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secondary 3 | A Maths
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Thxx
Date Posted: 3 years ago
Views: 303
Given :
sin a + sin b = x
sin a = x - sin b
Square both sides,
sin² a = (x - sin b)²
sin² a = x² - 2x sin b + sin² b ①
Given :
cos a + cos b = x
cos a = x - cos b
Square both sides,
cos² a = (x - cos b)²
cos² a = x² - 2x cos b + cos² b ②
① + ②
sin² a + cos² a = x² - 2x sin b + sin² b + x² - 2x cos b + cos² b
sin² a + cos² a = 2x² - 2x(sin b + cos b) + sin² b + cos² b
1 = 2x² - 2x(sin b + cos b) + 1
(Recall the property sin² θ + cos² θ = 1)
2x² - 2x(sin b + cos b) = 0
Since x ≠ 0, we can divide throughout by 2x.
x - (sin b + cos b) = 0
x = sin b + cos b
sin a + sin b = x
sin a = x - sin b
Square both sides,
sin² a = (x - sin b)²
sin² a = x² - 2x sin b + sin² b ①
Given :
cos a + cos b = x
cos a = x - cos b
Square both sides,
cos² a = (x - cos b)²
cos² a = x² - 2x cos b + cos² b ②
① + ②
sin² a + cos² a = x² - 2x sin b + sin² b + x² - 2x cos b + cos² b
sin² a + cos² a = 2x² - 2x(sin b + cos b) + sin² b + cos² b
1 = 2x² - 2x(sin b + cos b) + 1
(Recall the property sin² θ + cos² θ = 1)
2x² - 2x(sin b + cos b) = 0
Since x ≠ 0, we can divide throughout by 2x.
x - (sin b + cos b) = 0
x = sin b + cos b
See 1 Answer
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