Question

упростите выражение :
1) sin15°cos7°-cos11°cos79°
2)2sin2asina+cos3a
3)cos2b+2sin(b+30°)sin(b-30°)
помогите пожалуйста ​

Answer 1

$1) sin15^0cos7^0-cos11^0cos79^0=\\\\=sin15^0cos7^0-cos11^0cos(90^0-11^0)=\\\\=sin15^0cos7^0-cos11^0sin11^0=\\\\=sin15^0cos7^0-\frac{1}{2}sin22^0=\\\\=sin15^0cos7^0-\frac{1}{2}sin(15^0+7^0)=\\\\=\frac{2sin15^0cos7^0-(sin15^0cos7^0+cos15^0sin7^0)}{2}=\\\\=\frac{1}{2}(2sin15^0cos7^0-sin15^0cos7^0-cos15^0sin7^0)=\\\\=\frac{1}{2}(sin15^0cos7^0-cos15^0sin7^0)=\\\\=\frac{1}{2}sin(15^0-7^0)=\frac{1}{2}sin8^0$

$2)2sin2asina+cos3a=\\\\=2sin2asina+cos(2a+a)=\\\\=2sin2asina+cos2acosa-sin2asina=\\\\=sin2asina+cos2acosa=cos(2a-a)=cosa$

$3)cos2\beta +2sin(\beta+30^0)sin(\beta-30^0)=\\\\=cos2\beta+2(sin\beta cos30^0+cos\beta sin30^0)(sin\beta cos30^0-cos\beta sin30^0)=\\\\=cos2\beta +2(sin\beta *\sqrt{3}/2+cos\beta *1/2)(sin\beta *\sqrt{3}/2-cos\beta *1/2)=\\\\=cos2\beta +\frac{2}{2}((\sqrt{3}sin\beta)^2-cos^2\beta)=\\\\=cos^2\beta-sin^2\beta +3sin^2\beta-cos^2\beta=2sin^2\beta$