trigonometry relations | MathCelebrity Forum

trigonometry relations

  1. math_celebrity

    Simplify sin(x) * cos(x) * tan(x) * csc(x) * sec(x) * cot(x)

    We know the following: csc(x) = 1/sin(x) sec(x) = 1/cos(x) cot(x) = 1/tan(x) We can rewrite our original expression as: sin(x) * cos(x) * tan(x)/sin(x) * cos(x) * tan(x) Everything cancels and we are left with 1
  2. math_celebrity

    Simplify sin^2(x)/(1 - sin^2(x))

    We know from the pythagorean theorem: sin^2(x) + cos^2(x) = 1 Subtract sin^2(x) from each side and we get: cos^2(x) = 1 - sin^2(x) We can rewrite our original expression as: sin^2(x)/cos^2(x) But this expression equals tan^2(x)
  3. math_celebrity

    cscx/secx =cotx

    cscx/secx =cotx This is true Remember that: csc(x) = 1/sin(x) sec(x) = 1/cos(x) So we have: 1/sin(x)/1/cos(x) cos(x)/sin(x) cot(x)
  4. math_celebrity

    sin(x)cot(x)

    sin(x)cot(x) We know that cot(x) = cos(x)/sin(x), so we rewrite this as: sin(x)cos(x)/sin(x) The sin(x) terms cancel and we get: cos(x)
  5. math_celebrity

    cscx-cotx*cosx=sinx

    cscx-cotx*cosx=sinx A few transformations we can make based on trig identities: csc(x) = 1/sin(x) cot(x) = cos(x)/sin(x) So we have: 1/sin(x) - cos(x)/sin(x) * cos(x) = sin(x) (1 - cos^2(x))/sin(x) = sin(x) 1 - cos^2(x) = sin^2(x) This is true from the identity: sin^2(x) - cos^2(x) = 1
  6. math_celebrity

    cot(θ)=12 and θ is in Quadrant I, what is sin(θ)?

    cot(θ)=12 and θ is in Quadrant I, what is sin(θ)? cot(θ) = cos(θ)/sin(θ) 12 = cos(θ)/sin(θ) Cross multiply: 12sin(θ) = cos(θ) Divide each side by 12: sin(θ) = 12cos(θ)
  7. math_celebrity

    4 divided by sin60 degrees

    4 divided by sin60 degrees. We can write as 4/sin(60). Using our trigonometry calculator, we see sin(60) = sqrt(3)/2. So we have 4/sqrt(3)/2. Multiplying by the reciprocal we have: 4*2/sqrt(3) 8/sqrt(3)
  8. math_celebrity

    Express cos4θ and sin4θ in terms of sines and cosines of multiples of θ

    Express cos4θ and sin4θ in terms of sines and cosines of multiples of θ. Using a trignometric identity: cos (2θ) = cos^2(θ) - sin^2(θ) Since 4θ = 2*2θ, so we have: cos(4θ) = cos^2(2θ) - sin^2(2θ) Using another trignometric identity, we have: sin(2θ) = 2 sin(θ) cos(θ) Since 4θ = 2*2θ, so we...
  9. math_celebrity

    Find an angle (theta) with 0<(theta)<360° or 0<(theta)<(2*pi) that has the same sine value as 80°

    Find an angle (theta) with 0<(theta)<360° or 0<(theta)<(2*pi) that has the same sine value as 80°. The sine is positive in Quadrant I and Quadrant II. So we find the reference angle for 80°. It's 180 - 80 = 100°. This is our answer. Sin(80°) = Sin(100°)
  10. math_celebrity

    calculate cos(x) given tan(x)=8/15

    calculate cos(x) given tan(x)=8/15 tan(x) = sin(x)/cos(x) sin(x)/cos(x) = 8/15 Cross multiply: 15sin(x) = 8cos(x) Divide each side by 8 cos(x) = 15sin(x)/8
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