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View Full Version : Pre-Calculus Problems


tpolwort
Dec 1, 2009, 01:43 PM
Do which ever ones you can help me on please!

Identities
1. sin(x)[sec(x) - csc(x)] = tan(x) - 1

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Interval is 0 < Ѳ < 2π
1. 2cos^2(Ѳ) +cos(Ѳ) -1 = 0

2. sin(2Ѳ) - cos(Ѳ) - 2sin(Ѳ) + 1 = 0

3. sin(2Ѳ) = √2cos(Ѳ)
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Angles
csc(α) = 2, π/2 < α < π ; sec(β) = -3, π/2 < β < π
1. Find sin(2β)
2. Find cos(α+β)
3. Find sin(α/2)

s_cianci
Dec 1, 2009, 04:22 PM
Identities
1. sin(x)[sec(x) - csc(x)] = tan(x) - 1If you rewrite sec(x) and csc(x) in terms of sin(x) and cos(x), it'll fall right into place. Remember that tan(x) = sin(x)/cos(x).

2cos^2(Ѳ) +cos(Ѳ) -1 = 0Factor it, just like if it were 2x^2 + x - 1 = 0.

sin(2Ѳ) - cos(Ѳ) - 2sin(Ѳ) + 1 = 0Apply the double-angle formula to rewrite sin(2Ѳ) accordingly, then it'll fall right into place.

sin(2Ѳ) = √2cos(Ѳ)Same hint as above regarding the sin(2Ѳ). Then divide both sides by cos(Ѳ) and you're home free.

Angles
csc(α) = 2, π/2 < α < π ; sec(β) = -3, π/2 < β < π
1. Find sin(2β)
2. Find cos(α+β)
3. Find sin(α/2)Decide what quadrants α and β are located in. Then, since you know the csc of α you can find the sin of α. Likewise, since you know the sec of β, you can find the cos of β. Then apply the 1st. Pythagorean Identity to determine the sin of β and the cos of α. Finally, apply the double-angle formula for sin, the sum formula for cos and the half-angle formula for sin.