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PRECALCULUS B TEST #5 – TRIGONOMETRIC IDENTITIES, PRACTICE SECTION 5.1 – Fundamental Identities 5 and θ is in Quadrant IV, find the remaining trigonometric functions of θ. 13 1) If cos θ = 2) If tan θ = − 3) Simplify sec θ ⋅ cot θ ⋅ csc θ . Write your answer so it does not contain a fraction. 4) Simplify 5) Simplify ( sec θ + 1)( sec θ − 1) . 3 and θ is in Quadrant II, find the remaining trigonometric functions of θ. 4 1 − sin 2 θ . Write your answer so it does not contain a fraction. 1 + cot 2 θ SECTION 5.2 – Verifying Trigonometric Identities 6) Simplify sec x csc x . + csc x sec x 7) Simplify cos α ( sec α + csc α ) . 8) Simplify cos θ sin θ + . sin θ 1 + cos θ 9) Factor cos 2 θ − 1. 10) Factor 3sin 2θ + 4sin θ + 1. 11) Factor sin 4 x − cos 4 x. 12) Verify that tan θ = sin θ . sec θ 13) Verify that sin 2 β (1 + cot 2 β ) = 1. 14) Verify that sin 2 θ = sec θ − cos θ . cos θ 15) Verify that 16) Verify that 1 − cos a 2 = ( cot a − csc a ) . 1 + cos a 17) Verify that sin x − sin 3 x = sin x ⋅ cos 2 x. 1 + tan 2 θ = tan 2 θ . 2 1 + cot θ SECTION 5.3 – Sum and Difference Identities for Cosine 18) Find the exact value (NO DECIMALS) of cos 195°. HINT: 195° = 225° – 30° 19) Find the exact value (NO DECIMALS) of cos (–75°). HINT: –75° = (–30°) + (–45°) 20) Write sin 15° in terms of its cofunction. 21) Write cot 22) Find an angle θ such that sin θ = cos ( 2θ − 30° ) . 23) Find an angle θ such that cot (θ − 10° ) = tan ( 2θ + 20° ) . 24) sin x = 25) cos s = − 9π in terms of its cofunction. 10 2 1 and sin y = − , x in Quadrant II and y in Quadrant IV. 3 3 Find cos ( x + y ) and cos ( x − y ) . 8 3 and sin t = − , s and t in Quadrant III. Find cos ( s + t ) and cos ( s − t ) . 17 5 SECTION 5.4 – Sum and Difference Identities for Sine and Tangent 26) Find the exact value (NO DECIMALS) of sin 105°. HINT: 105° = 60° + 45° 27) Find the exact value (NO DECIMALS) of tan 285°. HINT: 285° = 330° – 45° 28) cos x = − 29) Verify that sin ( x + y ) + sin ( x − y ) = 2sin x ⋅ cos y. 30) Verify that 5 3 and sin y = , x and y in Quadrant II. Find sin ( x + y ) and sin ( x − y ) . 