我想問maths的functions.....

2007-07-09 5:00 am
Given the function p(x) = cos(x°+30°)
question: find the value of [p(11)]²+ [p(12)] ²+…+[p(19)] ²
(leave the answers in surd form)

回答 (2)

2007-07-09 5:23 am
✔ 最佳答案
[p(11)]²+ [p(12)] ²+…+[p(19)] ²

= cos2 41°+ cos2 42° + cos2 43° + cos2 44° + cos2 45° + cos2 46° + cos2 47°+ cos2 48° + cos2 49°

= cos2 41° + sin2 ( 90 – 49 )° + cos2 42° + sin2 ( 90 – 48 )° + cos2 43° + sin2 ( 90 – 47 )° + cos2 44° + sin2 ( 90 – 46 )° + cos2 45°

= 1 + 1 + 1 + 1 + ( √2 / 2 ) 2

= 4 + 1 / 2

= 9 / 2


參考: My Maths Knowledge
2007-07-09 5:20 am
Given the function p(x) = cos(x°+30°)
question: find the value of [p(11)]²+ [p(12)] ²+…+[p(19)] ²
[p(11)]²+ [p(12)] ²+…+[p(19)] ²
= cos²(11+30) + cos²(13+30) + cos²(15+30) + cos²(17+30) + cos²(19+30)
= cos²41 + cos²43 + cos²45 + cos²47 + cos²49
= sin²(90-41) + sin²(90-43) + cos²45 + cos²47 + cos²49
= sin²(90-41) + sin²(90-43) + cos²45 + cos²47 + cos²49
= sin²49 + sin²47 + cos²45 + cos²47 + cos²49
= (sin²49 + cos²49) + (sin²47 + cos²47) + cos²45
= 1 + 1 + (√2 / 2)²
= 2.5


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