"Solve sin⁡(x)+cos⁡(x)=1" was answered by our community

eozoischgc

Answered question

2021-12-13

Solve

\sin (x) + \cos (x) = 1

Jeffery Autrey

Beginner2021-12-14Added 35 answers

\sin (A + B) = \sin A \cos B + \sin B \cos A

\sin^{2} A + \cos^{2} A = 1

Compare this equation to

r \sin (x + a) = 1

r \sin x \cos a + r \cos x \sin a = 1

\sin x + \cos x = 1

∴ \cos^{2} a + \sin^{2} a = \frac{1}{r^{2}} + \frac{1}{r^{2}} = \frac{2}{r^{2}} = 1

r^{2} = 2 \Rightarrow r = \sqrt{2}

And

\tan a = 1 \Rightarrow a = \frac{π}{4}

Therefore,

2 \sin (x + \frac{π}{4}) = 1

\sin (x + \frac{π}{4}) = \frac{1}{\sqrt{2}}

x + \frac{π}{4} = \frac{π}{4} + 2 k π \Rightarrow x = 2 k π

Solutions:

S = {2 k π, \frac{π}{2} + 2 k π}, k \in Z

Laura Worden

Beginner2021-12-15Added 45 answers

{(\sin (x) + \cos (x))}^{2} = 1^{2}

\sin^{2} (x) + 2 \sin (x) \cos (x) + \cos^{2} (x) = 1

Use this identity

\sin^{2} θ + \cos^{θ} = 1

1 + 2 \sin (x) \cos (x) = 1

2 \sin (x) \cos (x) = 0

Use the identity

2 \sin θ \cos θ = \sin 2 θ

\sin 2 x = 0

2 x = 0, π

x = 2 π n, \frac{π}{2} + 2 π n

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