Getal & Ruimte (13e editie) - vwo wiskunde B

'Goniometrische vergelijkingen'.

vwo wiskunde B	8.3 Goniometrische vergelijkingen
Goniometrische vergelijkingen (10)	opgave 1 Bereken zo mogelijk exact de oplossingen in \([0 , 2 \pi ] \text{.}\) 3p a \(\sin(2 x - \frac{1}{6} \pi ) = 0\) ExacteWaarde (0) 004f - Goniometrische vergelijkingen - basis - basis - 41ms - dynamic variables a (Exacte waardencirkel) \(2 x - \frac{1}{6} \pi = k ⋅ \pi \) 1p ○ \(2 x = \frac{1}{6} \pi + k ⋅ \pi \) \(x = \frac{1}{12} \pi + k ⋅ \frac{1}{2} \pi \) 1p ○ \(x\) in \([0 , 2 \pi ]\) geeft \(x = \frac{1}{12} \pi ∨ x = \frac{7}{12} \pi ∨ x = 1\frac{1}{12} \pi ∨ x = 1\frac{7}{12} \pi \) 1p 4p b \(-2 \sin(\frac{3}{4} x + \frac{3}{4} \pi ) = 1\) ExacteWaarde (1) 004g - Goniometrische vergelijkingen - basis - midden - 0ms - dynamic variables b (Balansmethode) \(\sin(\frac{3}{4} x + \frac{3}{4} \pi ) = -\frac{1}{2} \text{.}\) 1p ○ (Exacte waardencirkel) \(\frac{3}{4} x + \frac{3}{4} \pi = -\frac{1}{6} \pi + k ⋅ 2 \pi ∨ \frac{3}{4} x + \frac{3}{4} \pi = -\frac{5}{6} \pi + k ⋅ 2 \pi \) 1p ○ \(\frac{3}{4} x = -\frac{11}{12} \pi + k ⋅ 2 \pi ∨ \frac{3}{4} x = -1\frac{7}{12} \pi + k ⋅ 2 \pi \) \(x = -1\frac{2}{9} \pi + k ⋅ 2\frac{2}{3} \pi ∨ x = -2\frac{1}{9} \pi + k ⋅ 2\frac{2}{3} \pi \) 1p ○ \(x\) in \([0 , 2 \pi ]\) geeft \(x = 1\frac{4}{9} \pi ∨ x = \frac{5}{9} \pi \) 1p 4p c \(-4 \cos(\frac{1}{2} x - \frac{1}{2} \pi ) = -2 \sqrt{2}\) ExacteWaarde (2) 004h - Goniometrische vergelijkingen - basis - midden - 0ms - dynamic variables c (Balansmethode) \(\cos(\frac{1}{2} x - \frac{1}{2} \pi ) = \frac{1}{2} \sqrt{2} \text{.}\) 1p ○ (Exacte waardencirkel) \(\frac{1}{2} x - \frac{1}{2} \pi = \frac{1}{4} \pi + k ⋅ 2 \pi ∨ \frac{1}{2} x - \frac{1}{2} \pi = 1\frac{3}{4} \pi + k ⋅ 2 \pi \) 1p ○ \(\frac{1}{2} x = \frac{3}{4} \pi + k ⋅ 2 \pi ∨ \frac{1}{2} x = 2\frac{1}{4} \pi + k ⋅ 2 \pi \) \(x = 1\frac{1}{2} \pi + k ⋅ 4 \pi ∨ x = 4\frac{1}{2} \pi + k ⋅ 4 \pi \) 1p ○ \(x\) in \([0 , 2 \pi ]\) geeft \(x = 1\frac{1}{2} \pi ∨ x = \frac{1}{2} \pi \) 1p 4p d \(5 \cos(\frac{1}{2} \pi x + \frac{1}{3} \pi ) = 2\frac{1}{2} \sqrt{3}\) ExacteWaarde (3) 006x - Goniometrische vergelijkingen - basis - midden - 0ms - dynamic variables d (Balansmethode) \(\cos(\frac{1}{2} \pi x + \frac{1}{3} \pi ) = \frac{1}{2} \sqrt{3} \text{.