Sec/Cosec/Cot Trigonometry Help!!

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:

$\mathrm{cosec(}$ $\frac{2pi}{3}$ $\mathrm{)}$


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!

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

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

Hey, @kkboyk maybe you could help? It was:

Prove this trig identity:

$(\sin \theta - \mathrm{cosec } \theta)^2 \equiv \sin^2 \theta + \mathrm{cot}^2 \theta -1$

I started with the left side, and got up to

$\sin^2 \theta - \frac{2 \sin \theta}{\sin \theta} + \frac{1}{\sin^2 \theta}$

But not really sure where to go from hear to end up with
$\sin^2 \theta + \mathrm{cot}^2 \theta -1$

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!

@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) $\mathrm{Prove\ that}$ $\tan 15^{\circ} = \ 2 - \sqrt 3$

(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) $\mathrm{Find\ the\ exact\ value\ of}$ $\left (\sin 22\frac{1}{2}^{\circ} + \cos 22\frac{1}{2}^{\circ} \right) ^2$

@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) $\mathrm{Prove\ that}$ $\tan 15^{\circ} = \ 2 - \sqrt 3$

(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) $\mathrm{Find\ the\ exact\ value\ of}$ $\left (\sin 22\frac{1}{2}^{\circ} + \cos 22\frac{1}{2}^{\circ} \right) ^2$

That is quite precisely what you should for (i) $\displaystyle \tan 15^{\circ} \equiv \tan \left(60^{\circ}- 45^{\circ} \right) = \ldots$ 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.

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 $sin^2{x} + cos^2{x} = 1$ so now you'll be focusing only on the middle value and use the half angle formula.

When you multiply it out you'll get something of the form $\sin^2 \alpha + 2\sin \alpha \cos \alpha + \cos^2 \alpha$, 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.

Thank you both for your help

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:

$\mathrm{cosec(}$ $\frac{2pi}{3}$ $\mathrm{)}$


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!