Is cos(330)=1-2sin^2(165) ?

The answer to whether cos(330)=1-2sin^2(165) is True

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prove cos(330)=1-2sin^2(165)

Solution

prove

cos (33 0^{\circ}) = 1 - 2 sin^{2} (16 5^{\circ})

Solution

True

Solution steps

cos (33 0^{\circ}) = 1 - 2 sin^{2} (16 5^{\circ})

Manipulating left side

cos (33 0^{\circ})

Simplify

cos (33 0^{\circ}) : \frac{3}{2}

cos (33 0^{\circ})

Rewrite using trig identities:

cos (18 0^{\circ}) cos (15 0^{\circ}) - sin (18 0^{\circ}) sin (15 0^{\circ})

cos (33 0^{\circ})

Write

cos (33 0^{\circ})

cos (18 0^{\circ} + 15 0^{\circ})

= cos (18 0^{\circ} + 15 0^{\circ})

Use the Angle Sum identity:

cos (s + t) = cos (s) cos (t) - sin (s) sin (t)

= cos (18 0^{\circ}) cos (15 0^{\circ}) - sin (18 0^{\circ}) sin (15 0^{\circ})

= cos (18 0^{\circ}) cos (15 0^{\circ}) - sin (18 0^{\circ}) sin (15 0^{\circ})

Use the following trivial identity:

cos (18 0^{\circ}) = (- 1)

cos (18 0^{\circ})

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)

Use the following trivial identity:

cos (15 0^{\circ}) = - \frac{3}{2}

cos (15 0^{\circ})

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}

= - \frac{3}{2}

Use the following trivial identity:

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

sin (18 0^{\circ})

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= 0

Use the following trivial identity:

sin (15 0^{\circ}) = \frac{1}{2}

sin (15 0^{\circ})

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= \frac{1}{2}

= (- 1) (- \frac{3}{2}) - 0 \cdot \frac{1}{2}

Simplify

= \frac{3}{2}

= \frac{3}{2}

Manipulating right side

1 - 2 sin^{2} (16 5^{\circ})

Expand

1 - 2 sin^{2} (16 5^{\circ}) : \frac{3}{2}

1 - 2 sin^{2} (16 5^{\circ})

2 sin^{2} (16 5^{\circ}) = \frac{2 - 3}{2}

2 sin^{2} (16 5^{\circ})

sin^{2} (16 5^{\circ}) = \frac{2 - 3}{4}

sin^{2} (16 5^{\circ})

sin (16 5^{\circ}) = \frac{6 - 2}{4}

sin (16 5^{\circ})

Rewrite using trig identities:

sin (13 5^{\circ}) cos (3 0^{\circ}) + cos (13 5^{\circ}) sin (3 0^{\circ})

sin (16 5^{\circ})

Write

sin (16 5^{\circ})

sin (13 5^{\circ} + 3 0^{\circ})

= sin (13 5^{\circ} + 3 0^{\circ})

Use the Angle Sum identity:

sin (s + t) = sin (s) cos (t) + cos (s) sin (t)

= sin (13 5^{\circ}) cos (3 0^{\circ}) + cos (13 5^{\circ}) sin (3 0^{\circ})

= sin (13 5^{\circ}) cos (3 0^{\circ}) + cos (13 5^{\circ}) sin (3 0^{\circ})

Use the following trivial identity:

sin (13 5^{\circ}) = \frac{2}{2}

sin (13 5^{\circ})

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= \frac{2}{2}

Use the following trivial identity:

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

cos (3 0^{\circ})

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}

= \frac{3}{2}

Use the following trivial identity:

cos (13 5^{\circ}) = - \frac{2}{2}

cos (13 5^{\circ})

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}

= - \frac{2}{2}

Use the following trivial identity:

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

sin (3 0^{\circ})

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= \frac{1}{2}

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

Simplify

\frac{2}{2} \cdot \frac{3}{2} + (- \frac{2}{2}) \frac{1}{2} : \frac{6 - 2}{4}

\frac{2}{2} \cdot \frac{3}{2} + (- \frac{2}{2}) \frac{1}{2}

Remove parentheses:

(- a) = - a

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

\frac{2}{2} \cdot \frac{3}{2} = \frac{6}{4}

\frac{2}{2} \cdot \frac{3}{2}

Multiply fractions:

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

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{2 3}{4}

Simplify

23 : 6

23

Apply radical rule:

a b = a \cdot b

23 = 2 \cdot 3

= 2 \cdot 3

Multiply the numbers:

2 \cdot 3 = 6

= 6

= \frac{6}{4}

\frac{2}{2} \cdot \frac{1}{2} = \frac{2}{4}

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

Multiply fractions:

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

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

Multiply:

2 \cdot 1 = 2

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{2}{4}

= \frac{6}{4} - \frac{2}{4}

Apply rule

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

= \frac{6 - 2}{4}

= \frac{6 - 2}{4}

= (\frac{6 - 2}{4})^{2}

Apply exponent rule:

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

= \frac{( 6 - 2 ) ^{2}}{4 ^{2}}

(6 - 2)^{2} = 8 - 43

(6 - 2)^{2}

Apply Perfect Square Formula:

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

a = 6, b = 2

= (6)^{2} - 262 + (2)^{2}

Simplify

(6)^{2} - 262 + (2)^{2} : 8 - 43

(6)^{2} - 262 + (2)^{2}

(6)^{2} = 6

(6)^{2}

Apply radical rule:

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

= (6^{\frac{1}{2}})^{2}

Apply exponent rule:

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

= 6^{\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

= 6

262 = 43

262

Factor integer

6 = 2 \cdot 3

= 2 2 \cdot 3 2

Apply radical rule:

n ab = n a n b

2 \cdot 3 = 23

= 2232

Apply radical rule:

a a = a

22 = 2

= 2 \cdot 23

Multiply the numbers:

2 \cdot 2 = 4

= 43

(2)^{2} = 2

(2)^{2}

Apply radical rule:

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

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

Apply exponent rule:

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

= 2^{\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

= 2

= 6 - 43 + 2

Add the numbers:

6 + 2 = 8

= 8 - 43

= 8 - 43

= \frac{8 - 4 3}{4 ^{2}}

Factor

8 - 43 : 4 (2 - 3)

8 - 43

Rewrite as

= 4 \cdot 2 - 43

Factor out common term

4

= 4 (2 - 3)

= \frac{4 ( 2 - 3 )}{4 ^{2}}

Cancel the common factor:

4

= \frac{2 - 3}{4}

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

Multiply fractions:

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

= \frac{( 2 - 3 ) \cdot 2}{4}

Cancel the common factor:

2

= \frac{2 - 3}{2}

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

Simplify

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

Apply the fraction rule:

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

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

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

Remove parentheses:

(a) = a, - (- a) = a

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

Apply rule

\frac{a}{a} = 1

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

1 - 1 = 0

= \frac{3}{2}

= \frac{3}{2}

= \frac{3}{2}

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

Is cos(330)=1-2sin^2(165) ?
The answer to whether cos(330)=1-2sin^2(165) is True