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secondary 4 | A Maths
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I need help with 9b. Thank you.
Date Posted: 3 years ago
Views: 325
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y = cos⁴ 2x
dy/dx = 4 cos³ 2x (-2sin 2x)
dy/dx = -8 cos³ 2x sin 2x
When x = π/8, dy/dx = -8 cos³ (2π/8) sin (2π/8)
= -8 cos³ (π/4) sin (π/4)
= -8 (√2 / 2)³ (√2 / 2)
= -8 (√2 / 2)⁴
= -8 (4/16)
= -8 (¼)
= -2
When x = -π/8, y = cos⁴ (2π/8)
= cos⁴ (π/4)
= (√2 / 2)⁴
= ¼
Equation of tangent : y = mx + c , where m = dy/dx at the respective value of x and c is the y-intercept.
Sub x = π/8, y = ¼, dy/dx = -2,
¼ = -2(π/8) + c
c = ¼ + π/4
So equation is : y = -2x + ¼ + π/4
dy/dx = 4 cos³ 2x (-2sin 2x)
dy/dx = -8 cos³ 2x sin 2x
When x = π/8, dy/dx = -8 cos³ (2π/8) sin (2π/8)
= -8 cos³ (π/4) sin (π/4)
= -8 (√2 / 2)³ (√2 / 2)
= -8 (√2 / 2)⁴
= -8 (4/16)
= -8 (¼)
= -2
When x = -π/8, y = cos⁴ (2π/8)
= cos⁴ (π/4)
= (√2 / 2)⁴
= ¼
Equation of tangent : y = mx + c , where m = dy/dx at the respective value of x and c is the y-intercept.
Sub x = π/8, y = ¼, dy/dx = -2,
¼ = -2(π/8) + c
c = ¼ + π/4
So equation is : y = -2x + ¼ + π/4
Your answer key has an error by the way. It should be ¼ and not ½
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