}\) 1p ○ (Exacte waardencirkel) \(\frac{1}{2} \pi x + \frac{1}{3} \pi = \frac{1}{6} \pi + k ⋅ 2 \pi ∨ \frac{1}{2} \pi x + \frac{1}{3} \pi = -\frac{1}{6} \pi + k ⋅ 2 \pi \) 1p ○ \(\frac{1}{2} \pi x = -\frac{1}{6} \pi + k ⋅ 2 \pi ∨ \frac{1}{2} \pi x = -\frac{1}{2} \pi + k ⋅ 2 \pi \) \(x = -\frac{1}{3} + k ⋅ 4 ∨ x = -1 + k ⋅ 4\) 1p ○ \(x\) in \([0 , 2 \pi ]\) geeft \(x = 3\frac{2}{3} ∨ x = 3\) 1p opgave 2 Bereken zo mogelijk exact de oplossingen in \([0 , 2 \pi ] \text{.}\) 4p \(3 - 4 \cos(3 x + \frac{4}{5} \pi ) = -1\) ExacteWaarde (4) 006y - Goniometrische vergelijkingen - basis - midden - 1ms - dynamic variables ○ (Balansmethode) \(-4 \cos(3 x + \frac{4}{5} \pi ) = -4\) dus \(\cos(3 x + \frac{4}{5} \pi ) = 1 \text{.}\) 1p ○ (Exacte waardencirkel) \(3 x + \frac{4}{5} \pi = k ⋅ 2 \pi \) 1p ○ \(3 x = -\frac{4}{5} \pi + k ⋅ 2 \pi \) \(x = -\frac{4}{15} \pi + k ⋅ \frac{2}{3} \pi \) 1p ○ \(x\) in \([0 , 2 \pi ]\) geeft \(x = \frac{2}{5} \pi ∨ x = 1\frac{1}{15} \pi ∨ x = 1\frac{11}{15} \pi \) 1p opgave 3 Los exact op. 3p a \(\sin^{2}(3 x - \frac{1}{6} \pi ) = 1\) Substitutie (1) 006z - Goniometrische vergelijkingen - basis - midden - 0ms - dynamic variables a \(\sin(3 x - \frac{1}{6} \pi ) = 1 ∨ \sin(3 x - \frac{1}{6} \pi ) = -1\) 1p ○ De exacte waardencirkel geeft \(3 x - \frac{1}{6} \pi = \frac{1}{2} \pi + k ⋅ 2 \pi ∨ 3 x - \frac{1}{6} \pi = 1\frac{1}{2} \pi + k ⋅ 2 \pi \) 1p ○ \(3 x = \frac{2}{3} \pi + k ⋅ 2 \pi ∨ 3 x = 1\frac{2}{3} \pi + k ⋅ 2 \pi \) \(x = \frac{2}{9} \pi + k ⋅ \frac{2}{3} \pi ∨ x = \frac{5}{9} \pi + k ⋅ \frac{2}{3} \pi \) 1p 3p b \(1\frac{1}{2} \sin(3 x + \frac{1}{2} \pi ) \cos(1\frac{1}{2} x + \frac{2}{3} \pi ) = 0\) Product 0070 - Goniometrische vergelijkingen - basis - midden - 1ms - dynamic variables b \(\sin(3 x + \frac{1}{2} \pi ) = 0 ∨ \cos(1\frac{1}{2} x + \frac{2}{3} \pi ) = 0\) 1p ○ (Exacte waardencirkel) \(3 x + \frac{1}{2} \pi = k ⋅ \pi ∨ 1\frac{1}{2} x + \frac{2}{3} \pi = \frac{1}{2} \pi + k ⋅ \pi \) 1p ○ \(3 x = -\frac{1}{2} \pi + k ⋅ \pi ∨ 1\frac{1}{2} x = -\frac{1}{6} \pi + k ⋅ \pi \) \(x = -\frac{1}{6} \pi + k ⋅ \frac{1}{3} \pi ∨ x = -\frac{1}{9} \pi + k ⋅ \frac{2}{3} \pi \) 1p 3p c \(\sin(2 \pi x+{2 \over 3} \pi) = \sin(\pi x+{1 \over 3} \pi)\) BasisAisASin 00sf - Goniometrische vergelijkingen - basis - midden - 0ms c (Eenheidscirkel) \(2 \pi x+{2 \over 3} \pi = (\pi x+{1 \over 3} \pi)+k \cdot 2 \pi ∨ 2 \pi x+{2 \over 3} \pi = \pi-(\pi x+{1 \over 3} \pi)+k \cdot 2 \pi\) 1p ○ \(2 \pi x+{2 \over 3} \pi = (\pi x+{1 \over 3} \pi)+k \cdot 2 \pi ∨ 2 \pi x+{2 \over 