What is the general solution for sin(θ)*csc(18)=1 ?

The general solution for sin(θ)*csc(18)=1 is θ=0.31415…+360n,θ=180-0.31415…+360n

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sin(θ)*csc(18)=1

Solution

sin (θ) \cdot csc (1 8^{\circ}) = 1

Solution

θ = 0.31415 \dots + 36 0^{\circ} n, θ = 18 0^{\circ} - 0.31415 \dots + 36 0^{\circ} n

Radians

θ = 0.31415 \dots + 2 πn, θ = π - 0.31415 \dots + 2 πn

Solution steps

sin (θ) csc (1 8^{\circ}) = 1

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

csc (1 8^{\circ})

Rewrite using trig identities:

\frac{1}{sin ( 1 8 ^{\circ} )}

csc (1 8^{\circ})

Use the basic trigonometric identity:

csc (x) = \frac{1}{sin ( x )}

= \frac{1}{sin ( 1 8 ^{\circ} )}

= \frac{1}{sin ( 1 8 ^{\circ} )}

Rewrite using trig identities:

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

sin (1 8^{\circ})

Rewrite using trig identities:

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

sin (1 8^{\circ})

Write

sin (1 8^{\circ})

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

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

Use the Half Angle identity:

sin (\frac{θ}{2}) = \frac{1 - cos ( θ )}{2}

Use the Double Angle identity

cos (2 θ) = 1 - 2 sin^{2} (θ)

Substitute

θ

with

\frac{θ}{2}

cos (θ) = 1 - 2 sin^{2} (\frac{θ}{2})

Switch sides

2 sin^{2} (\frac{θ}{2}) = 1 - cos (θ)

Divide both sides by

2

sin^{2} (\frac{θ}{2}) = \frac{( 1 - cos ( θ ) )}{2}

Square root both sides

Choose the root sign according to the quadrant of

\frac{θ}{2}

range [0, 9 0^{\circ}] [9 0^{\circ}, 18 0^{\circ}] [18 0^{\circ}, 27 0^{\circ}] [27 0^{\circ}, 36 0^{\circ}] quadrant I II III I V sin positive positive negative negative cos positive negative negative positive

sin (\frac{θ}{2}) = \frac{( 1 - cos ( θ ) )}{2}

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

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

Rewrite using trig identities:

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

cos (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}

= \frac{5 + 1}{4}

= \frac{1 - \frac{5 + 1}{4}}{2}

Simplify

\frac{1 - \frac{5 + 1}{4}}{2} : \frac{2 3 - 5}{4}

\frac{1 - \frac{5 + 1}{4}}{2}

\frac{1 - \frac{5 + 1}{4}}{2} = \frac{3 - 5}{8}

\frac{1 - \frac{5 + 1}{4}}{2}

Join

1 - \frac{5 + 1}{4} : \frac{3 - 5}{4}

1 - \frac{5 + 1}{4}

Convert element to fraction:

1 = \frac{1 \cdot 4}{4}

= \frac{1 \cdot 4}{4} - \frac{5 + 1}{4}

Since the denominators are equal, combine the fractions:

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

= \frac{1 \cdot 4 - ( 5 + 1 )}{4}

Multiply the numbers:

1 \cdot 4 = 4

= \frac{4 - ( 1 + 5 )}{4}

Expand

4 - (5 + 1) : 3 - 5

4 - (5 + 1)

- (5 + 1) : - 5 - 1

- (5 + 1)

Distribute parentheses

= - (5) - (1)

Apply minus-plus rules

+ (- a) = - a

= - 5 - 1

= 4 - 5 - 1

Subtract the numbers:

4 - 1 = 3

= 3 - 5

= \frac{3 - 5}{4}

= \frac{\frac{3 - 5}{4}}{2}

Apply the fraction rule:

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

= \frac{3 - 5}{4 \cdot 2}

Multiply the numbers:

4 \cdot 2 = 8

= \frac{3 - 5}{8}

= \frac{3 - 5}{8}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{3 - 5}{8}

8 = 22

8

Prime factorization of

8 : 2^{3}

8

8

divides by

28 = 4 \cdot 2

= 2 \cdot 4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2

2

is a prime number, therefore no further factorization is possible

= 2 \cdot 2 \cdot 2

= 2^{3}

= 2^{3}

Apply exponent rule:

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

= 2^{2} \cdot 2

Apply radical rule:

= 2 2^{2}

Apply radical rule:

2^{2} = 2

= 22

= \frac{3 - 5}{2 2}

Rationalize

\frac{3 - 5}{2 2} : \frac{2 3 - 5}{4}

\frac{3 - 5}{2 2}

Multiply by the conjugate

\frac{2}{2}

= \frac{3 - 5 2}{2 2 2}

222 = 4

222

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 3 - 5}{4}

= \frac{2 3 - 5}{4}

= \frac{2 3 - 5}{4}

= \frac{1}{\frac{2 3 - 5}{4}}

Simplify

\frac{1}{\frac{2 3 - 5}{4}} : \frac{2 ( 3 + 5 ) 3 - 5}{2}

\frac{1}{\frac{2 3 - 5}{4}}

Apply the fraction rule:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

= \frac{4}{2 3 - 5}

Factor

4 : 2^{2}

Factor

4 = 2^{2}

= \frac{2 ^{2}}{2 3 - 5}

Cancel

\frac{2 ^{2}}{2 3 - 5} : \frac{2 ^{\frac{3}{2}}}{3 - 5}

\frac{2 ^{2}}{2 3 - 5}

Apply radical rule:

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

= \frac{2 ^{2}}{2 ^{\frac{1}{2}} 3 - 5}

Apply exponent rule:

\frac{x ^{a}}{x ^{b}} = x^{a - b}

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

= \frac{2 ^{2 - \frac{1}{2}}}{3 - 5}

Subtract the numbers:

2 - \frac{1}{2} = \frac{3}{2}

= \frac{2 ^{\frac{3}{2}}}{3 - 5}

= \frac{2 ^{\frac{3}{2}}}{3 - 5}

2^{\frac{3}{2}} = 22

2^{\frac{3}{2}}

2^{\frac{3}{2}} = 2^{1 + \frac{1}{2}}

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

Apply exponent rule:

x^{a + b} = x^{a} x^{b}

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

Refine

= 22

= \frac{2 2}{3 - 5}

Rationalize

\frac{2 2}{3 - 5} : \frac{2 ( 3 + 5 ) 3 - 5}{2}

\frac{2 2}{3 - 5}

Multiply by the conjugate

\frac{3 - 5}{3 - 5}

= \frac{2 2 3 - 5}{3 - 5 3 - 5}

3 - 5 3 - 5 = 3 - 5

3 - 5 3 - 5

Apply radical rule:

a a = a

3 - 5 3 - 5 = 3 - 5

= 3 - 5

= \frac{2 2 3 - 5}{3 - 5}

Multiply by the conjugate

\frac{3 + 5}{3 + 5}

= \frac{2 2 3 - 5 ( 3 + 5 )}{( 3 - 5 ) ( 3 + 5 )}

(3 - 5) (3 + 5) = 4

(3 - 5) (3 + 5)

Apply Difference of Two Squares Formula:

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

a = 3, b = 5

= 3^{2} - (5)^{2}

Simplify

3^{2} - (5)^{2} : 4

3^{2} - (5)^{2}

3^{2} = 9

3^{2}

3^{2} = 9

= 9

(5)^{2} = 5

(5)^{2}

Apply radical rule:

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

= (5^{\frac{1}{2}})^{2}

Apply exponent rule:

(a^{b})^{c} = a^{b c}

= 5^{\frac{1}{2} \cdot 2}

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

\frac{1}{2} \cdot 2

Multiply fractions:

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

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

Cancel the common factor:

2

= 1

= 5

= 9 - 5

Subtract the numbers:

9 - 5 = 4

= 4

= 4

= \frac{2 2 ( 3 + 5 ) 3 - 5}{4}

Cancel the common factor:

2

= \frac{2 ( 3 + 5 ) 3 - 5}{2}

= \frac{2 ( 3 + 5 ) 3 - 5}{2}

= \frac{2 ( 3 + 5 ) 3 - 5}{2}

sin (θ) \frac{2 ( 3 + 5 ) 3 - 5}{2} = 1

Multiply both sides by

2

sin (θ) \frac{2 ( 3 + 5 ) 3 - 5}{2} = 1

Multiply both sides by

2

2 sin (θ) \frac{2 ( 3 + 5 ) 3 - 5}{2} = 1 \cdot 2

Simplify

2 (3 + 5) 3 - 5 sin (θ) = 2

2 (3 + 5) 3 - 5 sin (θ) = 2

Divide both sides by

2 (3 + 5) 3 - 5

2 (3 + 5) 3 - 5 sin (θ) = 2

Divide both sides by

2 (3 + 5) 3 - 5

\frac{2 ( 3 + 5 ) 3 - 5 sin ( θ )}{2 ( 3 + 5 ) 3 - 5} = \frac{2}{2 ( 3 + 5 ) 3 - 5}

