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 \(\cos(2 x + \frac{1}{6} \pi ) = 0\) ExacteWaarde (0) 004f - Goniometrische vergelijkingen - basis - basis - 41ms - dynamic variables a (Exacte waardencirkel) \(2 x + \frac{1}{6} \pi = \frac{1}{2} \pi + k ⋅ \pi \) 1p ○ \(2 x = \frac{1}{3} \pi + k ⋅ \pi \) \(x = \frac{1}{6} \pi + k ⋅ \frac{1}{2} \pi \) 1p ○ \(x\) in \([0 , 2 \pi ]\) geeft \(x = \frac{1}{6} \pi ∨ x = \frac{2}{3} \pi ∨ x = 1\frac{1}{6} \pi ∨ x = 1\frac{2}{3} \pi \) 1p 4p b \(-4 \cos(\frac{3}{4} \pi x - \frac{1}{2} \pi ) = 2\) ExacteWaarde (1) 004g - Goniometrische vergelijkingen - basis - midden - 0ms - dynamic variables b (Balansmethode) \(\cos(\frac{3}{4} \pi x - \frac{1}{2} \pi ) = -\frac{1}{2} \text{.}\) 1p ○ (Exacte waardencirkel) \(\frac{3}{4} \pi x - \frac{1}{2} \pi = \frac{2}{3} \pi + k ⋅ 2 \pi ∨ \frac{3}{4} \pi x - \frac{1}{2} \pi = -\frac{2}{3} \pi + k ⋅ 2 \pi \) 1p ○ \(\frac{3}{4} \pi x = 1\frac{1}{6} \pi + k ⋅ 2 \pi ∨ \frac{3}{4} \pi x = -\frac{1}{6} \pi + k ⋅ 2 \pi \) \(x = 1\frac{5}{9} + k ⋅ 2\frac{2}{3} ∨ x = -\frac{2}{9} + k ⋅ 2\frac{2}{3}\) 1p ○ \(x\) in \([0 , 2 \pi ]\) geeft \(x = 1\frac{5}{9} ∨ x = 4\frac{2}{9} ∨ x = 2\frac{4}{9} ∨ x = 5\frac{1}{9}\) 1p 4p c \(-2 \sin(\frac{2}{3} x + \frac{1}{6} \pi ) = -\sqrt{2}\) ExacteWaarde (2) 004h - Goniometrische vergelijkingen - basis - midden - 0ms - dynamic variables c (Balansmethode) \(\sin(\frac{2}{3} x + \frac{1}{6} \pi ) = \frac{1}{2} \sqrt{2} \text{.}\) 1p ○ (Exacte waardencirkel) \(\frac{2}{3} x + \frac{1}{6} \pi = \frac{1}{4} \pi + k ⋅ 2 \pi ∨ \frac{2}{3} x + \frac{1}{6} \pi = \frac{3}{4} \pi + k ⋅ 2 \pi \) 1p ○ \(\frac{2}{3} x = \frac{1}{12} \pi + k ⋅ 2 \pi ∨ \frac{2}{3} x = \frac{7}{12} \pi + k ⋅ 2 \pi \) \(x = \frac{1}{8} \pi + k ⋅ 3 \pi ∨ x = \frac{7}{8} \pi + k ⋅ 3 \pi \) 1p ○ \(x\) in \([0 , 2 \pi ]\) geeft \(x = \frac{1}{8} \pi ∨ x = \frac{7}{8} \pi \) 1p 4p d \(3 \sin(\frac{3}{5} x - \frac{2}{3} \pi ) = -1\frac{1}{2} \sqrt{3}\) ExacteWaarde (3) 006x - Goniometrische vergelijkingen - basis - midden - 0ms - dynamic variables d (Balansmethode) \(\sin(\frac{3}{5} x - \frac{2}{3} \pi ) = -\frac{1}{2} \sqrt{3} \text{.}\) 1p ○ (Exacte waardencirkel) \(\frac{3}{5} x - \frac{2}{3} \pi = -\frac{1}{3} \pi + k ⋅ 2 \pi ∨ \frac{3}{5} x - \frac{2}{3} \pi = -\frac{2}{3} \pi + k ⋅ 2 \pi \) 1p ○ \(\frac{3}{5} x = \frac{1}{3} \pi + k ⋅ 2 \pi ∨ \frac{3}{5} x = k ⋅ 2 \pi \) \(x = \frac{5}{9} \pi + k ⋅ 3\frac{1}{3} \pi ∨ x = k ⋅ 3\frac{1}{3} \pi \) 1p ○ \(x\) in \([0 , 2 \pi ]\) geeft \(x = \frac{5}{9} \pi ∨ x = 0\) 1p opgave 2 Bereken zo mogelijk exact de oplossingen in \([0 , 2 \pi ] \text{.