Need help with Trig!!!

[madoscar65]

Hi people,
I've been trying to solve this, but I end up no where near the answer... So I was just wondering if you guys can help  . The question is, prove that cosecx+cotx=cot(x/2)
Thanks

[cohen]

csc(x) + cot(x) = cot(x/2)
1/six(x) + cos(x)/sin(x) = (1 + cos(x))/sin(x)
as 1/sin(x) and cos(x)/sin(x) have the same denominator, we can simplify this to
(1 + cos(x))/sin(x) = (1 + cos(x))/sin(x)
LHS = RHS

For cot(x/2), i did 1/tan(x/2)

For useful trig identidies:
http://uppit.com/18umapw8cozi/TRIGONOMETRY_IDENTITIEs.pdf

[ninbam1k]

Wasn't sure about cohen's working out but heres how I did it. When I see proving questions, I keep in mind that you have to prove that the LHS = RHS or vice versa and you start off using one side only.

  cosec(x) +cot(x)

= 1/sin(x)   +   cos(x)/sin(x)  = 1+cos(x)/ sin(x)

cos(x) = 2(cos(x/2))^2 - 1
sin(x) = 2sin(x/2)cos(x/2)

= 1 + 2(cos(x/2))^2 - 1/ 2sin(x/2)cos(x/2)

= 2(cos(x/2))^2 / 2sin(x/2)cos(x/2)

cos(x/2)/sin(x/2)

=cot(x/2)

[pi]

For useful trig identidies:
http://uppit.com/18umapw8cozi/TRIGONOMETRY_IDENTITIEs.pdf

Holy crap... I wrote that sheet!

(This was my original link...)

[ivanivan0002]

I would also like help with a trig question.

It is possible to find the solutions between [-p,p] for sec x = 2.5 without using a calculator?

[brightsky]

You won't be able to get the numeric solutions, though you can probably make rough guesses based on a triangle.

[brightsky]

What about the exact value of sin1? (degrees)
Sounds rather bashy...

Theoretically, yes, using this: http://en.wikipedia.org/wiki/Trigonometric_identity#Double-.2C_triple-.2C_and_half-angle_formulae. But seeing as the exact value of sin(3) is already quite OMG, re: http://en.wikipedia.org/wiki/Exact_trigonometric_constants...

[ivanivan0002]

Next question, how would you do these questions

Solve each of the following equations [0,2p]

cos^2 x - cosxsinx = 0

sin2x = sinx

[Andiio]

For the first one, factorise the equation:

i.e.

cosx (cosx - sinx) = 0

Then utilising the null factor, you get:

cos x = 0, cos x = sin x.

I'm sure you can solve it from there


2. sin 2x = sin x

Use the identity of sin 2x!

[ivanivan0002]

Ahhh that's the thing Andiio, where do I from cos x = sin x? Haven't encountered a question like this before :S

[vea]

With cosx=sinx you must divide both sides of the equation by cosx which results in:

1=tanx

Then you can solve from there.

[ivanivan0002]

Ahh...thanks alot vea!

I'm getting pretty annoying but,

sin8x = cos4x, find all the solutions between [0,2p]

Would this involve compound formulas and factorising?

[kamil9876]

Ahh...thanks alot vea!

I'm getting pretty annoying but,

sin8x = cos4x, find all the solutions between [0,2p]

Would this involve compound formulas and factorising?

Yes double angle formula is enough. If it makes it any easier you may want to let , but then note that:

is equivalent to

So you need to solve over a big domain.

Then we get:





Now solve over the required domain as given above.