What is the general solution for sin(216)=sin(θ) ?

The general solution for sin(216)=sin(θ) is θ=-0.62831…+360n,θ=180+0.62831…+360n

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Popular >

sin(216)=sin(θ)

Solution

sin (21 6^{\circ}) = sin (θ)

Solution

θ = - 0.62831 \dots + 36 0^{\circ} n, θ = 18 0^{\circ} + 0.62831 \dots + 36 0^{\circ} n

Radians

θ = - 0.62831 \dots + 2 πn, θ = π + 0.62831 \dots + 2 πn

Solution steps

sin (21 6^{\circ}) = sin (θ)

sin (21 6^{\circ}) = - \frac{2 5 - 5}{4}

sin (21 6^{\circ})

Rewrite using trig identities:

- sin (14 4^{\circ})

sin (21 6^{\circ})

Use the following identity:

sin (x) = - sin (36 0^{\circ} - x)

sin (x)

Use the following property:

sin (θ) = - sin (- θ)

sin (x) = - sin (- x)

= - sin (- x)

Apply the periodicity of

sin

sin (36 0^{\circ} + θ) = sin (θ)

- sin (- x) = - sin (36 0^{\circ} - x)

= - sin (36 0^{\circ} - x)

= - sin (36 0^{\circ} - 21 6^{\circ})

Simplify:

36 0^{\circ} - 21 6^{\circ} = 14 4^{\circ}

36 0^{\circ} - 21 6^{\circ}

Convert element to fraction:

36 0^{\circ} = 36 0^{\circ}

= 36 0^{\circ} - 21 6^{\circ}

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{36 0 ^{\circ} 5 - 108 0 ^{\circ}}{5}

36 0^{\circ} 5 - 108 0^{\circ} = 72 0^{\circ}

36 0^{\circ} 5 - 108 0^{\circ}

Multiply the numbers:

2 \cdot 5 = 10

= 180 0^{\circ} - 108 0^{\circ}

Add similar elements:

180 0^{\circ} - 108 0^{\circ} = 72 0^{\circ}

= 72 0^{\circ}

= 14 4^{\circ}

= - sin (14 4^{\circ})

= - sin (14 4^{\circ})

Rewrite using trig identities:

sin (14 4^{\circ}) = \frac{2 5 - 5}{4}

sin (14 4^{\circ})

Rewrite using trig identities:

sin (3 6^{\circ})

sin (14 4^{\circ})

Use the basic trigonometric identity:

sin (x) = sin (18 0^{\circ} - x)

= sin (18 0^{\circ} - 14 4^{\circ})

Simplify:

18 0^{\circ} - 14 4^{\circ} = 3 6^{\circ}

18 0^{\circ} - 14 4^{\circ}

Convert element to fraction:

18 0^{\circ} = 18 0^{\circ}

= 18 0^{\circ} - 14 4^{\circ}

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{18 0 ^{\circ} 5 - 72 0 ^{\circ}}{5}

Add similar elements:

90 0^{\circ} - 72 0^{\circ} = 18 0^{\circ}

= 3 6^{\circ}

= sin (3 6^{\circ})

= sin (3 6^{\circ})

Rewrite using trig identities:

\frac{2 5 - 5}{4}

sin (3 6^{\circ})

Show that:

cos (3 6^{\circ}) - sin (1 8^{\circ}) = \frac{1}{2}

Use the following product to sum identity:

2 sin (x) cos (y) = sin (x + y) - sin (x - y)

2 cos (3 6^{\circ}) sin (1 8^{\circ}) = sin (5 4^{\circ}) - sin (1 8^{\circ})

Show that:

2 cos (3 6^{\circ}) sin (1 8^{\circ}) = \frac{1}{2}

Use the Double Angle identity:

sin (2 x) = 2 sin (x) cos (x)

sin (7 2^{\circ}) = 2 sin (3 6^{\circ}) cos (3 6^{\circ})

sin (7 2^{\circ}) sin (3 6^{\circ}) = 4 sin (3 6^{\circ}) sin (1 8^{\circ}) cos (3 6^{\circ}) cos (1 8^{\circ})

Divide both sides by

sin (3 6^{\circ})

sin (7 2^{\circ}) = 4 sin (1 8^{\circ}) cos (3 6^{\circ}) cos (1 8^{\circ})

Use the following identity:

sin (x) = cos (9 0^{\circ} - x)

sin (7 2^{\circ}) = cos (9 0^{\circ} - 7 2^{\circ})

cos (9 0^{\circ} - 7 2^{\circ}) = 4 sin (1 8^{\circ}) cos (3 6^{\circ}) cos (1 8^{\circ})

cos (1 8^{\circ}) = 4 sin (1 8^{\circ}) cos (3 6^{\circ}) cos (1 8^{\circ})

