Question

a) sin 37° 30' sin 7° 30'
b) cos 75°cos15°
c) sin 75° cos15°
e) cos 50° cos 20° -1/2 cos 70°
f) 2 sin 70° sin 10° + cos 80°
g) 2 cos a cos 2 a - cos 3 a​
Помогите , пожалуйста!!!

Answer 1

Ответ:

Объяснение:

a) sin 37° 30' sin 7° 30' =1/2(cos (37° 30' -7° 30')-cos (37° 30' +7° 30')) =1/2(sin30-sin45)=1/2(√3/2-√2/2)=(√3-√2)/4)

b) cos 75°cos15° =1/2(cos (75°+15° )+cos (75°-15° ))=1/2(cos 90+cos 60) =1/2(0+1/2)=1/4

c) sin 75° cos15°= 1/2(cos (75°-15° )-cos (75°+15° ))=1/2(cos 60-cos 90) = 1/2(1/2-0)=1/4

e) cos 50° cos 20° -1/2 cos 70° =1/2(cos (50°+20° )+cos (50°-20° ))-1/2 cos 70°= 1/2 cos 70°+1/2 cos30 -1/2 cos 70°=1/2 cos30=1/2*1/2=1/4

f) 2 sin 70° sin 10° + cos 80° =(cos (70-10)-cos (70+10))+ cos 80°= cos60-cos 80°+cos 80°=cos60=1/2

g) 2 cos a cos 2 a - cos 3 a​=cos (a+2a )-cos (2a-a)- cos 3 a​=cos 3 a​- cos  a- cos 3 a​=​cos a