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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).
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ԭ cos50 (tan10-tan60), cos50 (sin10cos60-sin60cos10), cos10cos60
cos50 sin(-50) split number: cos10cos60-sin 100 2cos10cos60-cos10 2cos10cos60
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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
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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).
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[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
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[(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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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
x should have a range, right? y=f(x)=(1+cos2x+8sin 2x) sin2x =(1+cos2x+4-4cos2x) sin2x =(5-3cos2x) sin2x =(5-3cos2x) [1-(cos2x) 2] and then solve according to the range Follow-up: When 0 is basically the same as the following.
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