Divide both sides by

cos (1 8^{\circ})

1 = 4 sin (1 8^{\circ}) cos (3 6^{\circ})

Divide both sides by

2

\frac{1}{2} = 2 sin (1 8^{\circ}) cos (3 6^{\circ})

Substitute

\frac{1}{2} = 2 sin (1 8^{\circ}) cos (3 6^{\circ})

\frac{1}{2} = sin (5 4^{\circ}) - sin (1 8^{\circ})

sin (5 4^{\circ}) = cos (9 0^{\circ} - 5 4^{\circ})

\frac{1}{2} = cos (9 0^{\circ} - 5 4^{\circ}) - sin (1 8^{\circ})

\frac{1}{2} = cos (3 6^{\circ}) - sin (1 8^{\circ})

Show that:

cos (3 6^{\circ}) + sin (1 8^{\circ}) = \frac{5}{4}

Use the factorization rule:

a^{2} - b^{2} = (a + b) (a - b)

a = cos (3 6^{\circ}) + sin (1 8^{\circ})

(cos (3 6^{\circ}) + sin (1 8^{\circ}))^{2} - (cos (3 6^{\circ}) - sin (1 8^{\circ}))^{2} = ((cos (3 6^{\circ}) + sin (1 8^{\circ})) + (cos (3 6^{\circ}) - sin (1 8^{\circ}))) ((cos (3 6^{\circ}) + sin (1 8^{\circ})) - (cos (3 6^{\circ}) - sin (1 8^{\circ})))

Refine

(cos (3 6^{\circ}) + sin (1 8^{\circ}))^{2} - (cos (3 6^{\circ}) - sin (1 8^{\circ}))^{2} = 2 (2 cos (3 6^{\circ}) sin (1 8^{\circ}))

Show that:

2 cos (3 6^{\circ}) sin (1 8^{\circ}) = \frac{1}{2}

Use the Double Angle identity:

sin (2 x) = 2 sin (x) cos (x)

sin (7 2^{\circ}) = 2 sin (3 6^{\circ}) cos (3 6^{\circ})

sin (7 2^{\circ}) sin (3 6^{\circ}) = 4 sin (3 6^{\circ}) sin (1 8^{\circ}) cos (3 6^{\circ}) cos (1 8^{\circ})

Divide both sides by

sin (3 6^{\circ})

sin (7 2^{\circ}) = 4 sin (1 8^{\circ}) cos (3 6^{\circ}) cos (1 8^{\circ})

Use the following identity:

sin (x) = cos (9 0^{\circ} - x)

sin (7 2^{\circ}) = cos (9 0^{\circ} - 7 2^{\circ})

cos (9 0^{\circ} - 7 2^{\circ}) = 4 sin (1 8^{\circ}) cos (3 6^{\circ}) cos (1 8^{\circ})

cos (1 8^{\circ}) = 4 sin (1 8^{\circ}) cos (3 6^{\circ}) cos (1 8^{\circ})

Divide both sides by

cos (1 8^{\circ})

1 = 4 sin (1 8^{\circ}) cos (3 6^{\circ})

Divide both sides by

2

\frac{1}{2} = 2 sin (1 8^{\circ}) cos (3 6^{\circ})

Substitute

2 cos (3 6^{\circ}) sin (1 8^{\circ}) = \frac{1}{2}

(cos (3 6^{\circ}) + sin (1 8^{\circ}))^{2} - (cos (3 6^{\circ}) - sin (1 8^{\circ}))^{2} = 1

Substitute

cos (3 6^{\circ}) - sin (1 8^{\circ}) = \frac{1}{2}

(cos (3 6^{\circ}) + sin (1 8^{\circ}))^{2} - (\frac{1}{2})^{2} = 1

Refine

(cos (3 6^{\circ}) + sin (1 8^{\circ}))^{2} - \frac{1}{4} = 1

Add

\frac{1}{4}

to both sides

(cos (3 6^{\circ}) + sin (1 8^{\circ}))^{2} - \frac{1}{4} + \frac{1}{4} = 1 + \frac{1}{4}

Refine

(cos (3 6^{\circ}) + sin (1 8^{\circ}))^{2} = \frac{5}{4}

Take the square root of both sides

cos (3 6^{\circ}) + sin (1 8^{\circ}) = \pm \frac{5}{4}

cos (3 6^{\circ})

cannot be negative

sin (1 8^{\circ})

cannot be negative

cos (3 6^{\circ}) + sin (1 8^{\circ}) = \frac{5}{4}

Add the following equations

cos (3 6^{\circ}) + sin (1 8^{\circ}) = \frac{5}{2}

((cos (3 6^{\circ}) + sin (1 8^{\circ})) + (cos (3 6^{\circ}) - sin (1 8^{\circ}))) = (\frac{5}{2} + \frac{1}{2})

