Theory question 110
Differentiation(a) The gradient of the tangent to the curve y = 4x3 at points P and Q is 108. Find the coordinates of P and Q.
(b) Given that  = 45°, B̂ = 30°, sin(A + B) = sin A cos B + sin B cos A and cos(A + B) = cos A cos B − sin A sin B
(i) show that: sin 15° = (√6 − √2)/4 and cos 15° = (√6 + √2)/4
(ii) hence find tan 15°
Show the model answer
(a) dy/dx = 12x² = 108, x² = 9, x = ±3
x = 3: y = 108; x = −3: y = −108
P(3, 108) and Q(−3, −108) (either order)
(b)(i) sin 15° = sin(45° − 30°) = sin 45° cos 30° − cos 45° sin 30° = (√2/2)(√3/2) − (√2/2)(½) = (√6 − √2)/4
cos 15° = cos 45° cos 30° + sin 45° sin 30° = (√6 + √2)/4
(ii) tan 15° = (√6 − √2)/(√6 + √2) = (√6 − √2)²/(6 − 2) = (8 − 4√3)/4 = 2 − √3
Set the derivative equal to 108; for 15°, use 45° − 30° (replace B by −B in the given formulas).