13 5 sin ( p + q ) = tan p + tan q. cos p ⋅ cos q SECTION 5.5 – Double-Angle Identities 31) cos 2 22.5° − sin 2 22.5° is equivalent to what single trigonometric function? 32) ⎛π ⎞ ⎛π ⎞ 2sin ⎜ ⎟ ⋅ cos ⎜ ⎟ is equivalent to what single trigonometric function? ⎝ 12 ⎠ ⎝ 12 ⎠ 33) 2 tan 75° is equivalent to what single trigonometric function? 1 − tan 2 75° 34) Verify that tan A ⋅ sin 2 A = 2sin 2 A. 35) Verify that 1 + cos 2 x = cot x. sin 2 x SECTION 5.6 – Half-Angle Identities 36) Find the exact value (NO DECIMALS) of sin 22.5°. 37) Find the exact value (NO DECIMALS) of cos 112.5°. 38) ⎛ 7π Find the exact value (NO DECIMALS) of tan ⎜ ⎝ 8 39) 40) 41) ⎞ ⎟. ⎠ 1 − cos 35° is equivalent to what single trigonometric function? 2 sin 50° is equivalent to what single trigonometric function? 1 + cos 50° 1 + cos 24° is equivalent to what single trigonometric function? 2 *******************ANSWERS********************* 2 1) 25 144 ⎛5⎞ Use sin 2 θ + cos 2 θ = 1 to determine sin θ → sin 2 θ + ⎜ ⎟ = 1 → sin 2 θ + = 1 → sin 2 θ = → 169 169 ⎝ 13 ⎠ 144 12 12 because θ is in Quadrant IV → sin θ = ± → sin θ = − 169 13 13 12 − sin θ 12 = → tan θ = 13 → Multiply top & bottom by 13 → tan θ = − 5 cos θ 5 13 13 equals reciprocal of sin θ → csc θ = − 12 13 equals reciprocal of cos θ → sec θ = 5 5 equals reciprocal of tan θ → cot θ = − 12 sin 2 θ = tan θ csc θ sec θ cot θ 2 2) 9 25 ⎛ 3⎞ Use tan 2 θ + 1 = sec 2 θ to determine sec θ → ⎜ − ⎟ + 1 = sec 2 θ → + 1 = sec 2 θ → = sec 2 θ → 4 16 16 ⎝ ⎠ 25 5 5 = sec 2 θ → ± = sec θ → sec θ = − because θ is in Quadrant II 16 4 4 4 cot θ equals reciprocal of tan θ → cot θ = − 3 4 cos θ equals reciprocal of secθ → cos θ = − 5 2 16 9 ⎛ 4⎞ Use sin 2 θ + cos 2 θ = 1 to determine sin θ → sin 2 θ + ⎜ − ⎟ = 1 → sin 2 θ + = 1 → sin 2 θ = → 25 25 ⎝ 5⎠ 9 3 3 → sin θ = ± → sin θ = because θ is in Quadrant II 25 5 5 5 csc θ equals reciprocal of sin θ → csc θ = 3 sin 2 θ = 3) 4) sec θ ⋅ cot θ ⋅ csc θ = 1 cos θ 1 1 1 1 with csc θ → ⋅ ⋅ = → Replace = csc 2 θ cos θ sin θ sin θ sin 2 θ sin θ sin 2 θ 1 − sin 2 θ cos 2 θ → Replace 1 − sin 2 θ with cos 2 θ & 1 + cot 2 θ with csc 2 θ → → 2 1 + cot θ csc 2 θ 1 cos 2 θ Replace csc2 θ with → → Multiply numerator by reciprocal of denominator → 1 sin 2 θ sin 2 θ 2 2 cos θ sin θ ⋅ = cos 2 θ ⋅ sin 2 θ 1 1 5) ( sec θ + 1)( secθ − 1) = sec2 θ − secθ + sec θ − 1 = sec2 θ − 1 → Transform tan 2 θ + 1 = sec2 θ to tan 2 θ = sec2 θ − 1 → Replace sec 2 θ − 1 with tan 2 θ → sec 2 θ − 1 = tan 2 θ 6) 7) 8) 1 1 sec x csc x 1 1 & csc x with + → Replace sec x with → cos x + sin x 1 1 csc x sec x cos x sin x sin x cos x 1 sin x 1 cos x sin x cos x ⋅ + ⋅ = + → Multiply each numerator by reciprocal of denominator → cos x 1 sin x 1 cos x sin x sin x ⋅ sin x cos x ⋅ cos x sin 2 x + cos 2 x Put over a common denominator → + = → cos x ⋅ sin x cos x ⋅ sin x cos x ⋅ sin x 1 1 1 Replace sin 2 x + cos 2 x with 1 → → Write as separate fractions → ⋅ → cos x ⋅ sin x cos x sin x 1 1 1 1 ⋅ = sec x ⋅ csc x Replace with sec x and with csc x → cos x sin x cos x sin x 1 1 1 ⎞ ⎛ 1 and csc α with → cos α ⎜ + ⎟→ cos α sin α ⎝ cos α sin α ⎠ cos α cos α cos α 1+ with cot α → 1 + → Replace = 1 + cot α sin α sin α sin α cos α ( sec α + csc α ) → Replace sec α with cos θ ⋅ (1 + cos θ ) cos θ sin θ sin θ ⋅ sin θ + → Put over a common denominator → + → sin θ 1 + cos θ sin θ ⋅ (1 + cos θ ) sin θ ⋅ (1 + cos θ ) cos θ + cos 2 θ sin 2 θ cos θ + cos 2 θ + sin 2 θ + → Combine into single fraction → → sin θ ⋅ (1 + cos θ ) sin θ ⋅ (1 + cos θ ) sin θ ⋅ (1 + cos θ ) Replace cos 2 θ + sin 2 θ with 1 → Reduce → 9) cos θ + 1 1 + cos θ → Rearrange numerator → → sin θ ⋅ (1 + cos θ ) sin θ ⋅ (1 + cos θ ) 1 + cos θ 1 1 1 = → Replace with csc θ → = csc θ sin θ ⋅ (1 + cos θ ) sin θ sin θ sin θ cos 2 θ − 1 → Replace cos θ with x → x 2 − 1 → Difference of Squares → x 2 − 1 = ( x + 1)( x − 1) → Replace x with cos θ → ( cos θ + 1)( cos θ − 1) 10) 3sin 2 θ + 4sin θ + 1 → Replace sin θ with x → 3 x 2 + 4 x + 1 → Un-FOIL → 3 x 2 + 4 x + 1 = ( 3x + 1)( x + 1) Replace x with