IV. Put the verb in the correct form, Past Continuous or Past Simple.
1. Carol broke (break) her arm last week.
2. It ...(happen) when she ... (paint) her room. She ... (fall)
of the larder. 3. The train ... (arrive) at the station when Paula
(get) off. 4. Two friends of hers, John and Jenny, ... (wait) to
meet her. 5. Yesterday Sue ... (walk) along the road when she
(meet) Jim. 6. He ... (go) to the station to catch a train and he ...
(carry) a bag. 7. They ... (stop) to talk for a few minutes.​

1. His name is Foster. You may have heard his name. 2. The woman in the blue suit standing at the window seems familiar to me. I must have met her somewhere. 3. I can't find this record anywhere. Had it been smashed? 4. I shouldn't have told him that. He probably took offense at me. 5. Everything will definitely be fine! Your daughter will certainly recover! 6. You returned late yesterday; you should go to bed early today. 7. You shouldn't have spoken to her like that; she didn't deserve it. Besides, she's older than you. 8. It is necessary that every student take part in this competition. 9. You can't talk about the same thing endlessly! 10. The Chairman suggested that everyone present express their opinion on this issue. 11. You don't have to invent anything. You will be told what to do. 12. Have you never heard of him? Well, you'll hear more! 13. Jane couldn't forget the day that was supposed to be her wedding day and which ended so tragically. 14. Now I don't have to get up early: I'm studying at

Сначала формулы . затем формулы двойного угла sin2α=2sinαcosα cos2α=cos²α-sin²α=cos²α-(1-cos²α)=2cos²α-1⇒  2cos²α=1+cos2α cos2α=cos²α-sin²α=1-sin²α-sin²α=1-2sin²α⇒  2sin²α=1-cos2α а) сos75°cos105°=cos(90°-15°)·cos(90°+15°)== sin15°(-sin15°)=-sin²15°=-(1-cos30°)/2=(cos30°-1)/2= ((√3/2)-1)/2=0,25√3-0,5б) sin75°sin15°=°sin(90°-15°)sin15°=cos15° sin15°=sin30°/2=1/4=0,25в) sin105°cos15°=sin(180°-75°)cos15°=sin75°cos15°=sin(90°-15°)cos15°=cos15°cos15°=(1+cos30°)/2=(1+(√3/2))/2=0,5 +0,25√3 2 способ формулы cosα·cosβ=0,5cos(α-β)+0,5cos(α+β) sinα·sinβ=0,5cos(α-β)-0,5cos(α+β) sinα·cosβ=0,5sin(α+β)+0,5sin(α-β) а) сos75°cos105°=0,5cos(75°-105°)+0,5cos(75°+105°)=0,5cos(-30°)+0,5 cos180°==0,5· √3/2+0,5·(-1)=0,25√3-0,5б) sin75°sin15°=0,5cos(75°-15°)-0,5cos(75°+15° )=0,5cos60°-0,5 cos90°=0,5·0,5=0,25в) sin105°cos15°=0,5sin(105°+15°)+0,5sin(105°-15°)= =0,5sin120°+0,5sin90°==0,5 sin(180°-60°)+0,5·1=0,5 sin 60°+0,5=0,25√3+0,5