Compound Angle Formulas (sin, cos) #1 sin ( A ± B ) = sin A cos B ± cos A sin B , cos ( A ± B ) = cos A cos B ∓ sin A sin B \sin(A \pm B) = \sin A \cos B \pm \cos A \sin B, \quad \cos(A \pm B) = \cos A \cos B \mp \sin A \sin B sin ( A ± B ) = sin A cos B ± cos A sin B , cos ( A ± B ) = cos A cos B ∓ sin A sin B 💡 For cos, the sign flips: cos(A+B) has minus, cos(A-B) has plus.
⚠ Using the same sign for both sin and cos compound angle formulas. TRAP: Students write sin(A+B) = sinA + sinB. This is WRONG. The correct expansion is sinA cosB + cosA sinB.
Compound Angle Formula (tan) #2 tan ( A ± B ) = tan A ± tan B 1 ∓ tan A tan B \tan(A \pm B) = \frac{\tan A \pm \tan B}{1 \mp \tan A \tan B} tan ( A ± B ) = 1 ∓ tan A tan B tan A ± tan B 💡 Undefined when denominator is zero, i.e., tan A tan B = 1 for A+B case.
⚠ Getting the sign wrong in the denominator. For tan(A+B), denominator has minus.
Double Angle Formulas #3 sin 2 A = 2 sin A cos A , cos 2 A = cos 2 A − sin 2 A = 2 cos 2 A − 1 = 1 − 2 sin 2 A , tan 2 A = 2 tan A 1 − tan 2 A \sin 2A = 2\sin A \cos A, \quad \cos 2A = \cos^2 A - \sin^2 A = 2\cos^2 A - 1 = 1 - 2\sin^2 A, \quad \tan 2A = \frac{2\tan A}{1 - \tan^2 A} sin 2 A = 2 sin A cos A , cos 2 A = cos 2 A − sin 2 A = 2 cos 2 A − 1 = 1 − 2 sin 2 A , tan 2 A = 1 − tan 2 A 2 tan A 💡 cos 2A has three equivalent forms. Pick the one that matches the unknowns in the problem.
⚠ Using sin 2A = sin A cos A (forgetting the factor of 2). TRAP: Students confuse cos 2A = cos^2A - sin^2A with cos 2A = cos^2A + sin^2A (which is just 1). Pick the form matching your unknowns.
Half Angle Formulas #4 sin A 2 = ± 1 − cos A 2 , cos A 2 = ± 1 + cos A 2 , tan A 2 = sin A 1 + cos A = 1 − cos A sin A \sin\frac{A}{2} = \pm\sqrt{\frac{1 - \cos A}{2}}, \quad \cos\frac{A}{2} = \pm\sqrt{\frac{1 + \cos A}{2}}, \quad \tan\frac{A}{2} = \frac{\sin A}{1 + \cos A} = \frac{1 - \cos A}{\sin A} sin 2 A = ± 2 1 − cos A , cos 2 A = ± 2 1 + cos A , tan 2 A = 1 + cos A sin A = sin A 1 − cos A 💡 The sign of the square root depends on the quadrant of A/2, not A. Derived from cos 2A = 1 - 2sin^2A (rearrange for sin A/2) and cos 2A = 2cos^2A - 1 (rearrange for cos A/2).
⚠ Dropping the absolute value or choosing the wrong sign for the square root form.
Product-to-Sum Formulas #5 2 sin A cos B = sin ( A + B ) + sin ( A − B ) , 2 cos A cos B = cos ( A − B ) + cos ( A + B ) , 2 sin A sin B = cos ( A − B ) − cos ( A + B ) 2\sin A \cos B = \sin(A+B) + \sin(A-B), \quad 2\cos A \cos B = \cos(A-B) + \cos(A+B), \quad 2\sin A \sin B = \cos(A-B) - \cos(A+B) 2 sin A cos B = sin ( A + B ) + sin ( A − B ) , 2 cos A cos B = cos ( A − B ) + cos ( A + B ) , 2 sin A sin B = cos ( A − B ) − cos ( A + B ) 💡 Useful for integrating products of trig functions and simplifying series. Derived by adding/subtracting compound angle formulas.
⚠ Confusing the signs: for 2 sin A sin B, it is cos(A-B) MINUS cos(A+B).
Sum-to-Product Formulas #6 sin C + sin D = 2 sin C + D 2 cos C − D 2 , cos C + cos D = 2 cos C + D 2 cos C − D 2 \sin C + \sin D = 2\sin\frac{C+D}{2}\cos\frac{C-D}{2}, \quad \cos C + \cos D = 2\cos\frac{C+D}{2}\cos\frac{C-D}{2} sin C + sin D = 2 sin 2 C + D cos 2 C − D , cos C + cos D = 2 cos 2 C + D cos 2 C − D 💡 For sin C - sin D, it becomes 2 cos((C+D)/2) sin((C-D)/2). For cos C - cos D, it becomes -2 sin((C+D)/2) sin((C-D)/2).
⚠ Mixing up product-to-sum and sum-to-product formulas, or getting the difference versions wrong.
General Solutions of Trigonometric Equations #7 sin x = sin α ⇒ x = n π + ( − 1 ) n α , cos x = cos α ⇒ x = 2 n π ± α , tan x = tan α ⇒ x = n π + α ( n ∈ Z ) \sin x = \sin\alpha \Rightarrow x = n\pi + (-1)^n \alpha, \quad \cos x = \cos\alpha \Rightarrow x = 2n\pi \pm \alpha, \quad \tan x = \tan\alpha \Rightarrow x = n\pi + \alpha \quad (n \in \mathbb{Z}) sin x = sin α ⇒ x = nπ + ( − 1 ) n α , cos x = cos α ⇒ x = 2 nπ ± α , tan x = tan α ⇒ x = nπ + α ( n ∈ Z ) 💡 Always express the general solution. For specific intervals, substitute integer values of n.
⚠ Writing sin x = sin alpha as x = n*pi + alpha (missing the (-1)^n factor).
Sine Rule #8 a sin A = b sin B = c sin C = 2 R \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R sin A a = sin B b = sin C c = 2 R 💡 R is the circumradius. Use when you know an angle and its opposite side, or need to find the circumradius.
⚠ Forgetting the 2R part, which is essential for circumradius problems.
Cosine Rule #9 cos A = b 2 + c 2 − a 2 2 b c , a 2 = b 2 + c 2 − 2 b c cos A \cos A = \frac{b^2 + c^2 - a^2}{2bc}, \quad a^2 = b^2 + c^2 - 2bc\cos A cos A = 2 b c b 2 + c 2 − a 2 , a 2 = b 2 + c 2 − 2 b c cos A 💡 Use when you know all three sides (SSS), or two sides and included angle (SAS).
⚠ Writing a^2 = b^2 + c^2 + 2bc cos A (wrong sign before the 2bc cos A term).
Area of Triangle Using Trig #10 Δ = 1 2 a b sin C = 1 2 b c sin A = 1 2 c a sin B \Delta = \frac{1}{2}ab\sin C = \frac{1}{2}bc\sin A = \frac{1}{2}ca\sin B Δ = 2 1 ab sin C = 2 1 b c sin A = 2 1 c a sin B 💡 Also equals abc/(4R) using the sine rule, and rs where r is the inradius and s is the semi-perimeter.
⚠ Using the wrong pair of sides for the included angle. The angle must be between the two sides used.