What is the value of sec(-(7pi)/(12)) ?

The value of sec(-(7pi)/(12)) is -sqrt(2)(1+sqrt(3))

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Popular >

sec(-(7pi)/(12))

Solution

sec (- \frac{7 π}{12})

Solution

- 2 (1 + 3)

Decimal Notation

- 3.86370 \dots

Solution steps

sec (- \frac{7 π}{12})

Use the following property:

sec (- x) = sec (x)

sec (- \frac{7 π}{12}) = sec (\frac{7 π}{12})

= sec (\frac{7 π}{12})

Rewrite using trig identities:

\frac{1}{cos ( \frac{7 π}{12} )}

sec (\frac{7 π}{12})

Use the basic trigonometric identity:

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

= \frac{1}{cos ( \frac{7 π}{12} )}

= \frac{1}{cos ( \frac{7 π}{12} )}

Rewrite using trig identities:

cos (\frac{7 π}{12}) = \frac{2 ( 1 - 3 )}{4}

cos (\frac{7 π}{12})

Rewrite using trig identities:

cos (\frac{π}{3}) cos (\frac{π}{4}) - sin (\frac{π}{3}) sin (\frac{π}{4})

cos (\frac{7 π}{12})

Write

cos (\frac{7 π}{12})

cos (\frac{π}{3} + \frac{π}{4})

= cos (\frac{π}{3} + \frac{π}{4})

Use the Angle Sum identity:

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

= cos (\frac{π}{3}) cos (\frac{π}{4}) - sin (\frac{π}{3}) sin (\frac{π}{4})

= cos (\frac{π}{3}) cos (\frac{π}{4}) - sin (\frac{π}{3}) sin (\frac{π}{4})

Use the following trivial identity:

cos (\frac{π}{3}) = \frac{1}{2}

cos (\frac{π}{3})

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= \frac{1}{2}

Use the following trivial identity:

cos (\frac{π}{4}) = \frac{2}{2}

cos (\frac{π}{4})

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} 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 (\frac{π}{3}) = \frac{3}{2}

sin (\frac{π}{3})

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= \frac{3}{2}

Use the following trivial identity:

sin (\frac{π}{4}) = \frac{2}{2}

sin (\frac{π}{4})

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} 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}

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

Simplify

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

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

Factor out common term

\frac{2}{2}

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

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

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

Apply rule

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

= \frac{1 - 3}{2}

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

Multiply fractions:

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

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{2 ( 1 - 3 )}{4}

= \frac{2 ( 1 - 3 )}{4}

= \frac{1}{\frac{2 ( 1 - 3 )}{4}}

Simplify

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

\frac{1}{\frac{2 ( 1 - 3 )}{4}}

Apply the fraction rule:

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

= \frac{4}{2 ( 1 - 3 )}

Factor

4 : 2^{2}

Factor

4 = 2^{2}

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

Cancel

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

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

Apply radical rule:

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

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

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}}}{1 - 3}

Subtract the numbers:

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

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

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

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}{1 - 3}

Rationalize

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

\frac{2 2}{1 - 3}

Multiply by the conjugate

\frac{1 + 3}{1 + 3}

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

(1 - 3) (1 + 3) = - 2

(1 - 3) (1 + 3)

Apply Difference of Two Squares Formula:

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

a = 1, b = 3

= 1^{2} - (3)^{2}

Simplify

1^{2} - (3)^{2} : - 2

1^{2} - (3)^{2}

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - (3)^{2}

(3)^{2} = 3

(3)^{2}

Apply radical rule:

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

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

Apply exponent rule:

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

= 3^{\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

= 3

= 1 - 3

Subtract the numbers:

1 - 3 = - 2

= - 2

= - 2

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

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

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

Divide the numbers:

\frac{2}{2} = 1

= - 2 (1 + 3)

= - 2 (1 + 3)

= - 2 (1 + 3)

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Frequently Asked Questions (FAQ)

What is the value of sec(-(7pi)/(12)) ?
The value of sec(-(7pi)/(12)) is -sqrt(2)(1+sqrt(3))