pre-calc 12 help on trig identities?
回答 (3)
2020-07-28 2:13 pm
tan[x - (π/2)] = 1/√3
tan[x - (π/2)] = tan(π/6)
x - (π/2) = nπ + (π/6)
x = nπ + (π/6) + (π/2)
x = nπ + (2π/3) where n ∊ Z
tan[x - (π/2)] = tan(π/6)
x - (π/2) = nπ + (π/6)
x = nπ + (π/6) + (π/2)
x = nπ + (2π/3) where n ∊ Z
2020-07-28 6:23 pm
tan(x - π/2) = 1/√3
so, x - π/2 = π/6 + nπ....for n ∈ ℤ
Hence, x = 2π/3 + nπ...general solution
Note: With n = 0 and n = 1 we have:
x = 2π/3 and 5π/3...for 0 < x < 2π
:)>
so, x - π/2 = π/6 + nπ....for n ∈ ℤ
Hence, x = 2π/3 + nπ...general solution
Note: With n = 0 and n = 1 we have:
x = 2π/3 and 5π/3...for 0 < x < 2π
:)>
2020-07-28 2:06 pm
tan(t) = 1/sqrt(3) when t = pi/6 + pi * k, where k is an integer. That's from the unit circle.
tan(x - pi/2) = 1/sqrt(3)
tan(x - pi/2) = tan(pi/6 + pi * k)
x - pi/2 = pi/6 + pi * k
x = pi/2 + pi/6 + pi * k
x = 3pi/6 + pi/6 + pi * k
x = 4pi/6 + pi * k
x = 2pi/3 + pi * k
x = (pi/3) * (2 + 3k)
Again, k is an integer
tan(x - pi/2) = 1/sqrt(3)
tan(x - pi/2) = tan(pi/6 + pi * k)
x - pi/2 = pi/6 + pi * k
x = pi/2 + pi/6 + pi * k
x = 3pi/6 + pi/6 + pi * k
x = 4pi/6 + pi * k
x = 2pi/3 + pi * k
x = (pi/3) * (2 + 3k)
Again, k is an integer
收錄日期: 2021-04-24 07:54:06
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