What is the value of 4cos(7.8*pi) ?

The value of 4cos(7.8*pi) is sqrt(5)+1

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4cos(7.8*pi)

Solution

4 cos (7.8 \cdot π)

Solution

5 + 1

Decimal Notation

3.23606 \dots

Solution steps

4 cos (7.8 π)

= 4 cos (\frac{39}{5} π)

Simplify:

\frac{39}{5} π = \frac{39 π}{5}

\frac{39}{5} π

Multiply fractions:

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

= \frac{39 π}{5}

= 4 cos (\frac{39 π}{5})

cos (\frac{39 π}{5}) = cos (\frac{9 π}{5})

cos (\frac{39 π}{5})

Rewrite

\frac{39 π}{5}

2 π \cdot 3 + \frac{9 π}{5}

= cos (2 π 3 + \frac{9 π}{5})

Apply the periodicity of

cos

cos (x + 2 π \cdot k) = cos (x)

cos (2 π \cdot 3 + \frac{9 π}{5}) = cos (\frac{9 π}{5})

= cos (\frac{9 π}{5})

= 4 cos (\frac{9 π}{5})

Rewrite using trig identities:

cos (\frac{9 π}{5}) = \frac{5 + 1}{4}

cos (\frac{9 π}{5})

Rewrite using trig identities:

cos (\frac{π}{5})

cos (\frac{9 π}{5})

Use the following identity:

cos (x) = cos (2 π - x)

cos (x)

Use the following property:

cos (θ) = cos (- θ)

cos (x) = cos (- x)

= cos (- x)

Apply the periodicity of

cos

cos (2 π + θ) = cos (θ)

cos (- x) = cos (2 π - x)

= cos (2 π - x)

= cos (2 π - \frac{9 π}{5})

Simplify:

2 π - \frac{9 π}{5} = \frac{π}{5}

2 π - \frac{9 π}{5}

Convert element to fraction:

2 π = \frac{2 π 5}{5}

= \frac{2 π 5}{5} - \frac{9 π}{5}

Since the denominators are equal, combine the fractions:

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

= \frac{2 π 5 - 9 π}{5}

2 π 5 - 9 π = π

2 π 5 - 9 π

Multiply the numbers:

2 \cdot 5 = 10

= 10 π - 9 π

Add similar elements:

10 π - 9 π = π

= π

= \frac{π}{5}

= cos (\frac{π}{5})

= cos (\frac{π}{5})

Rewrite using trig identities:

\frac{5 + 1}{4}

cos (\frac{π}{5})

Show that:

cos (\frac{π}{5}) - sin (\frac{π}{10}) = \frac{1}{2}

Use the following product to sum identity:

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

2 cos (\frac{π}{5}) sin (\frac{π}{10}) = sin (\frac{3 π}{10}) - sin (\frac{π}{10})

Show that:

2 cos (\frac{π}{5}) sin (\frac{π}{10}) = \frac{1}{2}

Use the Double Angle identity:

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

sin (\frac{2 π}{5}) = 2 sin (\frac{π}{5}) cos (\frac{π}{5})

sin (\frac{2 π}{5}) sin (\frac{π}{5}) = 4 sin (\frac{π}{5}) sin (\frac{π}{10}) cos (\frac{π}{5}) cos (\frac{π}{10})

Divide both sides by

sin (\frac{π}{5})

sin (\frac{2 π}{5}) = 4 sin (\frac{π}{10}) cos (\frac{π}{5}) cos (\frac{π}{10})

Use the following identity:

sin (x) = cos (\frac{π}{2} - x)

sin (\frac{2 π}{5}) = cos (\frac{π}{2} - \frac{2 π}{5})

cos (\frac{π}{2} - \frac{2 π}{5}) = 4 sin (\frac{π}{10}) cos (\frac{π}{5}) cos (\frac{π}{10})

cos (\frac{π}{10}) = 4 sin (\frac{π}{10}) cos (\frac{π}{5}) cos (\frac{π}{10})

Divide both sides by

cos (\frac{π}{10})

1 = 4 sin (\frac{π}{10}) cos (\frac{π}{5})

Divide both sides by

2

\frac{1}{2} = 2 sin (\frac{π}{10}) cos (\frac{π}{5})

Substitute

\frac{1}{2} = 2 sin (\frac{π}{10}) cos (\frac{π}{5})

\frac{1}{2} = sin (\frac{3 π}{10}) - sin (\frac{π}{10})

sin (\frac{3 π}{10}) = cos (\frac{π}{2} - \frac{3 π}{10})

