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secondary 4 | A Maths
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hello i need help idk How to do both parts!
Date Posted: 4 years ago
Views: 260
i)
cosA + sinA)² - (cosA - sinA)²
= cos²A + 2 cosAsinA + sin²A - (cos²A - 2 cosAsinA + sin²A)
= 4 cosAsinA
= 2 (2sinAcosA)
= 2 sin2A
Or
(cosA + sinA)² - (cosA - sinA)²
= [cosA + sinA + (cosA - sinA) ] [cosA + sinA - (cosA - sinA)]
(Using the property a² - b² = (a + b)(a - b) )
= (2cosA)(2sinA)
= 2(2 sinAcosA)
= 2 sin2A
cosA + sinA)² - (cosA - sinA)²
= cos²A + 2 cosAsinA + sin²A - (cos²A - 2 cosAsinA + sin²A)
= 4 cosAsinA
= 2 (2sinAcosA)
= 2 sin2A
Or
(cosA + sinA)² - (cosA - sinA)²
= [cosA + sinA + (cosA - sinA) ] [cosA + sinA - (cosA - sinA)]
(Using the property a² - b² = (a + b)(a - b) )
= (2cosA)(2sinA)
= 2(2 sinAcosA)
= 2 sin2A
ii)
(cosA + sinA)²/2 - (cosA - sinA)²/2 = -⅜
½ (2sin2A) = -⅜
sin 2A = -⅜
2A = sin-¹(-⅜)
If 0 ≤ A ≤ π,
Then 0 ≤ 2A ≤ 2π
Since sin 2A is negative, 2A is in the 3rd or 4th quadrant
Basic angle ≈ 0.3844 rad
2A ≈ (π + 0.3844) rad, (2π - 0.3844) rad
2A ≈ 3.526 rad, 5.898 rad
A ≈ 1.763 rad, 2.949 rad
A = 1.76 rad, 2.95 rad (3s.f)
(cosA + sinA)²/2 - (cosA - sinA)²/2 = -⅜
½ (2sin2A) = -⅜
sin 2A = -⅜
2A = sin-¹(-⅜)
If 0 ≤ A ≤ π,
Then 0 ≤ 2A ≤ 2π
Since sin 2A is negative, 2A is in the 3rd or 4th quadrant
Basic angle ≈ 0.3844 rad
2A ≈ (π + 0.3844) rad, (2π - 0.3844) rad
2A ≈ 3.526 rad, 5.898 rad
A ≈ 1.763 rad, 2.949 rad
A = 1.76 rad, 2.95 rad (3s.f)
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Date Posted:
4 years ago
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Hope this helps
Date Posted:
4 years ago
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