sin θ → ( 3sin θ + 1)( sin θ + 1) 11) sin 4 x − cos 4 x = ( sin 2 x + cos 2 x )( sin 2 x − cos 2 x ) = ( sin 2 x + cos 2 x ) ( sin x + cos x )( sin x − cos x ) → Replace sin 2 x + cos 2 x with 1 → 1⋅ ( sin x + cos x )( sin x − cos x ) = ( sin x + cos x )( sin x − cos x ) 12) sin θ tan θ sin θ 1 θ → cos → Replace tan θ with → and sec θ with 1 sec θ cos θ cos θ cos θ sin θ cos θ sin θ cos θ Multiply numerator by reciprocal of denominator → ⋅ → Reduce → ⋅ = sin θ cos θ 1 cos θ 1 13) 14) 15) 16) ⎛ cos 2 β cos 2 β 2 sin β → ⎜1 + 2 sin 2 β ⎝ sin β ⎞ ⎟ → Use Distributive property → ⎠ sin 2 β + cos 2 β → Replace sin 2 β + cos 2 β with 1 → sin 2 β + cos 2 β = 1 sin 2 β (1 + cot 2 β ) → Replace cot 2 β with sin 2 θ 1 − cos 2 θ → Replace sin 2 θ with 1 − cos 2 θ → → Write as two separate fractions → cos θ cos θ 1 cos 2 θ 1 1 with secθ → − → Simplify 2nd fraction → − cos θ → Replace cos θ cos θ cos θ cos θ sin 2 θ sec θ − cos θ → = sec θ − cos θ cos θ 1 + tan 2 θ sec2 θ → Replace 1 + tan 2 θ with sec 2 θ & replace 1 + cot 2 θ with csc 2 θ → → 2 1 + cot θ csc 2 θ 1 2 1 1 2 2 θ → cos → Change sec θ to & csc θ to 2 2 1 cos θ sin θ sin 2 θ 1 sin 2 θ sin 2 θ Multiply numerator by reciprocal of denominator → ⋅ = → 2 1 cos θ cos 2 θ sin θ sin 2 θ 1 + tan 2 θ 2 = tan θ → = tan → = tan 2 θ θ cos θ cos 2 θ 1 + cot 2 θ 1 − cos a → Multiply numerator and denominator by conjugate of denominator → 1 + cos a (1 − cos a ) ⋅ (1 − cos a ) 1 − cos a − cos a + cos2 a 1 − 2 cos a + cos 2 a → = → 1 − cos 2 a (1 + cos a ) ⋅ (1 − cos a ) 1 − cos a + cos a − cos2 a Replace 1 − cos 2 a with sin 2 a → 1 − 2 cos a + cos 2 a 1 − 2 cos a + cos 2 a = 1 − cos 2 a sin 2 a 2 cos a 1 1 ⎞ ⎛ cos a and csc a with →⎜ − ⎟ → sin a sin a sin sin a a⎠ ⎝ 1 ⎞ ⎛ cos a 1 ⎞ cos 2 a cos a cos a 1 ⎛ cos a Multiply with FOIL → ⎜ − ⋅ − − 2 − 2 + 2 → ⎟ ⎜ ⎟= 2 ⎝ sin a sin a ⎠ ⎝ sin a sin a ⎠ sin a sin a sin a sin a ( cot a − csc a ) → Replace cot a with 2 Combine into single fraction → cos 2 a − cos a − cos a + 1 → Combine like terms → sin 2 a cos 2 a − 2 cos a + 1 → Left-hand