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3.5. Double Angle Identities www.ck12.org 3.5 Double Angle Identities 1. If sin x = 4 5 and in Quadrant II, then cosine and tangent are negative. Also, by the Pythagorean Theorem, the p third side is 3(b = 52 − 42 ). So, cos x = − 53 and tan x = − 34 . Using this, we can find sin 2x, cos 2x, and tan 2x. 2 tan x 1 − tan2 x 2 · − 43 = 2 1 − − 34 cos 2x = 1 − sin2 x 2 4 = 1−2· 5 = 1−2· sin 2x = 2 sin x cos x 3 4 = 2· ·− 5 5 24 =− 25 = 1− =− tan 2x = 16 25 − 83 8 7 = − ÷− 3 9 1− 8 9 = − ·− 3 7 24 = 7 = 32 25 7 25 16 9 2. This is one of the forms for cos 2x. cos2 15◦ − sin2 15◦ = cos(15◦ · 2) = cos 30◦ √ 3 = 2 3. Step 1: Use the cosine sum formula cos 3θ = 4 cos3 θ − 3 cos θ cos(2θ + θ) = cos 2θ cos θ − sin 2θ sin θ Step 2: Use double angle formulas for cos 2θ and sin 2θ = (2 cos2 θ − 1) cos θ − (2 sin θ cos θ) sin θ Step 3: Distribute and simplify. = 2 cos3 θ − cos θ − 2 sin2 θ cos θ = − cos θ(−2 cos2 θ + 2 sin2 θ + 1) = − cos θ[−2 cos2 θ + 2(1 − cos2 θ) + 1] → Substitute 1 − cos2 θ for sin2 θ = − cos θ[−2 cos2 θ + 2 − 2 cos2 θ + 1] = − cos θ(−4 cos2 θ + 3) = 4 cos3 θ − 3 cos θ 4. Step 1: Expand sin 2t using the double angle formula. sin 2t − tant = tant cos 2t 2 sint cost − tant = tant cos 2t 50 www.ck12.org Chapter 3. Trigonometric Identities and Equations, Solution Key Step 2: change tant and find a common denominator. sint cost 2 sint cos2 t − sint cost sint(2 cos2 t − 1) cost sint · (2 cos2 t − 1) cost tant cos 2t 2 sint cost − 5. 9 If sin x = − 41 40 and in Quadrant III, then cos x = − 41 and tan x = 9 40 q (Pythagorean Theorem, b = 412 − (−9)2 ). So, cos 2x = 2 cos2 x − 1 40 2 =2 − −1 41 sin 2x = 2 sin x cos x = 2·− = 9 40 ·− 41 41 720 1681 tan 2x = sin 2x cos 2x = 3200 1681 − 1681 1681 = 720 1681 1519 1681 = 1519 1681 = 720 1519 6. Step 1: Expand sin 2x sin 2x + sin x = 0 2 sin x cos x + sin x = 0 sin x(2 cos x + 1) = 0 Step 2: Separate and solve each for x. 2 cos x + 1 = 0 1 2 2π 4π x= , 3 3 cos x = − sin x = 0 x = 0, π or 7. Expand cos 2x and simplify cos2 x − cos 2x = 0 cos2 x − (2 cos2 x − 1) = 0 − cos2 x + 1 = 0 cos2 x = 1 cos x = ±1 cos x = 1 when x = 0, and cos x = −1 when x = π. Therefore, the solutions are x = 0, π. 8. a. 3.429 b. 0.960 c. 0.280 51 3.5. Double Angle Identities www.ck12.org 9. a. 2 sin 2x 2 2 csc x 2x = 2 sin x cos x 1 2 csc x 2x = sin x cos x sin x 1 2 csc x 2x = sin x sin x cos x sin x 2 csc x 2x = 2 sin x cos x 1 sin x 2 csc x 2x = 2 · sin x cos x 2 csc x 2x = csc2 x tan x 2 csc x 2x = b. cos4 θ − sin4 θ = (cos2 θ + sin2 θ)(cos2 θ − sin2 θ) cos4 − sin4 θ = 1(cos2 θ − sin2 θ) cos 2θ = cos2 θ − sin2 θ ∴ cos4 θ − sin4 θ = cos 2θ c. sin 2x 1 + cos 2x sin 2x 1 + cos 2x sin 2x 1 + cos 2x sin 2x 1 + cos 2x sin 2x 1 + cos 2x sin 2x 1 + cos 2x = = = = = 2 sin x cos x 1 + (1 − 2 sin2 x) 2 sin x cos x 2 − 2 sin2 x 2 sin x cos x 2(1 − sin2 x) 2 sin x cos x 2 cos2 x sin x cos x = tan x 10. cos 2x − 1 = sin2 x (1 − 2 sin2 x) − 1 = sin2 x −2 sin2 x = sin2 x 0 = 3 sin2 x 0 = sin2 x 0 = sin x x = 0, π 52 www.ck12.org Chapter 3. Trigonometric Identities and Equations, Solution Key 11. cos 2x = cos x 2 2 cos x − 1 = cos x 2 2 cos x − cos x − 1 = 0 (2 cos x + 1)(cos x − 1) = 0 & & 2 cos x + 1 = 0 or cos x − 1 = 0 2 cos x = −1 1 cos x = − 2 cos x = 1 when x = 0 and cos x = − 12 when x = cos x = 1 2π 3 . 12. 2 csc 2x tan x = sec2 x 2 sin x 1 · = sin 2x cos x cos2 x 2 sin x 1 · = 2 sin x cos x cos x cos2 x 1 1 = 2 cos x cos2 x 13. sin 2x − cos 2x = 1 2 sin x cos x − (1 − 2 sin2 x) = 1 2 sin x cos x − 1 + 2 sin2 x = 1 2 sin x cos x + 2 sin2 x = 2 sin x cos x + sin2 x = 1 sin x cos x = 1 − sin2 x sin x cos x = cos2 x p ± 1 − cos2 x cos x = cos2 x 1 − cos2 x cos2 x = cos4 x cos2 x − cos4 x = cos4 x cos2 x − 2 cos4 x = 0 cos2 x(1 − 2 cos2 x) = 0 . & 1 − 2 cos2 x = 0 cos2 x = 0 cos x = 0 or π 3π x= , 2 2 − 2 cos2 x = −1 1 cos2 x = 2 √ 2 cos x = ± 2 π 5π x= , 4 4 Note: If we go back to the equation sin x cos x = cos2 x, we can see that sin x cos x must be positive or zero, since cos2 x is always positive or zero. For this reason, sin x and cos x must have the same sign (or one of them 53 3.5. Double Angle Identities www.ck12.org must be zero), which means that x cannot be in the second or fourth quadrants. This is why valid solutions. 14. Use the double angle identity for cos 2x. sin2 x − 2 = cos 2x sin2 x − 2 = cos 2x sin2 x − 2 = 1 − 2 sin2 x 3 sin2 x = 3 sin2 x = 1 sin x = ±1 π 3π x= , 2 2 54 3π 4 and 7π 4 are not