Question

Как сможете 1) sin82°30' cos52°30' 2) sin82°30' cos 37°30' 3) cos37°30' cos7°30' 4) cos82°30' cos37°30' 5) cos75° cos105° 6) cos45° cos75° 7) 2sinα sin2α+cos3α=cosα 8) 2sinα sin3α+2cos7α cos3α-cos10α=cos2α 9) cos(α+60°) + cos(α-60°) 10) cos(α+60°) - cos(α-60°) 11) 1+2sinα 12) 1-2sinα 13) 0.5+cosα 14) 0.5-cosα 15) cosα+1 16) 1-cosα 17) cos95°+cos94°+cos93°+cos85°+cos86°+cos87° 18) sin5x+sinx=0 19) cos2x+cosx=0

Answer 1

1) sin82°30' cos52°30' = $\frac{1}{2}$(sin(82°30' - 52°30') +sin(82°30' + 52°30') = $\frac{1}{2}$(sin30°+sin135°) = $\frac{1}{2}$($\frac{1}{2} + \frac{ \sqrt{2} }{2}$) = $\frac{1+ \sqrt{2} }{4}$
2) sin82°30' cos 37°30' = $\frac{1}{2}$(sin(82°30' - 37°30')+sin(82°30' + 37°30')) = $\frac{1}{2}$(sin45+sin120) = $\frac{1}{2} ( \frac{ \sqrt{2} }{2} + \frac{ \sqrt{3} }{2} )= \frac{ \sqrt{2}+ \sqrt{3} }{4}$
3) cos37°30' cos7°30' = $\frac{1}{2}$(cos(37°30' - 7°30')+cos(37°30' + 7°30'))=$\frac{1}{2}$(cos30+cos45)=$\frac{1}{2} ( \frac{ \sqrt{3} }{2}+ \frac{ \sqrt{2} }{2} )= \frac{ \sqrt{3}+ \sqrt{2} }{4}$
4) cos82°30' cos37°30' = $\frac{1}{2}$(cos(82°30' - 37°30')+cos(82°30' + 37°30') = $\frac{1}{2}$(cos45+cos120)=$\frac{1}{2} ( \frac{ \sqrt{2} }{2} - \frac{1}{2} )= \frac{ \sqrt{2}-1 }{4}$
5) cos75° cos105° = $\frac{1}{2}$(cos(75-105)+cos(75+105))=$\frac{1}{2}$(cos30+cos180) = $\frac{1}{2} ( \frac{ \sqrt{3} }{2} -1)= \frac{1}{2} ( \frac{ \sqrt{3}-2 }{2} )= \frac{ \sqrt{3}-2 }{4}$
6) cos45° cos75° = $\frac{1}{2}$(cos(45-75)+cos(45+75))=$\frac{1}{2}$(cos30+cos120)=$\frac{1}{2} ( \frac{ \sqrt{3} }{2} - \frac{1}{2} )= \frac{ \sqrt{3} -1}{4}$
17) cos95°+cos94°+cos93°+cos85°+cos86°+cos87° = (cos95+cos85)+(cos94+cos86)+(cos93+cos87)=2cos((95+85)/2)*cos((95-85)/2) + 2cos((94+86)/2)*cos((94-86)/2)+2cos((93+87)/2)*cos((93-87)/2) = 2cos90*cos5 + 2cos90*cos4 + 2cos90*cos3 = 0 (так как cos90=0
18) sin5x + sinx = 0
2sin((5x+x)/2)*cos((5x-x)/2) = 0
2sin3x*cos2x = 0
sin3x = 0   или cos2x = 0
3x = πn                2x = π/2+πn
x = πn/3               x = π/4 + πn/2
ответ: πn/3; x = π/4 + πn/2; n∈Z
19) cos2x + cosx = 0
2cos((2x+x)/2)*cos((2x-x)/2) = 0
2cos(3x/2)*cos(x/2)=0
cos(3x/2) = 0          или cos(x/2) = 0
3x/2 = π/2+πn                      x/2 = π/2+πn
3x = π+2πn                          x = π + 2πn
x=π/3 + 2πn/3
ответ: π/3 + 2πn/3; π + 2πn; n∈Z