Advertisement
If you have a new account but are having problems posting or verifying your account, please email us on hello@boards.ie for help. Thanks :)
Hello all! Please ensure that you are posting a new thread or question in the appropriate forum. The Feedback forum is overwhelmed with questions that are having to be moved elsewhere. If you need help to verify your account contact hello@boards.ie

Maths Trigonometry Question

Options
  • 08-06-2008 12:03am
    #1
    mike-gunners
    Closed Accounts Posts: 33 ✭


    How Would You Solve This:

    Q: (i) if tan (-)/2 = t , write sec (-) in terms of t.
    (ii) hence show that tan (-) + sec (-) = 1+t/1-t

    *(-) represents "theta"


Comments

  • Options
    #2
    Calorimeterman
    Closed Accounts Posts: 64 ✭✭Calorimeterman


    I even think sec is on the course....

    But just look at the log tables, stare at page 9, until you get an idea...

    It'll be there :)


  • Options
    #3
    fivetwenty
    Closed Accounts Posts: 172 ✭✭fivetwenty


    i) cos -/2 = sin -/2 / tan -/2

    ii) sec -/2 = 1 / cos -/2

    Sub in (i) to (ii)

    sec -/2 = tan -/2 / sin -/2
    = t / sin -/2 ?


  • Options
    #4
    boobookitty
    Closed Accounts Posts: 232 ✭✭boobookitty


    Is this OL Maths? Can someone give me a crash course on what I need for Trig (self learning) I need to get an A1 in Paper 2.

    I'm doing: Area, Line, Circle, Probability, Statistics and Linear Programming.
    Line, Circle and Linear are ok. Statistics can go either way but Prob and Area are a bitch.


  • Options
    #5
    starkinter
    Closed Accounts Posts: 59 ✭✭starkinter


    Start > Run > "charmap"

    All the θ you'll ever need.


  • Options
    #6
    Cokehead Mother
    Closed Accounts Posts: 773 ✭✭✭Cokehead Mother


    tan2A = [2tanA] / [1 - (tanA)^2] ... log tables, pg. 9

    So tanθ = (2t)/(1 - t^2)

    (secθ)^2 = 1 + (tanθ)^2 ... log tables, pg. 9

    You should be fine from here.


  • Advertisement
  • Options
    #7
    GallowsGhost
    Registered Users Posts: 93 ✭✭GallowsGhost


    Can someone help me with this, I seem to have forgotten all trig :eek:

    From the marking scheme

    sin | ∠bca | = 18.4 × sin 44 or
    14
    18.4 × 0.6947 or 0.9130
    ⇒ | ∠bca| = 65.9°
    = 66°

    How do you get it from 0.9130 to 65.9° ?

    Edit: I got it, it's 2ndF - I'm an idiot!


Advertisement