Trig math help please!?

2016-07-25 7:22 am

In each case find sin α, cos α, and tan α
(a)cos(2α) = −1/3 and π < 2α < 3π/2
(b) sin(α/2) = 4/5 and π/4 < α/2 < π/2

Thanks!

回答 (3)

wanszeto

2016-07-25 7:55 am

(a)
π < 2α < 3π/2
i.e. π/2 < α < 3π/4 (α in quadrant II)
Then, sin(α) > 0, cos(α) < 0 and tan(α) < 0

cos(2α) = -1/3
2cos²(α) - 1 = -1/3
2 cos²(α) = 2/3
cos²(α) = 1/3
As cos(α) < 0, then cos(α) = -1/√3
cos(α) = -(√3)/3

sin²(α) + cos²(α) = 1
sin²(α) + [(-√3)/3]² = 1
sin²(α) = 6/9
As sin(α) > 0, then sin(α) = (√6)/3

tan(α) = sin(α) / cos(α)
tan(α) = [(√6)/3] / [-(√3)/3]
tan(α) = -√2

====
(b)
π/4 < α/2 < π/2
i.e. π/2 < α < π (α in quadrant II)
Then, sin(α) > 0, cos(α) < 0 and tan(α) < 0

cos(α) = 1- 2 sin²(α/2)
cos(α) = 1 - 2 * (4/5)²
cos(α) = 1 - (32/25)
cos(α) = -7/25

sin²(α) + cos²(α) = 1
sin²(α) + (-7/25)² = 1
sin²(α) = 576/625
As sin(α) > 0, then sin(α) = √(576/25)
sin(α) = 24/25

tan(α) = sin(α) / cos(α)
tan(α) = (24/25) / (-7/25)
tan(α) = -24/7

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[匿名]

2016-07-25 7:32 am

IT IS 5

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用戶57802

2016-07-25 7:26 am

rule:
cos(2x) = 2 cos^2(x) - 1 = 1 - 2 sin^2(x) = (1 - tan^2(x))/(1 + tan^2(x))
or tan(x) = sin(x) / cos(x)

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