Refine

cos (3 6^{\circ}) = \frac{5 + 1}{4}

Square both sides

(cos (3 6^{\circ}))^{2} = (\frac{5 + 1}{4})^{2}

Use the following identity:

sin^{2} (x) = 1 - cos^{2} (x)

sin^{2} (3 6^{\circ}) = 1 - cos^{2} (3 6^{\circ})

Substitute

cos (3 6^{\circ}) = \frac{5 + 1}{4}

sin^{2} (3 6^{\circ}) = 1 - (\frac{5 + 1}{4})^{2}

Refine

sin^{2} (3 6^{\circ}) = \frac{5 - 5}{8}

Take the square root of both sides

sin (3 6^{\circ}) = \pm \frac{5 - 5}{8}

sin (3 6^{\circ})

cannot be negative

sin (3 6^{\circ}) = \frac{5 - 5}{8}

Refine

sin (3 6^{\circ}) = \frac{\frac{5 - 5}{2}}{2}

= \frac{\frac{5 - 5}{2}}{2}

\frac{\frac{5 - 5}{2}}{2} = \frac{2 5 - 5}{4}

\frac{\frac{5 - 5}{2}}{2}

\frac{5 - 5}{2} = \frac{5 - 5}{2}

\frac{5 - 5}{2}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{5 - 5}{2}

= \frac{\frac{5 - 5}{2}}{2}

Apply the fraction rule:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= \frac{5 - 5}{2 \cdot 2}

Rationalize

\frac{5 - 5}{2 2} : \frac{2 5 - 5}{4}

\frac{5 - 5}{2 2}

Multiply by the conjugate

\frac{2}{2}

= \frac{5 - 5 2}{2 \cdot 2 2}

2 \cdot 22 = 4

2 \cdot 22

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

222 = 2 \cdot 2^{\frac{1}{2}} \cdot 2^{\frac{1}{2}} = 2^{1 + \frac{1}{2} + \frac{1}{2}}

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

Add similar elements:

\frac{1}{2} + \frac{1}{2} = 2 \cdot \frac{1}{2}

= 2^{1 + 2 \cdot \frac{1}{2}}

2 \cdot \frac{1}{2} = 1

2 \cdot \frac{1}{2}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 2}{2}

Cancel the common factor:

2

= 1

= 2^{1 + 1}

Add the numbers:

1 + 1 = 2

= 2^{2}

2^{2} = 4

= 4

= \frac{2 5 - 5}{4}

= \frac{2 5 - 5}{4}

= \frac{2 5 - 5}{4}

= \frac{2 5 - 5}{4}

= - \frac{2 5 - 5}{4}

- \frac{2 5 - 5}{4} = sin (θ)

Switch sides

sin (θ) = - \frac{2 5 - 5}{4}

Apply trig inverse properties

sin (θ) = - \frac{2 5 - 5}{4}

General solutions for

sin (θ) = - \frac{2 5 - 5}{4}

sin (x) = - a \Rightarrow x = arcsin (- a) + 36 0^{\circ} n, x = 18 0^{\circ} + arcsin (a) + 36 0^{\circ} n

θ = arcsin (- \frac{2 5 - 5}{4}) + 36 0^{\circ} n, θ = 18 0^{\circ} + arcsin (\frac{2 5 - 5}{4}) + 36 0^{\circ} n

θ = arcsin (- \frac{2 5 - 5}{4}) + 36 0^{\circ} n, θ = 18 0^{\circ} + arcsin (\frac{2 5 - 5}{4}) + 36 0^{\circ} n

Show solutions in decimal form

θ = - 0.62831 \dots + 36 0^{\circ} n, θ = 18 0^{\circ} + 0.62831 \dots + 36 0^{\circ} n

Popular Examples

cos(θ)=-0.2

cos (θ) = - 0.2

sin^2(x)-5cos(x)=3

sin^{2} (x) - 5 cos (x) = 3

cos(x)=-9/41

cos (x) = - \frac{9}{41}

5sin(y)=3sin(y)+sqrt(3)

5 sin (y) = 3 sin (y) + 3

2sin^2(θ)+sin^2(θ)=0

2 sin^{2} (θ) + sin^{2} (θ) = 0

Frequently Asked Questions (FAQ)

What is the general solution for sin(216)=sin(θ) ?
The general solution for sin(216)=sin(θ) is θ=-0.62831…+360n,θ=180+0.62831…+360n