Is sin(15)=sqrt((1-cos(30))/2) ?

The answer to whether sin(15)=sqrt((1-cos(30))/2) is True

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prove sin(15)=sqrt((1-cos(30))/2)

Solution

prove

sin (1 5^{\circ}) = \frac{1 - cos ( 3 0 ^{\circ} )}{2}

Solution

True

Solution steps

sin (1 5^{\circ}) = \frac{1 - cos ( 3 0 ^{\circ} )}{2}

Manipulating left side

sin (1 5^{\circ})

Rewrite using trig identities

sin (1 5^{\circ})

= sin (\frac{3 0 ^{\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 0 ^{\circ} )}{2} = \frac{2 - 3}{2}

\frac{1 - cos ( 3 0 ^{\circ} )}{2}

\frac{1 - cos ( 3 0 ^{\circ} )}{2} = \frac{2 - 3}{4}

\frac{1 - cos ( 3 0 ^{\circ} )}{2}

1 - cos (3 0^{\circ}) = 1 - \frac{3}{2}

1 - cos (3 0^{\circ})

Use the following trivial identity:

cos (3 0^{\circ}) = \frac{3}{2}

cos (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= 1 - \frac{3}{2}

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

Join

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

1 - \frac{3}{2}

Convert element to fraction:

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

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

Multiply the numbers:

1 \cdot 2 = 2

= \frac{2 - 3}{2}

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

Apply the fraction rule:

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

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{2 - 3}{4}

= \frac{2 - 3}{4}

Simplify

\frac{2 - 3}{4}

Apply radical rule:

n \frac{a}{b} = \frac{n a}{n b},

assuming

a \geq 0, b \geq 0

= \frac{2 - 3}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

n a^{n} = a

2^{2} = 2

= 2

= \frac{2 - 3}{2}

= \frac{2 - 3}{2}

= \frac{2 - 3}{2}

= \frac{2 - 3}{2}

Manipulating right side

\frac{1 - cos ( 3 0 ^{\circ} )}{2}

Simplify

\frac{1 - cos ( 3 0 ^{\circ} )}{2} : \frac{2 - 3}{2}

\frac{1 - cos ( 3 0 ^{\circ} )}{2}

\frac{1 - cos ( 3 0 ^{\circ} )}{2} = \frac{2 - 3}{4}

\frac{1 - cos ( 3 0 ^{\circ} )}{2}

1 - cos (3 0^{\circ}) = 1 - \frac{3}{2}

1 - cos (3 0^{\circ})

Use the following trivial identity:

cos (3 0^{\circ}) = \frac{3}{2}

cos (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= 1 - \frac{3}{2}

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

Join

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

1 - \frac{3}{2}

Convert element to fraction:

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

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

Multiply the numbers:

1 \cdot 2 = 2

= \frac{2 - 3}{2}

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

Apply the fraction rule:

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

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{2 - 3}{4}

= \frac{2 - 3}{4}

Simplify

\frac{2 - 3}{4}

Apply radical rule:

n \frac{a}{b} = \frac{n a}{n b},

assuming

a \geq 0, b \geq 0

= \frac{2 - 3}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

n a^{n} = a

2^{2} = 2

= 2

= \frac{2 - 3}{2}

= \frac{2 - 3}{2}

= \frac{2 - 3}{2}

We showed that the two sides could take the same form

\Rightarrow True

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Is sin(15)=sqrt((1-cos(30))/2) ?
The answer to whether sin(15)=sqrt((1-cos(30))/2) is True