S.2 Trigo relations
2014-05-08 12:56 am
Given that cos theta = 5/6. By using simultaneousequations in two unknowns, find the value of 2 – 2 (sin theta)^2. I know how to use Pythagoras' Theorem and trigonometry identities (sin square theta + cos square theta = 1) to do. Buthow to use simultaneous equations to do?Thank you for answering me first.
回答 (3)
2014-05-08 10:40 pm
✔ 最佳答案
Let x = cosθ and y = sinθGive that : cosθ = 5/6 and (sinθ)² + (cosθ)² = 1
Hence, the two simultaneous equations are :
x = 5/6 ...... [1]
x² + y² = 1 ...... [2]
From [1] :
x² = (5/6)²
x² = 25/36 ...... [3]
[2] - [3] :
y² = 1- (25/36)
1 - y² = 25/36
2(1 - y²) = 2(25/36)
2 - 2y² = 25/18
Hence, 2 - 2(sinθ)² = 25/18
參考: 土扁
2014-05-08 7:07 pm
cos theta = 5 / 6 ------- (i)
x = 2 - 2sin square theta ------- (ii)
from (ii), x = 2(1 - sin square theta)
x = 2cos square theta ------- (iii)
substitute (i) into (iii), x = 2(5 / 6)^2 = 25 / 18
therefore, 2 - 2sin square theta = 25 / 18
x = 2 - 2sin square theta ------- (ii)
from (ii), x = 2(1 - sin square theta)
x = 2cos square theta ------- (iii)
substitute (i) into (iii), x = 2(5 / 6)^2 = 25 / 18
therefore, 2 - 2sin square theta = 25 / 18
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