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Aim: How do we solve trig equations involving more than one trig function?. Do Now:. Find all values of in the interval 0 < < 360 o that satisfy the equation 2 sin 2 + sin = 1. alternate form of Pythagorean ID. cos 2 = 1 – sin 2 . - PowerPoint PPT Presentation
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Aim: Trig Equations with Multiple Functions Course: Alg. 2 & Trig.
Aim: How do we solve trig equations involving more than one trig function?
Do Now:
Find all values of in the interval 0 < < 360o that satisfy the equation
2 sin2 + sin = 1.
Aim: Trig Equations with Multiple Functions Course: Alg. 2 & Trig.
Solving Multiple Trig Function Equations
Solve for in the interval 0º ≤ ≤ 360º:2cos2 – sin = 1
cos2 = 1 – sin2
alternate form of Pythagorean ID
substitute: 2(1 – sin2 ) – sin = 1
2 – 2sin2 – sin = 1
-2sin2 – sin + 1 = 0
2sin2 + sin – 1 = 0
(2sin – 1)(sin + 1) = 0
(2sin – 1) = 0 (sin + 1) = 0
sin = 1/2 sin = -1 = 30º or 150º = 270º{30º,150º, 270º}
standard form/factor/solve:
Aim: Trig Equations with Multiple Functions Course: Alg. 2 & Trig.
Solving Multiple Trig Function Equations
Solve for to the nearest degree in the interval 0º ≤ ≤ 360º:2sec2 – 3tan - 5 = 0
sec2 = 1 + tan2
alternate form of Pythagorean ID
substitute: 2(1 + tan2 ) – 3tan - 5 = 0
2 + 2tan2 – 3tan - 5 = 0
2tan2 – 3tan - 3 = 0
standard form/factor/solve:
x b b2 4ac
2aa = 2, b = -3, c = -3
2( 3) ( 3) 4(2)( 3)
2(2)x
3 33
4x
tan = 2.19, tan = -0.69
= 65o, 145o, 245o, 3250
Aim: Trig Equations with Multiple Functions Course: Alg. 2 & Trig.
Model Problems
2
tan 2cot 1
7cos 1 6sec
sec 4 2tan
Find to the nearest degree, all values of in the interval 00 < < 3600 that satisfy the given equation.
Aim: Trig Equations with Multiple Functions Course: Alg. 2 & Trig.
Model Problems
2
2sin 3csc 5
2sin 1 csc
13tan 2
cot
Find to the nearest degree, all values of in the interval 00 < < 3600 that satisfy the given equation.
Aim: Trig Equations with Multiple Functions Course: Alg. 2 & Trig.
Aim: Trig Equations with Multiple Functions Course: Alg. 2 & Trig.