Chapter Summary: Introduction to Trigonometry
A one-page recap of every key idea and formula in this chapter. Read this the night before the exam.
1. The Six Ratios
For an acute angle θ \theta θ sin θ = Opp Hyp , cos θ = Adj Hyp , tan θ = Opp Adj \sin\theta = \tfrac{\text{Opp}}{\text{Hyp}}, \; \cos\theta = \tfrac{\text{Adj}}{\text{Hyp}}, \; \tan\theta = \tfrac{\text{Opp}}{\text{Adj}} sin θ = Hyp Opp , cos θ = Hyp Adj , tan θ = Adj Opp
Reciprocals: csc θ = 1 sin θ \csc\theta = \tfrac{1}{\sin\theta} csc θ = s i n θ 1 sec θ = 1 cos θ \sec\theta = \tfrac{1}{\cos\theta} sec θ = c o s θ 1 cot θ = 1 tan θ \cot\theta = \tfrac{1}{\tan\theta} cot θ = t a n θ 1
Quotients: tan θ = sin θ cos θ \tan\theta = \tfrac{\sin\theta}{\cos\theta} tan θ = c o s θ s i n θ cot θ = cos θ sin θ \cot\theta = \tfrac{\cos\theta}{\sin\theta} cot θ = s i n θ c o s θ
Mnemonic: SOH-CAH-TOA.
2. Standard Values
0°
30°
45°
60°
90°
sin \sin sin 0
1 2 \tfrac12 2 1 1 2 \tfrac{1}{\sqrt2} 2 1 3 2 \tfrac{\sqrt3}{2} 2 3 1
cos \cos cos 1
3 2 \tfrac{\sqrt3}{2} 2 3 1 2 \tfrac{1}{\sqrt2} 2 1 1 2 \tfrac12 2 1 0
tan \tan tan 0
1 3 \tfrac{1}{\sqrt3} 3 1 1
3 \sqrt3 3 n.d.
3. The Three Identities
sin 2 θ + cos 2 θ = 1 \sin^2\theta + \cos^2\theta = 1 sin 2 θ + cos 2 θ = 1 1 + tan 2 θ = sec 2 θ 1 + \tan^2\theta = \sec^2\theta 1 + tan 2 θ = sec 2 θ 1 + cot 2 θ = csc 2 θ 1 + \cot^2\theta = \csc^2\theta 1 + cot 2 θ = csc 2 θ
Rearrangements: sec 2 θ − tan 2 θ = 1 \sec^2\theta - \tan^2\theta = 1 sec 2 θ − tan 2 θ = 1 csc 2 θ − cot 2 θ = 1 \csc^2\theta - \cot^2\theta = 1 csc 2 θ − cot 2 θ = 1 1 − cos 2 θ = sin 2 θ 1 - \cos^2\theta = \sin^2\theta 1 − cos 2 θ = sin 2 θ
Conjugate fact: 1 sec θ − tan θ = sec θ + tan θ \dfrac{1}{\sec\theta - \tan\theta} = \sec\theta + \tan\theta sec θ − tan θ 1 = sec θ + tan θ
4. Complementary Angles
sin ( 90 ° − θ ) = cos θ , tan ( 90 ° − θ ) = cot θ , sec ( 90 ° − θ ) = csc θ \sin(90° - \theta) = \cos\theta, \; \tan(90° - \theta) = \cot\theta, \; \sec(90° - \theta) = \csc\theta sin ( 90° − θ ) = cos θ , tan ( 90° − θ ) = cot θ , sec ( 90° − θ ) = csc θ
If sin A = cos B \sin A = \cos B sin A = cos B A + B = 90 ° A + B = 90° A + B = 90°
'Co' = complement: cosine, cotangent, cosecant.
Proof strategy
Start from the more complicated side; convert everything to sin θ \sin\theta sin θ cos θ \cos\theta cos θ
Use sin 2 θ + cos 2 θ = 1 \sin^2\theta + \cos^2\theta = 1 sin 2 θ + cos 2 θ = 1
For surds, use conjugate multiplication.
Common mistakes to avoid
sin 2 θ \sin^2\theta sin 2 θ ( sin θ ) 2 (\sin\theta)^2 ( sin θ ) 2 sin ( θ 2 ) \sin(\theta^2) sin ( θ 2 ) ( 3 2 ) 2 = 3 4 (\tfrac{\sqrt3}{2})^2 = \tfrac34 ( 2 3 ) 2 = 4 3 3 4 \tfrac{\sqrt3}{4} 4 3 tan 90 ° \tan 90° tan 90° sec 90 ° \sec 90° sec 90° csc 0 ° \csc 0° csc 0° undefined .For acute angles, sin θ \sin\theta sin θ cos θ \cos\theta cos θ
Final tip: Memorise the values table and the three identities cold; then most questions are a matter of clean substitution and conversion to sines and cosines.
