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secondary 4 | A Maths
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trigonometry
Date Posted: 4 years ago
Views: 368
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done
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clear
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Here. Quite hard if we do not use this method.
Date Posted:
4 years ago
In their syllabus they're expected to know the a³ + b³ and a³ - b³ nowadays
sin^6 x + cos^6 x
= (sin²x)³ + (cos²x)³
= (1 - cos²x)³ + (cos²x)³
= 1³ - 3(1²)cos²x + 3(1)(cos²x)² - (cos²x)³ + (cos²x)³
(useful property to remember here is (a - b)³ = a³ - 3a²b + 3ab² - b³. You can use binomial theorem to get the same answer)
= 1 - 3cos²x + 3cos⁴x
= 1 - 3cos²x(1 - cos²x)
= 1 - 3cos²xsin²x
= 1 - ¾(4sin²xcos²x)
= 1 - ¾ (2sinxcosx)²
= 1 - ¾ sin² 2x
= (sin²x)³ + (cos²x)³
= (1 - cos²x)³ + (cos²x)³
= 1³ - 3(1²)cos²x + 3(1)(cos²x)² - (cos²x)³ + (cos²x)³
(useful property to remember here is (a - b)³ = a³ - 3a²b + 3ab² - b³. You can use binomial theorem to get the same answer)
= 1 - 3cos²x + 3cos⁴x
= 1 - 3cos²x(1 - cos²x)
= 1 - 3cos²xsin²x
= 1 - ¾(4sin²xcos²x)
= 1 - ¾ (2sinxcosx)²
= 1 - ¾ sin² 2x
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