преобразуйте сумму в произведение :) и с объяснениями пожалуууйстаsin t + cost + 5*cos(t+ пи/4)
вычислить
(sin38 - cos38)/корень кв. из 2* sin 7
По формулам приведения заменим cost на sin(π/2-t).
sint + cost=sint+sin(π/2-t)=2*sin(t+π/2-t)/2 *cos(t-π/2+t)/2=2*sinπ/2 *cos(t-π/4)=2*cos(t-π/4)
2*cos(t-π/4) + 5*cos(t-π/4)=2*cos(t-π/4)+5*cos(t-π/4)=7*cos(t-π/4)
2)cos38=cos(90-52)=sin52
sin38-sin52=2sin45cos7
(sin38-sin52) /√2*sin7=2sin45cos7 /√2sin7 =2*(√2/2)*tg7/√2=tg7
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По формулам приведения заменим cost на sin(π/2-t).
sint + cost=sint+sin(π/2-t)=2*sin(t+π/2-t)/2 *cos(t-π/2+t)/2=2*sinπ/2 *cos(t-π/4)=2*cos(t-π/4)
2*cos(t-π/4) + 5*cos(t-π/4)=2*cos(t-π/4)+5*cos(t-π/4)=7*cos(t-π/4)
2)cos38=cos(90-52)=sin52
sin38-sin52=2sin45cos7
(sin38-sin52) /√2*sin7=2sin45cos7 /√2sin7 =2*(√2/2)*tg7/√2=tg7