2sin50 cos10 1 3 tan10 root number under 1 cos10 detailed steps

Updated on anime 2024-05-18
7 answers
  1. Anonymous users2024-02-10

    Molecule = 2sin50 + cos10 + root number 3 * sin102sin50 + 2 (sin10cos30 + cos10sin30) 2sin50 + 2sin40

    2sin40+2cos40

    2 roots, number: 2*sin(40+45).

    2 root number 2*sin85

    2 root number 2*cos5, and denominator = root number [1+2(cos5) 2-1] root number 2*cos5, so [2sin50+cos10(1+root number 3*tan10)] [root number(1+cos10)].

    2 root number 2*cos5) (root number 2*cos5).

  2. Anonymous users2024-02-09

    ԭ cos50 (tan10-tan60), cos50 (sin10cos60-sin60cos10), cos10cos60

    cos50 sin(-50) split number: cos10cos60-sin 100 2cos10cos60-cos10 2cos10cos60

  3. Anonymous users2024-02-08

    sin20=2sin10cos10

    cos20=1-2(sin10)^2

    Original = 2 (with bridge sin10) buried 2 [1-2(sin10) 2]-1+1 [1-2(sin10) 2].

    1+1 Argument Liquid Fierce Cos20

    sec20-1

  4. Anonymous users2024-02-07

    Solution: sin50° (1 + 3 tan10°)-cos20° cos80° (1-cos20°) = [sin50°(cos10°+ 3 sin10°) cos10°-cos20°] 2sin 10° = (1-cos20°) 2sin 10° = 2 (unity deformation and double angle formula).

  5. Anonymous users2024-02-06

    [2sin50+sin10(1+root:3*tan10)]root(sin 2(80))).

    2sin50+sin10(1+root:3*sin10 cos10)]sin80

    2sin50+sin10(1+root:3*sin10 cos10)]cos10

    2sin50cos10 + sin10 (cos10 + root number 3*sin10).

    2sin50cos10 + sin10

    2sin50cos10 + sin10

    2sin50cos10 + 2sin10 sin40

    2= 2= 2 sin(50+10)

    2sin60

    2 * root number 3 2

    Root number 3

  6. Anonymous users2024-02-05

    [(1+cos20°)/2sin20°]-sin10°(1/tan5°-tan5°)

    (1+cos20°)/2sin20°]-sin10°(cot5°-tan5°)

    (1+cos20°)/4sin10°cos10°]-sin10°(cot5°-tan5°)

    2cos10°/4sin10°)-2sin5°cos5°(cot5°-tan5°)

    cos10°/2sin10°)-2((cos5°)^2-(sin5°)^2)

    cos10°/2sin10°)-2cos10°

    cos10°-4sin10°cos10°)/2sin10°

    sin80°-2sin20°)/2sin10°

    (sin80°-sin20°)-sin20°)/2sin10°

    2cos50°sin30°-sin20°)/2sin10°

    sin40°-sin20°)/2sin10°

    2cos30°sin10°)/2sin10°

    cos30°

    Root No. 3) 2

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  7. Anonymous users2024-02-04

    Apparently 1 + root number 3tan10

    1+tan60tan10

    cos60cos10+sin60sin10)/cos60cos10

    cos(60-10)/cos60cos10=cos50/cos60cos10

    2cos50/cos10

    and 1+cos20=2(cos10) 2

    Therefore, the root number (1+cos20) = root number 2 * cos10 original formula = (2sin50 + 2sin10*cos50 cos10) * root number 2 *cos10).

    2 root number 2 * (sin50 * cos10 + sin10 * cos50) = 2 root number 2 * sin (50 + 10).

    Root number 6

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