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OCR C3 MATHS JANUARY 2007 help??!!

Can someone help me with question 2 part i?? (~~I love trig but I somehow just can't crack this one and I feel really stupid~~ :indiff:

) I wanna go about solving it using the identities but I can't get the right answer. Here is the question:

It is given that x is the acute angle such that sinx=12/13. Find the exact value of i) cotx

I've tried using 1+cot^2x=cosec^2x and rearranging that (with cosec^2x=1/sin^2x) but that doesn't work and I've also tried cotx=cosx/sinx. the answer is 5/12. I think I'm over complicating it with substitutions but i dont know. someone help please!!

Reply 1

Notnek

Original post by sgic31

Can someone help me with question 2 part i?? (~~I love trig but I somehow just can't crack this one and I feel really stupid~~ :indiff:

Using identites should work. Can you post some working?

Alternatively, draw a right-angled triangle with acute angle x, opposite length 12 and hypotenuse 13. What is tan x? What is cot x?

(edited 9 years ago)

Reply 2

Muttley79

The triangle method is much neater when it says 'exact value'.

Reply 3

sgic31

Original post by notnek

Using identites should work. Can you post some working?

Alternatively, draw a right-angled triangle with acute angle x, opposite length 12 and hypotenuse 13. What is tan x? What is cot x?

that's what I've done so far, excuse the crossing out, hopefully you can still read through it (It's probably all wrong and it's so embarrassing haha! >.<) Since the question asked for exact answers, I wondered why I was getting decimals (though I know I could leave it as an unsolved thing). And the triangle thing sounds good! thank you, but since we haven't really learned how to deal with these questions using a right angled triangle, I'd rather solve it using identities. *addition: I forgot to say I'm working in radians. I wonder if this has anything to do with it??

(edited 9 years ago)

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Reply 4

sgic31

Original post by Muttley79

The triangle method is much neater when it says 'exact value'.

yeah, I read on the mark scheme you could do it that way but I've never really been taught to do that on these kind of questions :/ sighhh, looks like I gotta learn it since it sounds so much simpler

Reply 5

Muttley79

Keep in fractions so cos^2 x = 1 - 144/169

Reply 6

sgic31

Original post by Muttley79

Keep in fractions so cos^2 x = 1 - 144/169

ok, that made it so much clearer! at first I was wondering where you got 144/169 from then I realised that's 12/13 squared. I've been so stupid. thanks guys, what I've been doing was entering sin(12/13)^2 the whole time when I should just be substituting where sinx is with 12/13. so I can just do the cosx/sinx now. sorry for being so stupid haha :colondollar:

Reply 7

Muttley79

No problem - once you've seen a question like this you'll be able to do the next one you see.

Reply 8

Notnek

Original post by sgic31

yeah, I read on the mark scheme you could do it that way but I've never really been taught to do that on these kind of questions :/ sighhh, looks like I gotta learn it since it sounds so much simpler

It's not really a method you need to learn since you should be familiar with trig ratios from GCSE.

Draw a right-angled triangle with angle x, opposite length 12 and hypotenuse 13. Then using pythagoras, the adjacent length is 5 (5^2 + 12^2 = 13^2)

Then using SOHCAHTOA, sin(x) = 12/13 (which you already know) and cos(x) = 5/13.

What's tan(x)?