Simplify

\frac{2 ( 3 + 5 ) 3 - 5 sin ( θ )}{2 ( 3 + 5 ) 3 - 5} = \frac{2}{2 ( 3 + 5 ) 3 - 5}

Simplify

\frac{2 ( 3 + 5 ) 3 - 5 sin ( θ )}{2 ( 3 + 5 ) 3 - 5} : sin (θ)

\frac{2 ( 3 + 5 ) 3 - 5 sin ( θ )}{2 ( 3 + 5 ) 3 - 5}

Cancel the common factor:

2

= \frac{( 3 + 5 ) 3 - 5 sin ( θ )}{( 3 + 5 ) 3 - 5}

Cancel the common factor:

3 + 5

= \frac{3 - 5 sin ( θ )}{3 - 5}

Cancel the common factor:

3 - 5

= sin (θ)

Simplify

\frac{2}{2 ( 3 + 5 ) 3 - 5} : \frac{2 3 - 5}{4}

\frac{2}{2 ( 3 + 5 ) 3 - 5}

Apply radical rule:

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

= \frac{2}{2 ^{\frac{1}{2}} ( 3 + 5 ) 3 - 5}

Apply exponent rule:

\frac{x ^{a}}{x ^{b}} = x^{a - b}

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

= \frac{2 ^{1 - \frac{1}{2}}}{( 3 + 5 ) 3 - 5}

Subtract the numbers:

1 - \frac{1}{2} = \frac{1}{2}

= \frac{2 ^{\frac{1}{2}}}{( 3 + 5 ) 3 - 5}

Apply radical rule:

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

= \frac{2}{( 3 + 5 ) 3 - 5}

Rationalize

\frac{2}{( 3 + 5 ) 3 - 5} : \frac{2 3 - 5}{4}

\frac{2}{( 3 + 5 ) 3 - 5}

Multiply by the conjugate

\frac{3 - 5}{3 - 5}

= \frac{2 ( 3 - 5 )}{( 3 + 5 ) 3 - 5 ( 3 - 5 )}

(3 + 5) 3 - 5 (3 - 5) = 3 (3 - 5)^{\frac{3}{2}} + 5 (3 - 5)^{\frac{3}{2}}

(3 + 5) 3 - 5 (3 - 5)

Apply exponent rule:

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

(3 - 5) 3 - 5 = (3 - 5) (3 - 5)^{\frac{1}{2}} = (3 - 5)^{1 + \frac{1}{2}}

= (3 + 5) (3 - 5)^{1 + \frac{1}{2}}

(3 - 5)^{1 + \frac{1}{2}} = (3 - 5)^{\frac{3}{2}}

(3 - 5)^{1 + \frac{1}{2}}

Join

1 + \frac{1}{2} : \frac{3}{2}

1 + \frac{1}{2}

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

1 \cdot 2 + 1 = 3

1 \cdot 2 + 1

Multiply the numbers:

1 \cdot 2 = 2

= 2 + 1

Add the numbers:

2 + 1 = 3

= 3

= \frac{3}{2}

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

= (3 + 5) (3 - 5)^{\frac{3}{2}}

Apply the distributive law:

a (b + c) = ab + a c

a = (3 - 5)^{\frac{3}{2}}, b = 3, c = 5

= (3 - 5)^{\frac{3}{2}} \cdot 3 + (3 - 5)^{\frac{3}{2}} 5

= 3 (3 - 5)^{\frac{3}{2}} + 5 (3 - 5)^{\frac{3}{2}}

= \frac{2 ( 3 - 5 )}{3 ( 3 - 5 ) ^{\frac{3}{2}} + 5 ( 3 - 5 ) ^{\frac{3}{2}}}

Factor out common term

(3 - 5)^{\frac{3}{2}}

= \frac{2 ( 3 - 5 )}{( 3 - 5 ) ^{\frac{3}{2}} ( 3 + 5 )}

Cancel

\frac{2 ( 3 - 5 )}{( 3 - 5 ) ^{\frac{3}{2}} ( 3 + 5 )} : \frac{2 3 - 5}{( 3 - 5 ) ( 3 + 5 )}

