What is the value of cos^2(18) ?

The value of cos^2(18) is (5+sqrt(5))/8

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Popular >

cos^2(18)

Solution

cos^{2} (1 8^{\circ})

Solution

\frac{5 + 5}{8}

Decimal Notation

0.90450 \dots

Solution steps

cos^{2} (1 8^{\circ})

Rewrite using trig identities:

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

cos^{2} (1 8^{\circ})

Use the following identity:

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

Use the Double Angle identity

cos (2 θ) = 2 cos^{2} (θ) - 1

Switch sides

2 cos^{2} (θ) - 1 = cos (2 θ)

Add

1

to both sides

2 sin^{2} (θ) = 1 + cos (2 θ)

Divide both sides by

2

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

= \frac{1 + cos ( 2 \cdot 1 8 ^{\circ} )}{2}

Simplify

= \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{5 + 5}{8}

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

Join

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

1 \cdot 4 + 5 + 1 = 5 + 5

1 \cdot 4 + 5 + 1

Multiply the numbers:

1 \cdot 4 = 4

= 4 + 5 + 1

Add the numbers:

4 + 1 = 5

= 5 + 5

= \frac{5 + 5}{4}

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

Apply the fraction rule:

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

= \frac{5 + 5}{4 \cdot 2}

Multiply the numbers:

4 \cdot 2 = 8

= \frac{5 + 5}{8}

= \frac{5 + 5}{8}

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Frequently Asked Questions (FAQ)

What is the value of cos^2(18) ?
The value of cos^2(18) is (5+sqrt(5))/8