What is the value of sqrt(sec(54)) ?

The value of sqrt(sec(54)) is sqrt((\sqrt{2)(5+sqrt(5))sqrt(5-\sqrt{5)})/(10)}

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Popular Trigonometry >

sqrt(sec(54))

Solution

sec (5 4^{\circ})

Solution

\frac{2 ( 5 + 5 ) 5 - 5}{10}

Decimal

1.30433 \dots

Solution steps

sec (5 4^{\circ})

Rewrite using trig identities:

sec (5 4^{\circ}) = \frac{2 ( 5 + 5 ) 5 - 5}{10}

sec (5 4^{\circ})

Rewrite using trig identities:

\frac{1}{cos ( 5 4 ^{\circ} )}

sec (5 4^{\circ})

Use the basic trigonometric identity:

sec (x) = \frac{1}{cos ( x )}

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

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

Rewrite using trig identities:

cos (5 4^{\circ}) = \frac{2 5 - 5}{4}

cos (5 4^{\circ})

Rewrite using trig identities:

sin (3 6^{\circ})

cos (5 4^{\circ})

Use the following identity:

cos (x) = sin (9 0^{\circ} - x)

= sin (9 0^{\circ} - 5 4^{\circ})

Simplify

= sin (3 6^{\circ})

= sin (3 6^{\circ})

Rewrite using trig identities:

\frac{2 5 - 5}{4}

sin (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}

Square both sides

(cos (3 6^{\circ}))^{2} = (\frac{5 + 1}{4})^{2}

Use the following identity:

sin^{2} (x) = 1 - cos^{2} (x)

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

Substitute

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

sin^{2} (3 6^{\circ}) = 1 - (\frac{5 + 1}{4})^{2}

Refine

sin^{2} (3 6^{\circ}) = \frac{5 - 5}{8}

Take the square root of both sides

sin (3 6^{\circ}) = \pm \frac{5 - 5}{8}

sin (3 6^{\circ})

cannot be negative

sin (3 6^{\circ}) = \frac{5 - 5}{8}

Refine

sin (3 6^{\circ}) = \frac{\frac{5 - 5}{2}}{2}

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

Simplify

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

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

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

Simplify

\frac{1}{\frac{2 5 - 5}{4}} : \frac{2 ( 5 + 5 ) 5 - 5}{10}

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

Apply the fraction rule:

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

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

Factor

4 : 2^{2}

Factor

4 = 2^{2}

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

Cancel

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

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

Apply radical rule:

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

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

Subtract the numbers:

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

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

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

Rationalize

\frac{2 2}{5 - 5} : \frac{2 ( 5 + 5 ) 5 - 5}{10}

\frac{2 2}{5 - 5}

Multiply by the conjugate

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

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

5 - 5 5 - 5 = 5 - 5

5 - 5 5 - 5

Apply radical rule:

a a = a

5 - 5 5 - 5 = 5 - 5

= 5 - 5

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

Multiply by the conjugate

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

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

(5 - 5) (5 + 5) = 20

(5 - 5) (5 + 5)

Apply Difference of Two Squares Formula:

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

a = 5, b = 5

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

Simplify

5^{2} - (5)^{2} : 20

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

5^{2} = 25

5^{2}

5^{2} = 25

= 25

(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

= 25 - 5

Subtract the numbers:

25 - 5 = 20

= 20

= 20

= \frac{2 2 ( 5 + 5 ) 5 - 5}{20}

Cancel the common factor:

2

= \frac{2 ( 5 + 5 ) 5 - 5}{10}

= \frac{2 ( 5 + 5 ) 5 - 5}{10}

= \frac{2 ( 5 + 5 ) 5 - 5}{10}

= \frac{2 ( 5 + 5 ) 5 - 5}{10}

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Frequently Asked Questions (FAQ)

What is the value of sqrt(sec(54)) ?
The value of sqrt(sec(54)) is sqrt((\sqrt{2)(5+sqrt(5))sqrt(5-\sqrt{5)})/(10)}