3} \pi = (-\pi x+{2 \over 3} \pi)+k \cdot 2 \pi\) \(\pi x = -{1 \over 3} \pi+k \cdot 2 \pi ∨ 3 \pi x = 0+k \cdot 2 \pi\) \(x = -{1 \over 3}+k \cdot 2 ∨ x = k \cdot {2 \over 3}\) 2p 3p d \(\cos(3 x+{1 \over 4} \pi) = \cos(2 x-{1 \over 3} \pi)\) BasisAisACos 00sg - Goniometrische vergelijkingen - basis - midden - 1184ms d (Eenheidscirkel) \(3 x+{1 \over 4} \pi = (2 x-{1 \over 3} \pi)+k \cdot 2 \pi ∨ 3 x+{1 \over 4} \pi = -(2 x-{1 \over 3} \pi)+k \cdot 2 \pi\) 1p ○ \(3 x+{1 \over 4} \pi = (2 x-{1 \over 3} \pi)+k \cdot 2 \pi ∨ 3 x+{1 \over 4} \pi = (-2 x+{1 \over 3} \pi)+k \cdot 2 \pi\) \(x = -{7 \over 12} \pi+k \cdot 2 \pi ∨ 5 x = {1 \over 12} \pi+k \cdot 2 \pi\) \(x = -{7 \over 12} \pi+k \cdot 2 \pi ∨ x = {1 \over 60} \pi+k \cdot {2 \over 5} \pi\) 2p opgave 4 Los exact op. 3p \(\tan(3 x+{2 \over 3} \pi) = \tan(-{2 \over 3} \pi)\) BasisTan 00sh - Goniometrische vergelijkingen - basis - midden - 0ms ○ (Eenheidscirkel) \(3 x+{2 \over 3} \pi = -{2 \over 3} \pi+k \pi\) 1p ○ \(3 x = -1\frac{1}{3} \pi+k \pi\) \(x = -{2 \over 9} \pi+k \cdot {1 \over 3} \pi\) 2p
vwo wiskunde B	12.1 Goniometrische formules bij vergelijkingen en herleidingen
Goniometrische vergelijkingen (6)	opgave 1 Los exact op. 4p a \(\cos(3 x+{3 \over 4} \pi) = \sin(x+{3 \over 4} \pi)\) BasisAisB 00si - Goniometrische vergelijkingen - basis - midden - 0ms a (\(\sin(A) = \cos(A + -{1 \over 2} \pi)\) geeft) \(\cos(3 x+{3 \over 4} \pi) = \cos(x+{3 \over 4} \pi-{1 \over 2} \pi)\) \(\cos(3 x+{3 \over 4} \pi) = \cos(x+{1 \over 4} \pi)\) 1p ○ (Eenheidscirkel) \(3 x+{3 \over 4} \pi = (x+{1 \over 4} \pi)+k \cdot 2 \pi ∨ 3 x+{3 \over 4} \pi = -(x+{1 \over 4} \pi)+k \cdot 2 \pi\) 1p ○ \(3 x+{3 \over 4} \pi = (x+{1 \over 4} \pi)+k \cdot 2 \pi ∨ 3 x+{3 \over 4} \pi = (-x-{1 \over 4} \pi)+k \cdot 2 \pi\) \(2 x = -{1 \over 2} \pi+k \cdot 2 \pi ∨ 4 x = -\pi+k \cdot 2 \pi\) \(x = -{1 \over 4} \pi+k \pi ∨ x = -{1 \over 4} \pi+k \cdot {1 \over 2} \pi\) 2p 4p b \(\sin(2 \pi x-{1 \over 2} \pi) = -\sin(3 \pi x-1\frac{1}{2} \pi)\) BasisAisMinA 00sj - Goniometrische vergelijkingen - basis - midden - 0ms b (\(-\sin(A) = \sin(A + \pi )\) geeft) \(\sin(2 \pi x-{1 \over 2} \pi) = \sin(3 \pi x-1\frac{1}{2} \pi+\pi)\) \(\sin(2 \pi x-{1 \over 2} \pi) = \sin(3 \pi x-{1 \over 2} \pi)\) 1p ○ (Eenheidscirkel) \(2 \pi x-{1 \over 2} \pi = (3 \pi x-{1 \over 2} \pi)+k \cdot 2 \pi ∨ 2 \pi x-{1 \over 2} \pi = \pi-(3 \pi x-{1 \over 2} \pi)+k \cdot 2 \pi\) 1p ○ \(2 \pi x-{1 \over 2} \pi = (3 \pi x-{1 \over 2} \pi)+k \cdot 2 \pi ∨ 2 \pi x-{1 \over 2} \pi = (-3 \pi x+1\frac{1}{2} \pi)+k \cdot 2 \pi\) \(-\pi x = 0+k \cdot 2 \pi ∨ 5 \pi x = 2 \pi+k \cdot 2 \pi\) \(x = k \cdot 2 ∨ x = {2 \over 5}+k \cdot {2 \over 5}\) 2p 5p c \(\cos(2 x+{1 \over 2} \pi) = -\sin(x-{1 \over 4} \pi)\) BasisAisMinB 00sk - Goniometrische vergelijkingen - basis - eind - 0ms c (\(-\sin(A) = \sin(A + \pi )\) geeft) \(\cos(2 x+{1 \over 2} \pi) = \sin(x-{1 \over 4} \pi+\pi)\) \(\cos(2 x+{1 \over 2} \pi) = \sin(x+{3 \over 4} \pi)\) 1p ○ (\(\sin(A) = \cos(A + -{1 \over 2} \pi)\) geeft) \(\cos(2 x+{1 \over 2} \pi) = \cos(x+{3 \over 4} \pi-{1 \over 2} \pi)\) \(\cos(2 x+{1 \over 2} \pi) = \cos(x+{1 \over 4} \pi)\) 1p ○ (Eenheidscirkel) \(2 x+{1 \over 2} \pi = (x+{1 \over 4} \pi)+k \cdot 2 \pi ∨ 2 x+{1 \over 2} \pi = -(x+{1 \over 4} \pi)+k \cdot 2 \pi\) 1p ○ \(2 x+{1 \over 2} \pi = (x+{1 \over 4} \pi)+k \cdot 2 \pi ∨ 2 x+{1 \over 2} \pi = (-x-{1 \over 4} \pi)+k \cdot 2 \pi\) \(x = -{1 \over 4} \pi+k \cdot 2 \pi ∨ 3 x = -{3 \over 4} \pi+k \cdot 2 \pi\) \(x = -{1 \over 4} \pi+k \cdot 2 \pi ∨ x = -{1 \over 4} \pi+k \cdot {2 \over 3} \pi\) 2p 3p d \(\sin^{2}(x) - \frac{1}{4} = 0\) Substitutie (2) 00sl - Goniometrische vergelijkingen - basis - midden - 9ms d \(u = \sin(x)\) geeft \(u^{2} - \frac{1}{4}\) 1p ○ \(4 u^{2} - 1 = 0\) \(4 u^{2} = 1\) \(u^{2} = \frac{1}{4}\) \(u = -\frac{1}{2} ∨ u = \frac{1}{2}\) 1p ○ \(\sin(x) = -\frac{1}{2} ∨ \sin(x) = \frac{1}{2}\) geeft \(x = -{1 \over 6} \pi+k \cdot 2 \pi ∨ x = -{5 \over 6} \pi+k \cdot 2 \pi ∨ x = {1 \over 6} \pi+k \cdot 2 \pi ∨ x = {5 \over 6} \pi+k \cdot 2 \pi\) 1p opgave 2 Los exact op. 3p a \(\cos^{2}(x) - 2 \cos(x) = -1\) Substitutie (3) 00sm - Goniometrische vergelijkingen - basis - midden - 0ms a \(u = \cos(x)\) geeft \(u^{2} - 2 u = -1\) 1p ○ \(u^{2} - 2 u + 1 = 0\) \((u - 1) (u - 1) = 0\) \(u = 1\) 1p ○ \(\cos(x) = 1\) geeft \(x = k \cdot 2 \pi\) 1p 4p b \(3 \sin^{2}(x) + 4 \cos^{2}(x) - 2 \cos(x) - 2 = 0\) Substitutie (4) 00sn - Goniometrische vergelijkingen - basis - midden - 1ms b (\(\sin^{2}(A) + \cos^{2}(A) = 1\) geeft) \(3 (1 - \cos^{2}(x)) + 4 \cos^{2}(x) - 2 \cos(x) - 2 = 0\) ofwel \(\cos^{2}(x) - 2 \cos(x) + 1 = 0\) 1p ○ \(u = \cos(x)\) geeft \(u^{2} - 2 u + 1 = 0\) 1p ○ \((u - 1) (u - 1) = 0\) \(u = 1\) 1p ○ \(\cos(x) = 1\) geeft \(x = k \cdot 2 \pi\) 1p

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