\frac{1}{2} = cos (\frac{π}{2} - \frac{3 π}{10}) - sin (\frac{π}{10})

\frac{1}{2} = cos (\frac{π}{5}) - sin (\frac{π}{10})

Show that:

cos (\frac{π}{5}) + sin (\frac{π}{10}) = \frac{5}{4}

Use the factorization rule:

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

a = cos (\frac{π}{5}) + sin (\frac{π}{10})

(cos (\frac{π}{5}) + sin (\frac{π}{10}))^{2} - (cos (\frac{π}{5}) - sin (\frac{π}{10}))^{2} = ((cos (\frac{π}{5}) + sin (\frac{π}{10})) + (cos (\frac{π}{5}) - sin (\frac{π}{10}))) ((cos (\frac{π}{5}) + sin (\frac{π}{10})) - (cos (\frac{π}{5}) - sin (\frac{π}{10})))

Refine

(cos (\frac{π}{5}) + sin (\frac{π}{10}))^{2} - (cos (\frac{π}{5}) - sin (\frac{π}{10}))^{2} = 2 (2 cos (\frac{π}{5}) sin (\frac{π}{10}))

Show that:

2 cos (\frac{π}{5}) sin (\frac{π}{10}) = \frac{1}{2}

Use the Double Angle identity:

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

sin (\frac{2 π}{5}) = 2 sin (\frac{π}{5}) cos (\frac{π}{5})

sin (\frac{2 π}{5}) sin (\frac{π}{5}) = 4 sin (\frac{π}{5}) sin (\frac{π}{10}) cos (\frac{π}{5}) cos (\frac{π}{10})

Divide both sides by

sin (\frac{π}{5})

sin (\frac{2 π}{5}) = 4 sin (\frac{π}{10}) cos (\frac{π}{5}) cos (\frac{π}{10})

Use the following identity:

sin (x) = cos (\frac{π}{2} - x)

sin (\frac{2 π}{5}) = cos (\frac{π}{2} - \frac{2 π}{5})

cos (\frac{π}{2} - \frac{2 π}{5}) = 4 sin (\frac{π}{10}) cos (\frac{π}{5}) cos (\frac{π}{10})

cos (\frac{π}{10}) = 4 sin (\frac{π}{10}) cos (\frac{π}{5}) cos (\frac{π}{10})

Divide both sides by

cos (\frac{π}{10})

1 = 4 sin (\frac{π}{10}) cos (\frac{π}{5})

Divide both sides by

2

\frac{1}{2} = 2 sin (\frac{π}{10}) cos (\frac{π}{5})

Substitute

2 cos (\frac{π}{5}) sin (\frac{π}{10}) = \frac{1}{2}

(cos (\frac{π}{5}) + sin (\frac{π}{10}))^{2} - (cos (\frac{π}{5}) - sin (\frac{π}{10}))^{2} = 1

Substitute

cos (\frac{π}{5}) - sin (\frac{π}{10}) = \frac{1}{2}

(cos (\frac{π}{5}) + sin (\frac{π}{10}))^{2} - (\frac{1}{2})^{2} = 1

Refine

(cos (\frac{π}{5}) + sin (\frac{π}{10}))^{2} - \frac{1}{4} = 1

Add

\frac{1}{4}

to both sides

(cos (\frac{π}{5}) + sin (\frac{π}{10}))^{2} - \frac{1}{4} + \frac{1}{4} = 1 + \frac{1}{4}

Refine

(cos (\frac{π}{5}) + sin (\frac{π}{10}))^{2} = \frac{5}{4}

Take the square root of both sides

cos (\frac{π}{5}) + sin (\frac{π}{10}) = \pm \frac{5}{4}

cos (\frac{π}{5})

cannot be negative

sin (\frac{π}{10})

cannot be negative

cos (\frac{π}{5}) + sin (\frac{π}{10}) = \frac{5}{4}

Add the following equations

cos (\frac{π}{5}) + sin (\frac{π}{10}) = \frac{5}{2}

((cos (\frac{π}{5}) + sin (\frac{π}{10})) + (cos (\frac{π}{5}) - sin (\frac{π}{10}))) = (\frac{5}{2} + \frac{1}{2})

Refine

cos (\frac{π}{5}) = \frac{5 + 1}{4}

= \frac{5 + 1}{4}

= \frac{5 + 1}{4}

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

Simplify

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

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

Multiply fractions:

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

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

Cancel the common factor:

4

= 5 + 1

= 5 + 1

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Frequently Asked Questions (FAQ)

What is the value of 4cos(7.8*pi) ?
The value of 4cos(7.8*pi) is sqrt(5)+1