}\) 4p \(-1 - 3 \sin(4 x + \frac{1}{2} \pi ) = -4\) ExacteWaarde (4) 006y - Goniometrische vergelijkingen - basis - midden - 1ms - dynamic variables ○ (Balansmethode) \(-3 \sin(4 x + \frac{1}{2} \pi ) = -3\) dus \(\sin(4 x + \frac{1}{2} \pi ) = 1 \text{.}\) 1p ○ (Exacte waardencirkel) \(4 x + \frac{1}{2} \pi = \frac{1}{2} \pi + k ⋅ 2 \pi \) 1p ○ \(4 x = k ⋅ 2 \pi \) \(x = k ⋅ \frac{1}{2} \pi \) 1p ○ \(x\) in \([0 , 2 \pi ]\) geeft \(x = 0 ∨ x = \frac{1}{2} \pi ∨ x = \pi ∨ x = 1\frac{1}{2} \pi ∨ x = 2 \pi \) 1p opgave 3 Los exact op. 3p a \(\cos^{2}(4 x + \frac{2}{5} \pi ) = 1\) Substitutie (1) 006z - Goniometrische vergelijkingen - basis - midden - 0ms - dynamic variables a \(\cos(4 x + \frac{2}{5} \pi ) = 1 ∨ \cos(4 x + \frac{2}{5} \pi ) = -1\) 1p ○ De exacte waardencirkel geeft \(4 x + \frac{2}{5} \pi = k ⋅ 2 \pi ∨ 4 x + \frac{2}{5} \pi = \pi + k ⋅ 2 \pi \) 1p ○ \(4 x = -\frac{2}{5} \pi + k ⋅ 2 \pi ∨ 4 x = \frac{3}{5} \pi + k ⋅ 2 \pi \) \(x = -\frac{1}{10} \pi + k ⋅ \frac{1}{2} \pi ∨ x = \frac{3}{20} \pi + k ⋅ \frac{1}{2} \pi \) 1p 3p b \(2\frac{1}{4} \cos(4 x + \frac{1}{6} \pi ) \cos(\frac{3}{4} x) = 0\) Product 0070 - Goniometrische vergelijkingen - basis - midden - 1ms - dynamic variables b \(\cos(4 x + \frac{1}{6} \pi ) = 0 ∨ \cos(\frac{3}{4} x) = 0\) 1p ○ (Exacte waardencirkel) \(4 x + \frac{1}{6} \pi = \frac{1}{2} \pi + k ⋅ \pi ∨ \frac{3}{4} x = \frac{1}{2} \pi + k ⋅ \pi \) 1p ○ \(4 x = \frac{1}{3} \pi + k ⋅ \pi ∨ \frac{3}{4} x = \frac{1}{2} \pi + k ⋅ \pi \) \(x = \frac{1}{12} \pi + k ⋅ \frac{1}{4} \pi ∨ x = \frac{2}{3} \pi + k ⋅ 1\frac{1}{3} \pi \) 1p 3p c \(\sin(3 x+{3 \over 4} \pi) = \sin(2 x+{2 \over 3} \pi)\) BasisAisASin 00sf - Goniometrische vergelijkingen - basis - midden - 0ms c (Eenheidscirkel) \(3 x+{3 \over 4} \pi = (2 x+{2 \over 3} \pi)+k \cdot 2 \pi ∨ 3 x+{3 \over 4} \pi = \pi-(2 x+{2 \over 3} \pi)+k \cdot 2 \pi\) 1p ○ \(3 x+{3 \over 4} \pi = (2 x+{2 \over 3} \pi)+k \cdot 2 \pi ∨ 3 x+{3 \over 4} \pi = (-2 x+{1 \over 3} \pi)+k \cdot 2 \pi\) \(x = -{1 \over 12} \pi+k \cdot 2 \pi ∨ 5 x = -{5 \over 12} \pi+k \cdot 2 \pi\) \(x = -{1 \over 12} \pi+k \cdot 2 \pi ∨ x = -{1 \over 12} \pi+k \cdot {2 \over 5} \pi\) 2p 3p d \(\cos(x-{1 \over 4} \pi) = \cos(2 x+{1 \over 4} \pi)\) BasisAisACos 00sg - Goniometrische vergelijkingen - basis - midden - 1184ms d (Eenheidscirkel) \(x-{1 \over 4} \pi = (2 x+{1 \over 4} \pi)+k \cdot 2 \pi ∨ x-{1 \over 4} \pi = -(2 x+{1 \over 4} \pi)+k \cdot 2 \pi\) 1p ○ \(x-{1 \over 4} \pi = (2 x+{1 \over 4} \pi)+k \cdot 2 \pi ∨ x-{1 \over 4} \pi = (-2 x-{1 \over 4} \pi)+k \cdot 2 \pi\) \(-x = {1 \over 2} \pi+k \cdot 2 \pi ∨ 3 x = 0+k \cdot 2 \pi\) \(x = -{1 \over 2} \pi+k \cdot 2 \pi ∨ x = k \cdot {2 \over 3} \pi\) 2p opgave 4 Los exact op. 3p \(\tan(2 x+{1 \over 3} \pi) = \tan(3 x+{1 \over 2} \pi)\) BasisTan 00sh - Goniometrische vergelijkingen - basis - midden - 0ms ○ (Eenheidscirkel) \(2 x+{1 \over 3} \pi = (3 x+{1 \over 2} \pi)+k \pi\) 1p ○ \(-x = {1 \over 6} \pi+k \pi\) \(x = -{1 \over 6} \pi+k \pi\) 2p
vwo wiskunde B	12.1 Goniometrische formules bij vergelijkingen en herleidingen
Goniometrische vergelijkingen (6)	opgave 1 Los exact op. 4p a \(\cos(3 x-{1 \over 2} \pi) = \sin(1\frac{1}{2} \pi)\) BasisAisB 00si - Goniometrische vergelijkingen - basis - midden - 0ms a (\(\sin(A) = \cos(A + -{1 \over 2} \pi)\) geeft) \(\cos(3 x-{1 \over 2} \pi) = \cos(1\frac{1}{2} \pi-{1 \over 2} \pi)\) \(\cos(3 x-{1 \over 2} \pi) = \cos(\pi)\) 1p ○ (Eenheidscirkel) \(3 x-{1 \over 2} \pi = \pi+k \cdot 2 \pi ∨ 3 x-{1 \over 2} \pi = -\pi+k \cdot 2 \pi\) 1p ○ \(3 x-{1 \over 2} \pi = \pi+k \cdot 2 \pi ∨ 3 x-{1 \over 2} \pi = -\pi+k \cdot 2 \pi\) \(3 x = 1\frac{1}{2} \pi+k \cdot 2 \pi ∨ 3 x = -{1 \over 2} \pi+k \cdot 2 \pi\) \(x = {1 \over 2} \pi+k \cdot {2 \over 3} \pi ∨ x = -{1 \over 6} \pi+k \cdot {2 \over 3} \pi\) 2p 4p b \(\sin(\pi x-{3 \over 4} \pi) = -\sin(3 \pi x-1\frac{1}{4} \pi)\) BasisAisMinA 00sj - Goniometrische vergelijkingen - basis - midden - 0ms b (\(-\sin(A) = \sin(A + \pi )\) geeft) \(\sin(\pi x-{3 \over 4} \pi) = \sin(3 \pi x-1\frac{1}{4} \pi+\pi)\) \(\sin(\pi x-{3 \over 4} \pi) = \sin(3 \pi x-{1 \over 4} \pi)\) 1p ○ (Eenheidscirkel) \(\pi x-{3 \over 4} \pi = (3 \pi x-{1 \over 4} \pi)+k \cdot 2 \pi ∨ \pi x-{3 \over 4} \pi = \pi-(3 \pi x-{1 \over 4} \pi)+k \cdot 2 \pi\) 1p ○ \(\pi x-{3 \over 4} \pi = (3 \pi x-{1 \over 4} \pi)+k \cdot 2 \pi ∨ \pi x-{3 \over 4} \pi = (-3 \pi x+1\frac{1}{4} \pi)+k \cdot 2 \pi\) \(-2 \pi x = {1 \over 2} \pi+k \cdot 2 \pi ∨ 4 \pi x = 2 \pi+k \cdot 2 \pi\) \(x = -{1 \over 4}+k \cdot 1 ∨ x = {1 \over 2}+k \cdot {1 \over 2}\) 2p 5p c \(\sin(3 x+{1 \over 3} \pi) = -\cos(x-\pi)\) BasisAisMinB 00sk - Goniometrische vergelijkingen - basis - eind - 0ms c (\(-\cos(A) = \cos(A + \pi )\) geeft) \(\sin(3 x+{1 \over 3} \pi) = \cos(x-\pi+\pi)\) \(\sin(3 x+{1 \over 3} \pi) = \cos(x)\) 1p ○ (\(\cos(A) = \sin(A + {1 \over 2} \pi)\) geeft) \(\sin(3 x+{1 \over 3} \pi) = \sin(x+{1 \over 2} \pi)\) \(\sin(3 x+{1 \over 3} \pi) = \sin(x+{1 \over 2} \pi)\) 1p ○ (Eenheidscirkel) \(3 x+{1 \over 3} \pi = (x+{1 \over 2} \pi)+k \cdot 2 \pi ∨ 3 x+{1 \over 3} \pi = \pi-(x+{1 \over 2} \pi)+k \cdot 2 \pi\) 1p ○ \(3 x+{1 \over 3} \pi = (x+{1 \over 2} \pi)+k \cdot 2 \pi ∨ 3 x+{1 \over 3} \pi = (-x+{1 \over 2} \pi)+k \cdot 2 \pi\) \(2 x = {1 \over 6} \pi+k \cdot 2 \pi ∨ 4 x = {1 \over 6} \pi+k \cdot 2 \pi\) \(x = {1 \over 12} \pi+k \pi ∨ x = {1 \over 24} \pi+k \cdot {1 \over 2} \pi\) 2p 3p d \(\cos^{2}(x) = \frac{3}{4}\) Substitutie (2) 00sl - Goniometrische vergelijkingen - basis - midden - 9ms d \(u = \cos(x)\) geeft \(u^{2} = \frac{3}{4}\) 1p ○ \(u^{2} - \frac{3}{4} = 0\) \(4 u^{2} - 3 = 0\) \(4 u^{2} = 3\) \(u^{2} = \frac{3}{4}\) \(u = -{1 \over 2} \sqrt{3} ∨ u = {1 \over 2} \sqrt{3}\) 1p ○ \(\cos(x) = -{1 \over 2} \sqrt{3} ∨ \cos(x) = {1 \over 2} \sqrt{3}\) geeft \(x = {5 \over 6} \pi+k \cdot 2 \pi ∨ x = -{5 \over 6} \pi+k \cdot 2 \pi ∨ x = {1 \over 6} \pi+k \cdot 2 \pi ∨ x = -{1 \over 6} \pi+k \cdot 2 \pi\) 1p opgave 2 Los exact op. 3p a \(\sin^{2}(x) = -\frac{1}{2} \sin(x)\) Substitutie (3) 00sm - Goniometrische vergelijkingen - basis - midden - 0ms a \(u = \sin(x)\) geeft \(u^{2} = -\frac{1}{2} u\) 1p ○ \(u^{2} + \frac{1}{2} u = 0\) \(2 u^{2} + u = 0\) \(u (2 u + 1) = 0\) \(u = 0 ∨ 2 u = -1\) \(u = 0 ∨ u = -\frac{1}{2}\) 1p ○ \(\sin(x) = 0 ∨ \sin(x) = -\frac{1}{2}\) geeft \(x = k \pi ∨ x = -{1 \over 6} \pi+k \cdot 2 \pi ∨ x = -{5 \over 6} \pi+k \cdot 2 \pi\) 1p 4p b \(3 \cos^{2}(x) + 4 \sin^{2}(x) + -2\frac{3}{4} = -\sin(x)\) Substitutie (4) 00sn - Goniometrische vergelijkingen - basis - midden - 1ms b (\(\sin^{2}(A) + \cos^{2}(A) = 1\) geeft) \(3 (1 - \sin^{2}(x)) + 4 \sin^{2}(x) + -2\frac{3}{4} = -\sin(x)\) ofwel \(\sin^{2}(x) + \sin(x) + \frac{1}{4} = 0\) 1p ○ \(u = \sin(x)\) geeft \(u^{2} + u + \frac{1}{4} = 0\) 1p ○ \(4 u^{2} + 4 u + 1 = 0\) \(D = 4^{2} - 4 ⋅ 4 ⋅ 1 = 0\) \(u = {-4 - \sqrt{0} \over 2 ⋅ 4} ∨ u = {-4 + \sqrt{0} \over 2 ⋅ 4}\) \(u = -\frac{1}{2}\) 1p ○ \(\sin(x) = -\frac{1}{2}\) geeft \(x = -{1 \over 6} \pi+k \cdot 2 \pi ∨ x = -{5 \over 6} \pi+k \cdot 2 \pi\) 1p

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