side now equals right-hand side sin 2 a 17) sin x − sin 3 x → Factor out sin x → sin x (1 − sin 2 x ) → Replace 1 − sin 2 x with cos 2 x → sin x (1 − sin 2 x ) = sin x ⋅ cos 2 x 18) 19) ⎛ 2⎞ ⎛ 3⎞ ⎛ 2 ⎞ ⎛1⎞ cos195° = cos ( 225° − 30° ) = cos 225° ⋅ cos 30° + sin 225° ⋅ sin 30° = ⎜⎜ − ⎟ ⋅ ⎜⎜ ⎟ + ⎜⎜ − ⎟ ⎟ ⎟⎟ ⋅ ⎜ ⎟ → ⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎝2⎠ ⎛ ⎛ 6⎞ ⎛ 2⎞ 6⎞ ⎛ 2⎞ − 6− 2 ⎜⎜ − ⎟⎟ + ⎜⎜ − ⎟⎟ → Combine into single fraction → ⎜⎜ − ⎟⎟ + ⎜⎜ − ⎟⎟ = 4 4 4 4 4 ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ cos ( −75° ) = cos ( −45° + −30° ) = cos ( −45° ) ⋅ cos ( −30° ) − sin ( −45° ) ⋅ sin ( −30° ) → ⎛ 2⎞ ⎛ 3⎞ ⎛ 2 ⎞ ⎛ 1⎞ 6 2 6− 2 − → Combine into single fraction → ⎜⎜ ⎟⎟ ⋅ ⎜⎜ ⎟⎟ − ⎜⎜ − ⎟⎟ ⋅ ⎜ − ⎟ = 4 4 ⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2⎠ 4 20) 21) 22) sin θ = cos ( 90° − θ ) → sin15° = cos ( 90° − 15° ) = cos 75° ⎛π ⎞ ⎛ 9π ⎞ ⎛ π 9π ⎞ cot θ = tan ⎜ − θ ⎟ when θ is expressed in radians → cot ⎜ ⎟ = tan ⎜ − ⎟→ ⎝2 ⎠ ⎝ 10 ⎠ ⎝ 2 10 ⎠ ⎛ 5π 9π ⎞ ⎛ 4π ⎞ ⎛ 2π ⎞ Put over common denominator → tan ⎜ − ⎟ = tan ⎜ − ⎟ = tan ⎜ − ⎟ ⎝ 10 10 ⎠ ⎝ 10 ⎠ ⎝ 5 ⎠ sin θ = cos ( 90° − θ ) → Replace sin θ with cos ( 90° − θ ) → sin θ = cos ( 2θ − 30° ) → cos ( 90° − θ ) = cos ( 2θ − 30° ) → 90° − θ = 2θ − 30° → 120° = 3θ → 40° = θ 23) cot x = tan ( 90° − x ) → Replace cot (θ − 10° ) with tan ⎡⎣90° − (θ − 10° ) ⎤⎦ → cot (θ − 10° ) = tan ( 2θ + 20° ) → tan ⎡⎣90° − (θ − 10° ) ⎤⎦ = tan ( 2θ + 20° ) → 90° − (θ − 10° ) = 2θ + 20° → 100° − θ = 2θ + 20° → 80 = 3θ → 80° 2 = θ OR 26 ° = θ 3 3 2 24) 4 5 ⎛2⎞ Use sin 2 x + cos 2 x = 1 to determine cos x → ⎜ ⎟ + cos 2 x = 1 → + cos 2 x = 1 → cos 2 x = → 3 9 9 ⎝ ⎠ cos 2 x = 5 5 5 since x is in Quadrant II → cos x = ± → cos x = − 9 3 3 2 1 8 ⎛ 1⎞ Use sin y + cos y = 1 to determine cos y → ⎜ − ⎟ + cos 2 y = 1 → + cos 2 y = 1 → cos 2 y = → 3 9 9 ⎝ ⎠ 2 cos 2 y = 2 8 8 2 2 2 2 since y is in Quadrant IV → cos y = ± =± → cos y = 9 3 3 3 ⎛ 5 ⎞ ⎛ 2 2 ⎞ ⎛ 2 ⎞ ⎛ 1 ⎞ ⎛ 2 10 ⎞ ⎛ 2 ⎞ cos ( x + y ) = cos x ⋅ cos y − sin x ⋅ sin y = ⎜⎜ − ⎟⎟ ⋅ ⎜⎜ ⎟⎟ − ⎜ ⎟ ⋅ ⎜ − ⎟ = ⎜⎜ − ⎟−⎜− ⎟ → 9 ⎟⎠ ⎝ 9 ⎠ ⎝ 3 ⎠ ⎝ 3 ⎠ ⎝ 3⎠ ⎝ 3⎠ ⎝ ⎛ 2 10 ⎞ ⎛ 2 ⎞ −2 10 + 2 ⎜⎜ − ⎟⎟ + ⎜ ⎟ → Combine into single fraction → cos ( x + y ) = 9 ⎠ ⎝9⎠ 9 ⎝ ⎛ 5 ⎞ ⎛ 2 2 ⎞ ⎛ 2 ⎞ ⎛ 1 ⎞ ⎛ 2 10 ⎞ ⎛ 2 ⎞ cos ( x − y ) = cos x ⋅ cos y + sin x ⋅ sin y = ⎜⎜ − ⎟⎟ ⋅ ⎜⎜ ⎟⎟ + ⎜ ⎟ ⋅ ⎜ − ⎟ = ⎜⎜ − ⎟+⎜− ⎟ → 9 ⎟⎠ ⎝ 9 ⎠ ⎝ 3 ⎠ ⎝ 3 ⎠ ⎝ 3⎠ ⎝ 3⎠ ⎝ Combine into single fraction → −2 10 − 2 −2 10 − 2 → cos ( x − y ) = 9 9 2 25) 64 225 ⎛ 8⎞ Use sin 2 s + cos 2 s = 1 to determine sin s → sin 2 s + ⎜ − ⎟ = 1 → sin 2 s + = 1 → sin 2 s = → 17 289 289 ⎝ ⎠ sin 2 s = 225 15 15 since s is in Quadrant III → sin s = ± → sin s = − 289 17 17 2 9 16 ⎛ 3⎞ Use sin t + cos t = 1 to determine cos t → ⎜ − ⎟ + cos 2 t = 1 → + cos 2 t = 1 → cos 2 t = → 5 25 25 ⎝ ⎠ 2 2 16 4 4 → cos t = ± → cos t = − since t is in Quadrant III 25 5 5 13 ⎛ 8 ⎞ ⎛ 4 ⎞ ⎛ 15 ⎞ ⎛ 3 ⎞ ⎛ 32 ⎞ ⎛ 45 ⎞ cos ( s + t ) = cos s ⋅ cos t − sin s ⋅ sin t = ⎜ − ⎟ ⋅ ⎜ − ⎟ − ⎜ − ⎟ ⋅ ⎜ − ⎟ = ⎜ ⎟ − ⎜ ⎟ = − 17 5 17 5 85 85 85 ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎛ 8 ⎞ ⎛ 4 ⎞ ⎛ 15 ⎞ ⎛ 3 ⎞ ⎛ 32 ⎞ ⎛ 45 ⎞ 77 cos ( s − t ) = cos s ⋅ cos t + sin s ⋅ sin t = ⎜ − ⎟ ⋅ ⎜ − ⎟ − ⎜ − ⎟ ⋅ ⎜ − ⎟ = ⎜ ⎟ + ⎜ ⎟ = ⎝ 17 ⎠ ⎝ 5 ⎠ ⎝ 17 ⎠ ⎝ 5 ⎠ ⎝ 85 ⎠ ⎝ 85 ⎠ 85 cos 2 t = 26) sin105° = sin ( 60° + 45° ) = sin 60° ⋅ cos 45° + cos 60° ⋅ sin 45° = Combine into single fraction → 27) 3 2 1 2 6 2 ⋅ + ⋅ = + → 2 2 2 2 4 4 6+ 2 4 tan 330° − tan 45° = tan 285° = tan ( 330° − 45° ) = 1 + tan 330° ⋅ tan 45° 3 3 −1 − −1 3 3 = → ⎛ 3 3⎞ 1− 1 + ⎜⎜ − ⎟⎟ ⋅1 3 ⎝ 3 ⎠ − Multiply each term in numerator and denominator by 3 → − 3 −3 3− 3 → ( ) Multiply numerator and denominator by conjugate of denominator 3 + 3 → −3 3 − 3 − 9 − 3 3 9+3 3 −3 3 −3 = − 3 −3 3+ 3 ⋅ → 3− 3 3+ 3 −12 − 6 3 −12 − 6 3 → Reduce → = −2 − 3 6 6 2 28) 25 144 ⎛ 5⎞ Use sin 2 x + cos 2 x = 1 to determine sin x → sin 2 x + ⎜ − ⎟ = 1 → sin 2 x + = 1 → sin 2 x = → 169 169 ⎝ 13 ⎠ sin 2 x = 144 12 12 since x is in Quadrant II → sin x = ± → sin x = 169 13 13 2 9 16 ⎛3⎞ Use sin 2 y + cos 2 y = 1 to determine cos y → ⎜ ⎟ + cos 2 y = 1 → + cos 2 y = 1 → cos 2 y = → 25 25 ⎝5⎠ 16 4 4 → cos y = ± → cos y = − since y is in Quadrant II 25 5 5 12 ⎛ 4 ⎞ ⎛ 5 ⎞ 3 48 ⎛ 3 ⎞ 63 sin ( x + y ) = sin x ⋅ cos y + cos x ⋅ sin y = ⋅ ⎜ − ⎟ + ⎜ − ⎟ ⋅ = − + ⎜ − ⎟ = − 13 ⎝ 5 ⎠ ⎝ 13 ⎠ 5 65 ⎝ 13 ⎠ 65 12 ⎛ 4 ⎞ ⎛ 5 ⎞ 3 48 ⎛ 3 ⎞ 33 sin ( x − y ) = sin x ⋅ cos y − cos x ⋅ sin y = ⋅ ⎜ − ⎟ − ⎜ − ⎟ ⋅ = − − ⎜ − ⎟ = − 13 ⎝ 5 ⎠ ⎝ 13 ⎠ 5 65 ⎝ 13 ⎠ 65 cos 2 y = 29) sin ( x + y ) + sin ( x − y ) → Replace sin ( x + y ) with sin x ⋅ cos y + cos x ⋅ sin y → Replace sin ( x − y ) with sin x ⋅ cos y − cos x ⋅ sin y → sin ( x + y ) + sin ( x − y ) = sin x ⋅ cos y + cos x ⋅ sin y + sin x ⋅ cos y − cos x ⋅ sin y → Combine like terms → sin x ⋅ cos y + cos x ⋅ sin y + sin x ⋅ cos y − cos x ⋅ sin y = 2 ⋅ sin x ⋅ cos y 30) sin ( p + q ) cos p ⋅ cos q Write as separate fractions → Replace sin p ⋅ cos q + cos p ⋅ sin q → cos p ⋅ cos q → Replace sin ( p + q ) with sin p ⋅ cos q + cos p ⋅ sin q → sin p ⋅ cos q cos p ⋅ sin q sin p sin q + → Reduce → + → cos p ⋅ cos q cos p ⋅ cos q cos p cos q sin p sin q sin p sin q with tan p and with tan q → + = tan p + tan q cos p cos q cos p cos q 31) cos 2 A = cos 2 A − sin 2 A → cos 2 22.5° − sin 2 22.5° = cos ( 2 ⋅ 22.5° ) = cos 45° 32) ⎛π ⎞ ⎛π ⎞ ⎛ π ⎞ ⎛π ⎞ sin 2 A = 2sin A ⋅ cos A → 2sin ⎜ ⎟ ⋅ cos ⎜ ⎟ = sin ⎜ 2 ⋅ ⎟ = sin ⎜ ⎟ ⎝ 12 ⎠ ⎝ 12 ⎠ ⎝ 12 ⎠ ⎝6⎠ 33) tan 2 A = 34) tan A ⋅ sin 2 A → Replace tan A with 2 tan A 2 tan 75° → = tan ( 2 ⋅ 75° ) = tan150° 2 1 − tan A 1 − tan 2 75° Simplify → 35) sin A sin A and sin 2 A with 2 ⋅ sin A ⋅ cos A → ⋅ ( 2 ⋅ sin A ⋅ cos A) → cos A cos A sin A ⋅ ( 2 ⋅ sin A ⋅ cos A ) = 2 ⋅ sin A ⋅ sin A = 2sin 2 A cos A 1 + cos 2 x → Replace cos 2 x with ( cos 2 x − sin 2 x ) and sin 2 x with ( 2sin x cos x ) → sin 2 x 1 + cos 2 x − (1 − cos 2 x ) 1 + cos 2 x − sin 2 x 2 2 → Replace sin x with (1 − cos x ) → → Simplify → 2sin x cos x 2sin x cos x 1 + cos 2 x − 1 + cos 2 x 2 cos 2 x cos x cos x cos x = → Reduce → → Replace with cot x → = cot x 2sin x cos x 2sin x cos x sin x sin x sin x 36) 1 − cos A 1 − cos 45° ⎛ A⎞ ⎛ 45° ⎞ sin ⎜ ⎟ = ± → sin 22.5° = sin ⎜ =± ⎟=± 2 2 ⎝2⎠ ⎝ 2 ⎠ Multiply each term under