A brief note on Special Angles and Trigonometric Identities . Read it before a test. Try each question at the end before you open its answer.
The special angles Half a square, sides 1 , 1 , 2 1, 1, \sqrt{2} 1 , 1 , 2 : sin 45 ∘ = cos 45 ∘ = 1 2 \sin 45^\circ = \cos 45^\circ = \dfrac{1}{\sqrt{2}} sin 4 5 ∘ = cos 4 5 ∘ = 2 1 and tan 45 ∘ = 1 \tan 45^\circ = 1 tan 4 5 ∘ = 1 . Half an equilateral triangle, sides 1 , 3 , 2 1, \sqrt{3}, 2 1 , 3 , 2 : sin 30 ∘ = cos 60 ∘ = 1 2 \sin 30^\circ = \cos 60^\circ = \dfrac{1}{2} sin 3 0 ∘ = cos 6 0 ∘ = 2 1 and sin 60 ∘ = cos 30 ∘ = 3 2 \sin 60^\circ = \cos 30^\circ = \dfrac{\sqrt{3}}{2} sin 6 0 ∘ = cos 3 0 ∘ = 2 3 . tan 30 ∘ = 1 3 \tan 30^\circ = \dfrac{1}{\sqrt{3}} tan 3 0 ∘ = 3 1 and tan 60 ∘ = 3 \tan 60^\circ = \sqrt{3} tan 6 0 ∘ = 3 . sin 0 ∘ = cos 90 ∘ = 0 \sin 0^\circ = \cos 90^\circ = 0 sin 0 ∘ = cos 9 0 ∘ = 0 and sin 90 ∘ = cos 0 ∘ = 1 \sin 90^\circ = \cos 0^\circ = 1 sin 9 0 ∘ = cos 0 ∘ = 1 . tan 90 ∘ \tan 90^\circ tan 9 0 ∘ is not defined, because cos 90 ∘ = 0 \cos 90^\circ = 0 cos 9 0 ∘ = 0 . The sines of 0 ∘ , 30 ∘ , 45 ∘ , 60 ∘ , 90 ∘ 0^\circ, 30^\circ, 45^\circ, 60^\circ, 90^\circ 0 ∘ , 3 0 ∘ , 4 5 ∘ , 6 0 ∘ , 9 0 ∘ are 0 2 , 1 2 , 2 2 , 3 2 , 4 2 \dfrac{\sqrt{0}}{2}, \dfrac{\sqrt{1}}{2}, \dfrac{\sqrt{2}}{2}, \dfrac{\sqrt{3}}{2}, \dfrac{\sqrt{4}}{2} 2 0 , 2 1 , 2 2 , 2 3 , 2 4 . The identities sin 2 A + cos 2 A = 1 \sin^2 A + \cos^2 A = 1 sin 2 A + cos 2 A = 1 , for every angle. 1 + tan 2 A = sec 2 A 1 + \tan^2 A = \sec^2 A 1 + tan 2 A = sec 2 A , for A < 90 ∘ A < 90^\circ A < 9 0 ∘ . 1 + cot 2 A = cosec 2 A 1 + \cot^2 A = \text{cosec}^2 A 1 + cot 2 A = cosec 2 A , for A > 0 ∘ A > 0^\circ A > 0 ∘ . For an acute angle, take the positive square root. sin ( 90 ∘ − A ) = cos A \sin(90^\circ - A) = \cos A sin ( 9 0 ∘ − A ) = cos A and tan ( 90 ∘ − A ) = cot A \tan(90^\circ - A) = \cot A tan ( 9 0 ∘ − A ) = cot A . Proving an identity Start with the harder side and work on it alone. Write everything in sin A \sin A sin A and cos A \cos A cos A when stuck. Look for 1 − sin 2 A 1 - \sin^2 A 1 − sin 2 A , sec 2 A − 1 \sec^2 A - 1 sec 2 A − 1 and cosec 2 A − 1 \text{cosec}^2 A - 1 cosec 2 A − 1 : each is a single term. One angle where a claim fails shows it is not an identity. A worked example Prove that cos A 1 − sin A = sec A + tan A \dfrac{\cos A}{1 - \sin A} = \sec A + \tan A 1 − sin A cos A = sec A + tan A .
Start with the right side. sec A + tan A = 1 cos A + sin A cos A = 1 + sin A cos A \sec A + \tan A = \dfrac{1}{\cos A} + \dfrac{\sin A}{\cos A} = \dfrac{1 + \sin A}{\cos A} sec A + tan A = cos A 1 + cos A sin A = cos A 1 + sin A . Multiply top and bottom by 1 − sin A 1 - \sin A 1 − sin A . ( 1 + sin A ) ( 1 − sin A ) cos A ( 1 − sin A ) = 1 − sin 2 A cos A ( 1 − sin A ) \dfrac{(1 + \sin A)(1 - \sin A)}{\cos A (1 - \sin A)} = \dfrac{1 - \sin^2 A}{\cos A(1 - \sin A)} cos A ( 1 − sin A ) ( 1 + sin A ) ( 1 − sin A ) = cos A ( 1 − sin A ) 1 − sin 2 A . Use the identity. cos 2 A cos A ( 1 − sin A ) = cos A 1 − sin A \dfrac{\cos^2 A}{\cos A(1 - \sin A)} = \dfrac{\cos A}{1 - \sin A} cos A ( 1 − sin A ) cos 2 A = 1 − sin A cos A , the left side. Practice with new numbers
A program to try The program prints the sine and cosine of 30 ∘ 30^\circ 3 0 ∘ , 45 ∘ 45^\circ 4 5 ∘ and 60 ∘ 60^\circ 6 0 ∘ , and checks the first identity for each.
Run it. Does the last column always show 1 1 1 ?
1. Add 15 ∘ 15^\circ 1 5 ∘ and 75 ∘ 75^\circ 7 5 ∘ to the list. Does the identity still hold?
Mistakes to avoid Swapping sin 30 ∘ \sin 30^\circ sin 3 0 ∘ and cos 30 ∘ \cos 30^\circ cos 3 0 ∘ . Writing tan 90 ∘ = 0 \tan 90^\circ = 0 tan 9 0 ∘ = 0 . Writing sin A + cos A = 1 \sin A + \cos A = 1 sin A + cos A = 1 . Only the squares add to 1 1 1 . Pairing tan \tan tan with cosec \text{cosec} cosec . It is tan \tan tan with sec \sec sec , and cot \cot cot with cosec \text{cosec} cosec . Check yourself Find sin 2 45 ∘ + tan 2 30 ∘ \sin^2 45^\circ + \tan^2 30^\circ sin 2 4 5 ∘ + tan 2 3 0 ∘ . If cot A = 7 24 \cot A = \dfrac{7}{24} cot A = 24 7 for an acute angle, find cosec A \text{cosec}\,A cosec A . Simplify tan 2 A cos 2 A + cos 2 A \tan^2 A \cos^2 A + \cos^2 A tan 2 A cos 2 A + cos 2 A . Is tan A + cot A = sec A cosec A \tan A + \cot A = \sec A\,\text{cosec}\,A tan A + cot A = sec A cosec A an identity? 
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