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How to do these questions? (using CAST/congruent triangles etc)
Tan2(squared)β = tanβ for 0°≤β≤360°
2sin2(squared) β – 3sin β +2 = 0 for -π≤ β ≤π
Thanks in advance~

2007-03-03 18:19:52 · 4 answers · asked by Carly L 2 in Science & Mathematics Mathematics

hmm tried algbrahelp website... would be helpful but this is calculus.

2007-03-05 19:49:14 · update #1

4 answers

In each case, put in x for the trig function.

x^2 = x
x^2 - x = 0
x=0,1

Now put the trig function back in
tanβ = 0,1
tanβ= 0 when β= 0,180 degrees
tanβ=1 when β=45,225 degrees

2x^2-3x+2 = 0
I hope you meant -2 for that last term. Let's pretend you did. If not, the technique is the same, but you need a calculator

2x^2-3x-2 =0
(2x-1)(x+1)=0
x=1/2, -1
sinβ= 1/2 at pi/6 and -5pi/6
sinβ=-1 at -pi/2

2007-03-03 18:35:39 · answer #1 · answered by Anonymous · 0 0

Tan²β = tanβ for 0°≤β≤360°
shift the tan β to the left hand side
tan²β-tan²β=0
then factorise the eqn, take out the common factor tanβ
tanβ(tanβ-1)=0
tanβ=0 or tan β-1=0

tan β=0
β=0, 180°, 360° (this must be known or you can key into your calculator to verify)

As for tanβ-1=0
tanβ=1 (1 is positive)
So, β=45°(1st quad) or β=180°+45° (cause tan is positive in the 3rd quadrant also)

When you mean CAST, do you mean that for the
1st quadrant(0°-90°) all sin, cos and tan are positive,
2nd quadrant(90°-180°) only sin is positive, cos and tan negative,
3rd quadrant (180°-270°) only tan positive, sin and cos negative,
4th quadrant (270°-360°) only cos positive, sin and tan negative.
A handy mnemonic to remember would be "All Science Teachers are Crazy". (ASTC they stand for the trigo that is positive-in order from anticlockwise direction)

Hope this helps.

Try the 2nd qns yourself. I think you might have made typed the 2nd qns wrongly, should it be 2sin2(squared) β – 3sin β -2 = 0 instead of 2sin2(squared) β – 3sin β +2 = 0? (but the others have solved it for you using -2 instead).

2007-03-03 18:55:18 · answer #2 · answered by tabletennisrulez 2 · 0 0

Question 1
tan²ß - tanß = 0
tanß.( tanß - 1 ) = 0
tanß = 0,tanß = 1
ß = 0°,45°,180° and 225°

Question 2
2sin²ß - 3sinß - 2 = 0
(2sinß + 1).(sinß - 2) = 0
sinß = -1/2 (is only acceptable answer)
ß = 7π/6,11π/6 (3rd and 4th quadrants)

2007-03-03 18:48:32 · answer #3 · answered by Como 7 · 0 0

Try this http://www.algebrahelp.com/

2007-03-03 18:24:17 · answer #4 · answered by Joe 3 · 0 1

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