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Double/Half Angle formulas...:(?
Could some one please explain how to do these problems using Half/Double angle formulas please? Any help will be greatly appreciated.
cos(φ) = 4/5 (0° < φ < 90°)
Need to find cos(2φ) tan(2φ) and sin(2φ)
Same type of problem but with tan.
How to use a half angle formula on sin(7π/8)
Thank you for your help or any point in the right direction :P
EDIT- FOUND MY NOTES, I think I got it now. If anyone could still show a problem with tan it would be great.
1 Answer
- Anonymous1 decade agoFavorite Answer
(1) Since 0° < θ < 90°, θ is in Quadrant I; this means that:
sinθ > 0 and tanθ > 0.
By constructing a 3-4-5 triangle in Quadrant I, we see that:
sinθ = opp/hyp = 3/5.
Hence:
(a) cos(2θ) = 2cos^2θ - 1 = 2(4/5)^2 - 1 = 7/25
(b) sin(2θ) = 2sinθcosθ = 2(3/5)(4/5) = 24/25
(c) tan(2θ) = sin(2θ)/cos(2θ) = (24/25)/(7/25) = 24/7.
(2) By the half-angle formula for sine:
sin(θ/2) = ±√[(1 - cosθ)/2].
(The sign of √[(1 - cosθ)/2] depends on the sign of sin(θ/2).)
Since 7π/8 is located in Quadrant II where sine is positive, pick the positive square root and use θ = 7π/4 to yield:
sin(7π/8) = √[(1 - cos 7π/4)/2]
= √[(1 + √2/2)/2]
= √[(2 + √2)/4]
= √(2 + √2)/2.
I hope this helps!