AP CALCULUS BC STUDY GUIDE
COMPLETE REVIEW WITH PRACTICE
QUESTIONS AND VERIFIED ANSWERS|
GRADED A+ | GUARANTEED SUCCESS
Updated 2026 Questions and Answers | 100% Verified
Exam Prep and Comprehensive Rationales Included
, sin(-θ) = -sin θ
cos(-θ) = cos θ
tan(-θ) = -tan θ
sin(A + B) = sinAcosB + sinBcosA
sin(A - B) = sinAcosB - sinBcosA
cos(A + B) = cosAcosB - sinAsinB
cos(A - B) = cosAcosB + sinAsinB
sin 2θ = 2sinθcosθ
cos 2θ = cos² θ - sin² θ
= 2cos² θ - 1
= 1 - 2sin² θ
tan θ = sin θ / cos θ
= 1 / cot θ
cot θ = cos θ / sin θ
= 1 / tan θ
sec θ = 1 / cos θ
csc θ = 1 / sin θ
cos² θ = ()(1 + cos 2θ)
sin² θ = ()(1 - cos 2θ)
d/dx (x^n) = nx^(n - 1)
d/dx (fg) = fg' + gf'
d/dx (f / g) = (gf' - fg') / g^2
d/dx f(g(x)) = f'(g(x))g'(x)
d/dx (sin x) = cos x
d/dx (cos x) = -sin x
d/dx (tan x) = sec² x
COMPLETE REVIEW WITH PRACTICE
QUESTIONS AND VERIFIED ANSWERS|
GRADED A+ | GUARANTEED SUCCESS
Updated 2026 Questions and Answers | 100% Verified
Exam Prep and Comprehensive Rationales Included
, sin(-θ) = -sin θ
cos(-θ) = cos θ
tan(-θ) = -tan θ
sin(A + B) = sinAcosB + sinBcosA
sin(A - B) = sinAcosB - sinBcosA
cos(A + B) = cosAcosB - sinAsinB
cos(A - B) = cosAcosB + sinAsinB
sin 2θ = 2sinθcosθ
cos 2θ = cos² θ - sin² θ
= 2cos² θ - 1
= 1 - 2sin² θ
tan θ = sin θ / cos θ
= 1 / cot θ
cot θ = cos θ / sin θ
= 1 / tan θ
sec θ = 1 / cos θ
csc θ = 1 / sin θ
cos² θ = ()(1 + cos 2θ)
sin² θ = ()(1 - cos 2θ)
d/dx (x^n) = nx^(n - 1)
d/dx (fg) = fg' + gf'
d/dx (f / g) = (gf' - fg') / g^2
d/dx f(g(x)) = f'(g(x))g'(x)
d/dx (sin x) = cos x
d/dx (cos x) = -sin x
d/dx (tan x) = sec² x