radical sign by 2 → ± sin 22.5° = 1− 2 2 2 → 2− 2 2− 2 2− 2 → Simplify → ± =± 4 2 4 2− 2 because 22.5° is in Quadrant I 2 37) ⎛ 2⎞ 1 + ⎜⎜ − ⎟ 2 ⎟⎠ 1 + cos A 1 + cos 225° ⎛ A⎞ ⎛ 225° ⎞ ⎝ cos ⎜ ⎟ = ± → cos112.5° = cos ⎜ =± → ⎟=± 2 2 2 ⎝2⎠ ⎝ 2 ⎠ Multiply each term under radical sign by 2 → ± cos112.5° = − 38) 2− 2 2− 2 2− 2 → Simplify → ± =± 4 2 4 2− 2 because 112.5° is in Quadrant II 2 1 − cos A ⎛ A⎞ ⎛ 7π tan ⎜ ⎟ = ± → tan ⎜ 1 + cos A ⎝2⎠ ⎝ 8 ⎛ 7π 1 − cos ⎜ ⎞ ⎛ 7π /4 ⎞ ⎝ 4 ⎟ = tan ⎜ ⎟=± ⎛ 7π ⎠ ⎝ 2 ⎠ 1 + cos ⎜ ⎝ 4 Multiply each term under radical sign by 2 → ± ( ) conjugate of denominator 2 − 2 → ± ⎛ 7π Reduce → ± 3 − 2 2 → tan ⎜ ⎝ 4 2− 2 2+ 2 ⎞ 2 1− ⎟ ⎠ =± 2 → ⎞ 2 1+ ⎟ ⎠ 2 → Multiply numerator and denominator by (2 − 2 ) ⋅(2 − 2 ) = ± (2 + 2 ) ⋅(2 − 2 ) 4−2 2 −2 2 +2 4−2 2 +2 2 −2 7π ⎞ is in Quadrant IV ⎟ = − 3 − 2 2 because 4 ⎠ 39) 1 − cos A 1 − cos 35° ⎛ A⎞ ⎛ 35° ⎞ sin ⎜ ⎟ = → = sin ⎜ ⎟ = sin17.5° 2 2 ⎝2⎠ ⎝ 2 ⎠ 40) sin A sin 50° ⎛ A⎞ ⎛ 50° ⎞ tan ⎜ ⎟ = → = tan ⎜ ⎟ = tan 25° ⎝ 2 ⎠ 1 + cos A 1 + cos 50° ⎝ 2 ⎠ 41) 1 + cos A 1 + cos 24° ⎛ A⎞ ⎛ 24° ⎞ cos ⎜ ⎟ = → = cos ⎜ ⎟ = cos12° 2 2 ⎝2⎠ ⎝ 2 ⎠ =± 6−4 2 2 Pythagorean Identities Negative-Angle Identities 1) 1) sin ( −θ ) = − sin θ 2) tan 2 θ + 1 = sec 2 θ 2) cos ( −θ ) = cos θ 3) 1 + cot 2 θ = csc2 θ 3) tan ( −θ ) = − tan θ Sum/Difference Identities Cofunction Identities 1) cos ( A + B ) = cos A ⋅ cos B − sin A ⋅ sin B 1) sin θ = cos ( 90° − θ ) 2) cos ( A − B ) = cos A ⋅ cos B + sin A ⋅ sin B 2) sec θ = csc ( 90° − θ ) 3) sin ( A + B ) = sin A ⋅ cos B + cos A ⋅ sin B 3) tan θ = cot ( 90° − θ ) 4) sin ( A − B ) = sin A ⋅ cos B − cos A ⋅ sin B tan A + tan B 1 − tan A ⋅ tan B tan A − tan B 6) tan ( A − B ) = 1 + tan A ⋅ tan B 5) tan ( A + B ) = Double-Angle Identities Half-Angle Identities 1) cos 2 A = cos 2 A − sin 2 A 1) cos A 1 + cos A =± 2 2 2) cos 2 A = 1 − 2sin 2 A 2) sin A 1 − cos A =± 2 2 3) cos 2 A = 2 cos 2 A − 1 3) tan 4) sin 2 A = 2 ⋅ sin A ⋅ cos A 5) tan 2 A = 2 tan A 1 − tan 2 A A 1 − cos A =± 2 1 + cos A A sin A 4) tan = 2 1 + cos A A 1 − cos A 5) tan = 2 sin A