator and then plug

\alpha

into

\displaystyle \cos \left(\frac{90-\alpha}{2}\right)

i didnt find the value of a first, so i was struggling

Reply 10

Notnek

Original post by alesha98

i didnt find the value of a first, so i was struggling

Yes it's common to use exact trig values for these kinds of questions so I can see why this question caused you problems.

Reply 11

Zacken

Original post by notnek

I really should have read the question first - I thought this was a standard question with a given angle.

But finding the angle using arctan is a simple way to do this, as you say.

The markscheme (June 2015) uses the exact value method, presumably from double-angle formulae and ugly manipulation, thought that was quite weird. :erm:

Original post by alesha98

...

No worries. Even if a picture wasn't possible, a simple link to the June 2015 paper would have sufficed.

Reply 12

alesha98

Original post by notnek

Yes it's common to use exact trig values for these kinds of questions so I can see why this question caused you problems.

Thanks, i will remember to find the angle first for this kind of questions

Reply 13

alesha98

Original post by Zacken

The markscheme (June 2015) uses the exact value method, presumably from double-angle formulae and ugly manipulation, thought that was quite weird. :erm:

No worries. Even if a picture wasn't possible, a simple link to the June 2015 paper would have sufficed.

yeah that is why i was trying to solve the equation instead of finding the value of a first.

Reply 14

Ayman!

Original post by Zacken

The markscheme (June 2015) uses the exact value method, presumably from double-angle formulae and ugly manipulation, thought that was quite weird. :erm: