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# Sec/Cosec/Cot Trigonometry Help!! watch

1. Hi there,

I wonder if someone could help me with this question - I'm sure it's simple but it's driving me mad! I have to express this term:

as the trig ratio of an acute angle, using the same trig function.

If you could please explain how to go about working this out, I'd be most grateful! Thanks!
2. Cosec(2π/3) = 1/sin(2π/3).
and remember that the sine function is symmetric about the the line x=π/2 or π/2 meaning at sin(60) = sin(120) both in degrees
3. (Original post by Funky_Giraffe)
Hi there,

I wonder if someone could help me with this question - I'm sure it's simple but it's driving me mad! I have to express this term:

as the trig ratio of an acute angle, using the same trig function.

If you could please explain how to go about working this out, I'd be most grateful! Thanks!
Find the sine of that value and divide it from one.
4. (Original post by B_9710)
Cosec(2π/3) = 1/sin(2π/3).
and remember that the sine function is symmetric about the the line x=π/2 or π/2 meaning at sin(60) = sin(120) both in degrees
(Original post by kkboyk)
Find the sine of that value and divide it from one.
Hey! Thanks for the help - I just wanted to check: is cot(490) the same as -cot(50)

[both angles in degrees]

And for that last question ^^ I got cosec(1/3π) - is that right?! Thanks so much
5. Yes cot(490) = - cot(50)
And the answer you got is right
6. (Original post by Funky_Giraffe)
Hey! Thanks for the help - I just wanted to check: is cot(490) the same as -cot(50)

[both angles in degrees]

And for that last question ^^ I got cosec(1/3π) - is that right?! Thanks so much
Yes, and what last question?
7. Hey, kkboyk maybe you could help? It was:

Prove this trig identity:

I started with the left side, and got up to

But not really sure where to go from hear to end up with
8. What you can do here is combine sinθ and cosecθ and make it as one fraction all over a common denominator of sinθ
Then squaring this term is easy and you should then be able to get to the RHS
9. (Original post by Funky_Giraffe)
Hey, kkboyk maybe you could help? It was:

Prove this trig identity:

I started with the left side, and got up to

But not really sure where to go from hear to end up with
Expand the first bracket to get sin^2 -2 +cosec^2 (2sinxcosec is is equal to 2).

use other identities such as cosec^2 = 1+cot^2 and simplify.
10. (Original post by kkboyk)
Expand the first bracket to get sin^2 -2 +cosec^2 (2sinxcosec is is equal to 2).

use other identities such as cosec^2 = 1+cot^2 and simplify.
Thank you!
11. (Original post by Funky_Giraffe)
Thank you!
No problem
12. B_9710 kkboyk Hey, these two questions are linked and I just can't seem to work out how to go about answering!! Can you help me in any way??

i)

(I tried to split up the fraction into more obvious numbers (ie tan(60-45) but I don't think that's allowed?)

and then:

ii)
13. (Original post by Funky_Giraffe)
B_9710 kkboyk Hey, these two questions are linked and I just can't seem to work out how to go about answering!! Can you help me in any way??

i)

(I tried to split up the fraction into more obvious numbers (ie tan(60-45) but I don't think that's allowed?)

and then:

ii)
That is quite precisely what you should for (i) I'm sure you can take it from there.
14. (Original post by Funky_Giraffe)
B_9710 kkboyk Hey, these two questions are linked and I just can't seem to work out how to go about answering!! Can you help me in any way??

i)

(I tried to split up the fraction into more obvious numbers (ie tan(60-45) but I don't think that's allowed?)

and then:

ii)
Thats precisely what you dofor (i). For (ii) you can try to expand the brackets, and dividng by either cos or sine.
15. (Original post by Zacken)
That is quite precisely what you should for (i) I'm sure you can take it from there.
Thanks very much! Problem solved... !! With the second question, which kkboyk kindly I suggested I multiply out, I'm having trouble. Is there another method or is it easier than I'm making it? I'm confused about dealing with:

sin22.5cos22.5 ... + etc. when I multiply out.

16. (Original post by Funky_Giraffe)
Thanks very much! Problem solved... !! With the second question, which kkboyk kindly I suggested I multiply out, I'm having trouble. Is there another method or is it easier than I'm making it? I'm confused about dealing with:

sin22.5cos22.5 ... + etc. when I multiply out.

Ignore what I've said, sorry. I'm bad at trig myself

Here's a big hint: use so now you'll be focusing only on the middle value and use the half angle formula.
17. (Original post by Funky_Giraffe)
Thanks very much! Problem solved... !! With the second question, which kkboyk kindly I suggested I multiply out, I'm having trouble. Is there another method or is it easier than I'm making it? I'm confused about dealing with:

sin22.5cos22.5 ... + etc. when I multiply out.

When you multiply it out you'll get something of the form , which I'm sure are forms you recognise - one is a double angle formula for something and the other is the well-known Pythagorean identity.
Spoiler:
Show

is a well known trig value.
18. (Original post by kkboyk)
Ignore what I've said, sorry. I'm bad at trig myself Here's a big hint: use so now you'll be focusing only on the middle value and use the half angle formula.
(Original post by Zacken)
When you multiply it out you'll get something of the form , which I'm sure are forms you recognise - one is a double angle formula for something and the other is the well-known Pythagorean identity.
Spoiler:
Show

is a well known trig value.

Thank you both for your help
19. (Original post by Funky_Giraffe)
Thank you both for your help
Glad I could be of service!
20. (Original post by Funky_Giraffe)
Hi there,

I wonder if someone could help me with this question - I'm sure it's simple but it's driving me mad! I have to express this term:

as the trig ratio of an acute angle, using the same trig function.

If you could please explain how to go about working this out, I'd be most grateful! Thanks!
Sorry for being random, but I just LOVE your avatar

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