What is the value of sec(195) ?

The value of sec(195) is sqrt(2)(1-sqrt(3))

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Popular >

sec(195)

Solution

sec (19 5^{\circ})

Solution

2 (1 - 3)

Decimal Notation

- 1.03527 \dots

Solution steps

sec (19 5^{\circ})

Rewrite using trig identities:

\frac{1}{cos ( 19 5 ^{\circ} )}

sec (19 5^{\circ})

Use the basic trigonometric identity:

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

= \frac{1}{cos ( 19 5 ^{\circ} )}

= \frac{1}{cos ( 19 5 ^{\circ} )}

Rewrite using trig identities:

cos (19 5^{\circ}) = \frac{2 ( - 3 - 1 )}{4}

cos (19 5^{\circ})

Rewrite using trig identities:

cos (15 0^{\circ}) cos (4 5^{\circ}) - sin (15 0^{\circ}) sin (4 5^{\circ})

cos (19 5^{\circ})

Write

cos (19 5^{\circ})

cos (15 0^{\circ} + 4 5^{\circ})

= cos (15 0^{\circ} + 4 5^{\circ})

Use the Angle Sum identity:

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

= cos (15 0^{\circ}) cos (4 5^{\circ}) - sin (15 0^{\circ}) sin (4 5^{\circ})

= cos (15 0^{\circ}) cos (4 5^{\circ}) - sin (15 0^{\circ}) sin (4 5^{\circ})

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:

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

cos (4 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 (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}

Use the following trivial identity:

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

sin (4 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}

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

Simplify

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

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

Remove parentheses:

(- a) = - a

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

Factor out common term

\frac{2}{2}

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

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

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

Apply rule

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

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

Multiply the numbers:

2 \cdot 2 = 4

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

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

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

Simplify

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

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

Apply the fraction rule:

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

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

Factor

4 : 2^{2}

Factor

4 = 2^{2}

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

Cancel

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

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

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}}}{- 3 - 1}

Subtract the numbers:

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

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

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

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}{- 3 - 1}

Rationalize

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

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

Multiply by the conjugate

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

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

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

(- 3 - 1) (- 3 + 1)

Apply FOIL method:

(a + b) (c + d) = a c + a d + b c + b d

a = - 3, b = - 1, c = - 3, d = 1

= (- 3) (- 3) + (- 3) \cdot 1 + (- 1) (- 3) + (- 1) \cdot 1

Apply minus-plus rules

(- a) (- b) = ab, + (- a) = - a

= 33 - 1 \cdot 3 + 1 \cdot 3 - 1 \cdot 1

Simplify

33 - 1 \cdot 3 + 1 \cdot 3 - 1 \cdot 1 : 2

33 - 1 \cdot 3 + 1 \cdot 3 - 1 \cdot 1

Add similar elements:

- 1 \cdot 3 + 1 \cdot 3 = 0

= 33 - 1 \cdot 1

Apply radical rule:

a a = a

33 = 3

= 3 - 1 \cdot 1

Multiply the numbers:

1 \cdot 1 = 1

= 3 - 1

Subtract the numbers:

3 - 1 = 2

= 2

= 2

= \frac{2 2 ( - 3 + 1 )}{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(195) ?
The value of sec(195) is sqrt(2)(1-sqrt(3))