\frac{2 ( 3 - 5 )}{( 3 - 5 ) ^{\frac{3}{2}} ( 3 + 5 )}

3 - 5 = (3 - 5)^{\frac{1}{2} + \frac{1}{2}}, (3 - 5)^{\frac{3}{2}} = (3 - 5)^{\frac{1}{2} + 1}

= \frac{2 ( 3 - 5 ) ^{\frac{1}{2} + \frac{1}{2}}}{( 3 + 5 ) ( 3 - 5 ) ^{\frac{1}{2} + 1}}

Apply exponent rule:

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

(3 - 5)^{\frac{1}{2} + \frac{1}{2}} = (3 - 5)^{\frac{1}{2}} (3 - 5)^{\frac{1}{2}}, (3 - 5)^{\frac{1}{2} + 1} = (3 - 5)^{\frac{1}{2}} (3 - 5)^{1}

= \frac{2 ( 3 - 5 ) ^{\frac{1}{2}} ( 3 - 5 ) ^{\frac{1}{2}}}{( 3 - 5 ) ^{1} ( 3 + 5 ) ( 3 - 5 ) ^{\frac{1}{2}}}

Cancel the common factor:

(3 - 5)^{\frac{1}{2}}

= \frac{2 ( 3 - 5 ) ^{\frac{1}{2}}}{( 3 - 5 ) ( 3 + 5 )}

(3 - 5)^{\frac{1}{2}} = 3 - 5, (3 - 5)^{\frac{1}{2}} = 3 - 5, (3 - 5)^{\frac{1}{2}} = 3 - 5

= \frac{2 3 - 5}{( 3 - 5 ) ( 3 + 5 )}

= \frac{2 3 - 5}{( 3 - 5 ) ( 3 + 5 )}

Factor

3 - 5 : - (5 - 3)

3 - 5

Factor out common term

- 1

= - (5 - 3)

= - \frac{2 3 - 5}{( 5 - 3 ) ( 3 + 5 )}

Cancel

- \frac{2 3 - 5}{( 5 - 3 ) ( 3 + 5 )} : \frac{2 3 - 5}{( 3 + 5 ) ( 3 - 5 )}

- \frac{2 3 - 5}{( 5 - 3 ) ( 3 + 5 )}

5 - 3 = - (3 - 5)

= - \frac{2 3 - 5}{- ( 3 + 5 ) ( 3 - 5 )}

Refine

= \frac{2 3 - 5}{( 3 + 5 ) ( 3 - 5 )}

= \frac{2 3 - 5}{( 3 + 5 ) ( 3 - 5 )}

Expand

(3 + 5) (3 - 5) : 4

(3 + 5) (3 - 5)

Apply Difference of Two Squares Formula:

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

a = 3, b = 5

= 3^{2} - (5)^{2}

Simplify

3^{2} - (5)^{2} : 4

3^{2} - (5)^{2}

3^{2} = 9

3^{2}

3^{2} = 9

= 9

(5)^{2} = 5

(5)^{2}

Apply radical rule:

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

= (5^{\frac{1}{2}})^{2}

Apply exponent rule:

(a^{b})^{c} = a^{b c}

= 5^{\frac{1}{2} \cdot 2}

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

\frac{1}{2} \cdot 2

Multiply fractions:

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

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

Cancel the common factor:

2

= 1

= 5

= 9 - 5

Subtract the numbers:

9 - 5 = 4

= 4

= 4

= \frac{2 3 - 5}{4}

= \frac{2 3 - 5}{4}

sin (θ) = \frac{2 3 - 5}{4}

sin (θ) = \frac{2 3 - 5}{4}

sin (θ) = \frac{2 3 - 5}{4}

Apply trig inverse properties

sin (θ) = \frac{2 3 - 5}{4}

General solutions for

sin (θ) = \frac{2 3 - 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 3 - 5}{4}) + 36 0^{\circ} n, θ = 18 0^{\circ} - arcsin (\frac{2 3 - 5}{4}) + 36 0^{\circ} n

θ = arcsin (\frac{2 3 - 5}{4}) + 36 0^{\circ} n, θ = 18 0^{\circ} - arcsin (\frac{2 3 - 5}{4}) + 36 0^{\circ} n

Show solutions in decimal form

θ = 0.31415 \dots + 36 0^{\circ} n, θ = 18 0^{\circ} - 0.31415 \dots + 36 0^{\circ} n

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Frequently Asked Questions (FAQ)

What is the general solution for sin(θ)*csc(18)=1 ?
The general solution for sin(θ)*csc(18)=1 is θ=0.31415…+360n,θ=180-0.31415…+360n