I am doing practice questions before my test on monday and these are the ones I am having trouble with. If you could please explain how to get to the answer, that would be much appreciated!
1.Find cot θ if csc θ =√17/4 and tan θ > 0.
2.Simplify the expression: cot x sin x - sin (pi/2 - x) + cos x
3. Find tan θ if sec θ = √37/6 and sin θ < 0.
4. Find all solutions in the interval [0, 2π):
sec^2x - 2 = tan^2x
5.Find all solutions to the equation:
sin x = √3/2
6. Write the expression as the sine, cosine, or tangent of an angle.
sin 9x cos x - cos 9x sin x
7. Write the expression as the sine, cosine, or tangent of an angle.
cos 112° cos 45° + sin 112° sin 45°
8. Rewrite with only sin x and cos x.
sin 2x - cos 2x
9.Find the exact value by using a half-angle identity:
sin 22.5°
10. Verify the identity (Cosx)/(1+sinx) + (1+sinx)/(cosx) = 2 secx
Answers & Comments
Verified answer
1) given cot θ = 1/tan θ and csc θ = 1/ sin θ
also hypotenuse² = opposite² + adjacent²
=> (√17)² = 4² + adjacent²
also as tan θ > 0 then θ is in either quadrant 1 or 3
also as csc θ > 0 then θ is in either quadrant 1 or 2
=> θ is in quadrant 1
=> adjacent is positive
=> adjacent = √(17 - 16) = 1
as cot θ = adjacent/opposite
=> cot θ = 1/4
2.) cot(x) sin(x) - sin (π/2 - x) + cos (x)
given cot(x) = cos(x)/sin(x)
=> cos(x) - sin (π/2 - x) + cos (x)
=> 2cos(x) - sin (π/2 - x)
given cos(x) = sin (π/2 - x)
=> cos(x)
3.) given sec (θ) = 1/cos(θ)
=> 37 = 6² + opposite²
also as sin θ < 0 then θ is in either quadrant 3 or 4
also as sec θ > 0 then θ is in either quadrant 1 or 4
=> θ is in quadrant 4
=> opposite is negative
=> opposite = -√(37 - 36)
=> opposite = -1
as tan(θ) = opposite/adjacent
=> tan(θ) = -1/6
4) give sec²(x) = 1 + tan²(x)
=> 1 + tan²(x) -2 = tan²(x)
=> tan²(x) -1 = tan²(x)
=> -1 = 0
There are no solution for x
5) x = 2(3πn+π)/3 and x = (6πn+π)/3 where n element of the set of integers
6) given sin(a - b) = sin(a)cos(b) - cos(a)sin(b)
=> sin(9x - x)
=> sin(8x)
7) given cos(a - b) = cos (a) cos (b) + sin (a) sin (b)
=> cos(112° - 45°)
=> cos(67°)
8) given sin(2x) = 2sin(x)cos(x) and cos(2x) = cos²(x) – sin²(x) = 2cos²(x) – 1 = 1 – 2 sin²(x)
=> 2sin(x)cos(x) - (cos²(x) – sin²(x))
=> sin²(x) + 2sin(x)cos(x) - cos²(x)
9) the half-angle identy for sin is sin(t/2) = ±√((1 – cos(t)) / 2)
=> sin(22.5°) = ±√((1 – cos(45°)) / 2)
given cos(45°) = 1/√2 = √2/2
=> sin(22.5°) = ±√((1 – √2/2) / 2)
=> sin(22.5°) = ±√((2 – √2) / 4)
given sin(22.5°) is in quadrant 1 it is positive
=> sin(22.5°) = √(2 – √2) / 2
10) (cos(x))/(1+sin(x)) + (1+sin(x))/(cos(x)) = 2sec(x)
=> [cos²(x) + (1+sin(x))²]/[(1+sin(x))cos(x)] = 2sec(x)
=> [cos²(x) + sin²(x) + 2sin(x) +1]/[(1+sin(x))cos(x)] = 2sec(x)
=> [2sin(x) +2]/[(1+sin(x))cos(x)] = 2sec(x)
=>[ 2(sin(x)+1)]]/[(1+sin(x))cos(x)] = 2sec(x)
=> 2/cos(x) = 2sec(x)
=> 2sec(x) = 2sec(x)