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Beliebt Trigonometrie >

cos(θ)= 1/(sqrt(1^2+(4)^2+8^2))

Lösung

cos (θ) = \frac{1}{1 ^{2} + ( 4 ) ^{2} + 8 ^{2}}

Lösung

θ = 1.45945 \dots + 2 πn, θ = 2 π - 1.45945 \dots + 2 πn

+1

Radianten

θ = 1.45945 \dots + 6.28318 \dots n, θ = 4.82372 \dots + 6.28318 \dots n

Schritte zur Lösung

cos (θ) = \frac{1}{1 ^{2} + ( 4 ) ^{2} + 8 ^{2}}

Vereinfache

\frac{1}{1 ^{2} + 4 ^{2} + 8 ^{2}} : \frac{1}{9}

\frac{1}{1 ^{2} + 4 ^{2} + 8 ^{2}}

Wende Regel an

1^{a} = 1

1^{2} = 1

= \frac{1}{1 + 4 ^{2} + 8 ^{2}}

1 + 4^{2} + 8^{2} = 81

1 + 4^{2} + 8^{2}

4^{2} = 16

= 1 + 16 + 8^{2}

8^{2} = 64

= 1 + 16 + 64

Addiere die Zahlen:

1 + 16 + 64 = 81

= 81

= \frac{1}{81}

81 = 9

81

Faktorisiere die Zahl:

81 = 9^{2}

= 9^{2}

Wende Radikal Regel an:

n a^{n} = a

9^{2} = 9

= 9

= \frac{1}{9}

Wende die Eigenschaften der Trigonometrie an

cos (θ) = \frac{1}{9}

Allgemeine Lösung für

cos (θ) = \frac{1}{9}

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

θ = arccos (\frac{1}{9}) + 2 πn, θ = 2 π - arccos (\frac{1}{9}) + 2 πn

θ = arccos (\frac{1}{9}) + 2 πn, θ = 2 π - arccos (\frac{1}{9}) + 2 πn

Zeige Lösungen in Dezimalform

θ = 1.45945 \dots + 2 πn, θ = 2 π - 1.45945 \dots + 2 πn

Graph

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