Trigonometry (1 Viewer)

[lyounamu]

Time for another question;

A triangle PQR is right-angled at Q. S is the point on PR such that QS is perpendicular to PR. Let angle RPQ = x. You are given that 2PS - QR = PR.

a) Show that 2cosx - tanx = secx.

b) Deduce that 2sin² + sinx - 1 = 0.

c) Find x.

a) L.H.S. = 2cosx - tanx = 2PS/PQ - QR/PQ = (2PS - QR)/PQ = PR/PQ = secx = R.H.S.

 

[Fortian09]

hey iyounamu havent seen u for a while

 

[lyounamu]

hey iyounamu havent seen u for a while

hey, haven't seen you a while too. how you doing in maths?

EDIT: OFF TO ENG, will come back in few hours.

 

[Aerath]

Started getting annoyed that I didn't specify sinx = s and cosx = c.

What's with your fraction bars?

 

[Aplus]

It's called a ruler

 

[Aerath]

It's called a ruler

Screw that - I never use a ruler.

 

[lyounamu]

Shit, I nearly slept while doing this quesiton. I am off to bed. To the question, I was solving, first get the sin^2(x) + sinx -1 to the simplified version and get the sin in to prove that it is 0 on one bracket.

Answer to part 3 is pi/6 = 30 degrees.

 

[lyounamu]

I got more problems for trig.


Given tanA= 1/2 and tanB=-2, where 180<270<B<360. p value the find of<>
(secA-secB)/(cscA-cscB)


given 3sec² + 6tan² = 4 and 90<X<180. p value the of Find sinx+tanx.<>

given sinx = 2ab/ a²+b², where a<0

Given sin²x/(s+7cos²x = 1/9 where 270<X<360. p value the of Find tanx.<>
Thanks for help

I will make sure I do this q at 3am. (unless someone comes here and kindly take my place. I am currently in a state of emergency where immediate sleep is required.... I am off.

 

[lyounamu]

I got more problems for trig.


Given tanA= 1/2 and tanB=-2, where 180<270<B<360. p value the find of<>
(secA-secB)/(cscA-cscB)


given 3sec² + 6tan² = 4 and 90<X<180. p value the of Find sinx+tanx.<>

given sinx = 2ab/ a²+b², where a<0

Given sin²x/(s+7cos²x = 1/9 where 270<X<360. p value the of Find tanx.<>
Thanks for help

What am I supposed to find here?

 

[Wassup?]

Bleah anything to avoid doing English -

Cancels out to make (1+tan³x)/(1-tan³x)

I'll do the next one if you're still struggling, but think about the box a little bit.

Yo, just to let you know, you cannot just cross out parts of a question, you'll lose marks. You can put a very faint line through it, but other than that, and you'll be penalised. I found out the hard way when my teacher took a mark off my paper for doing that.

 

[Fortian09]

I got more problems for trig.

Given tanA= 1/2 and tanB=-2, where 180<270<B<360.
find the value of (secA-secB)/(cscA-cscB)

given 3sec² + 6tan² = 4 and 90<x<180.
Find the value of sinx+tanx.

given sinx = 2ab/ a²+b², where a<0<b and x lies in Quadrant III.
Express cosx and tanx in terms of a and b.

Given sin²x/(s+7cos²x = 1/9 where 270<x<360.
Find the value of tanx.

Thanks for help

 

[Fortian09]

there we go

 

[Aerath]

Yo, just to let you know, you cannot just cross out parts of a question, you'll lose marks. You can put a very faint line through it, but other than that, and you'll be penalised. I found out the hard way when my teacher took a mark off my paper for doing that.

Er...are you sure?

 

[Aplus]

I always do that. Never lost marks. Nazi teacher please.

 

[Aerath]

I always do that. Never lost marks.

Yeah, same.

 

[Aplus]

In fact, it displays a methodology of application of logical thinking in which you present to your marker the way in which you went about answering the question and reaching your end result.

 

[tommykins]

回复: Re: Trigonometry

Yo, just to let you know, you cannot just cross out parts of a question, you'll lose marks. You can put a very faint line through it, but other than that, and you'll be penalised. I found out the hard way when my teacher took a mark off my paper for doing that.

No. Just no.

 

[lyounamu]

Re: 回复: Re: Trigonometry

It may sound irrelevant but I cannot bother creating another thread.

When you find the point of inflexion, do you have to test each side?

 

[tommykins]

Re: 回复: Re: Trigonometry

Yes, for inflextion the concavity changes. Also take note that inflection points don't necessarily mean f'(x) = 0 at that point either.

 

[lyounamu]

Re: 回复: Re: Trigonometry

Yes, for inflextion the concavity changes. Also take note that inflection points don't necessarily mean f'(x) = 0 at that point either.

Isn't it f''(x)?

So what the hell is the point like if its f''(x) = 0 but the concavity does not change?