T-Formulae, Auxilary Method and General Solutions (1 Viewer)

[ForbiddenND]

Hi

Would anyone mind explaining to me how to properly use the t - formulae, auxilary method and general solutions. I have a general idea of how to use them but sometimes my answers don't match the answers wahhh...

Step by step method would be nice. :mad1::spzz

Cheers guys​

 

[gurmies]

If you wouldn't mind giving an example of each type and how you've gone wrong, we can fix the error and provide other solutions.

 

[randomnessss]

Example: 3cosX + 4sinX = -1 (0° =< X =< 360°)

T-method:
1. Find the appropriate trig expansion (e.g. sin(a+b), sin(a-b), cos(a+b), cos(a-b).
For this example, sin(a+b) or cos(a-b) would be appropriate.
2. Equate one of the trig expansions with the current equation, whilst ignoring the RHS.
3cosX + 4sinX = Rcos(X-a)
= RcosXcosa + RsinXsina
3. Equate the coefficients.
5cosX = RcosXcosa => Rcosa = 3
2sinX = RsinXsina => Rsina = 4
4. Find R and a.
i. R can be found by drawing a right-angled triangle and letting a be the angle for which sina and cos a are the appropriate sides.
R^2 = 3^2 + 4^2
r = 5 (R>0)
ii. a can be found by:
tana = Rsina/Rcosa
= 4/3
a = 53° 7' 48.3672"
5.
Rcos(X-a)= -1
5cos(X-53° 8') =-1
cos(X-53° 8') =-1/5
X-53° 8' = 101° 32' 13", (360 - 101° 32' 13")
X = (101° 32' 13" + 53° 8'), (360 - 101° 32' 13"+53° 8' )

 

[randomnessss]

T-method:
Basic info:
Let tan(X/2) = t.
sinX = 2t / 1 + t^2
cosX = 1 - t^2 / 1 + t^2

Example: 3cosX + 4sinX = -1 (0° =< X =< 360°)
1. Simplify where possible (this is an easy question so you won't be able to simplify it much further).
2. Substitute and solve. Sometimes it will be necessary to use the quadratic formula if factorisation is not applicable.
3(1 - t^2 / 1 + t^2) + 4(2t / 1 + t^2) = -1
3 - 3t^2 + 8t / 1 + t^2 = -1
3 - 3t^2 + 8t = -1 - t^2
4 - 4t^2 +8t = 0
4t^2 - 8t - 4 = 0
t^2 - 2t - 2 = 0
. . .

4. Use result from quadratic formula/factorisation and replace t with tan(x/2).
5. Solve using the result(s).
6. Test for x=180° in the actual equation.
LHS = 3cosX + 4sinX
= 3cos180° + 4sin180°
= -1
Therefore x = 180° is not a solution (LHS =/= RHS)
Therefore X = 154° 40', 206° 60'

 

[GUSSSSSSSSSSSSS]

basically t formula is a way of expressing TRIG identities in terms of normal "x, y etc" variables

since t = tan(@/2)
through use of double angle expansion we can express tan@ = (2t)/(1-t^2)
and then by drawing a triangle of angle @ we can express
sin@ = (2t)/(1+t^2)
cos@ = (1-t^2)/(1+t^2)

then you are able to substitute these values of sin@, cos@ and tan@ (naturally its easy to work out sec@, cosec@ and cot@) and u can then work out the question WITHOUT having to deal with trigonometry at all (except at the end when u sub it back xD)

 

[Drongoski]

Can also do it this way:




This is a quadratic eqn in sin x ... solve for sin x and find possible values of 'x' and verify valid; some as per usual are extraneous (not correct)

 

[studentcheese]

General solutions. (where n is an integer)













e.g. Find the General Solution for:




Find the related angle, and then substitute it into n.

For trig equations, I would try to solve them by factorising. I wouldn't use the t formula unless told. Auxilary method cannot be used if the equation is equal to zero (correct me if I am wrong). General solutions - only use when told.

 

[ForbiddenND]

cool thanks people