What is the value of 4cot^3(135)+7tan^4(150)-csc^2(240) ?

The value of 4cot^3(135)+7tan^4(150)-csc^2(240) is -41/9

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4cot^3(135)+7tan^4(150)-csc^2(240)

Solution

4 cot^{3} (13 5^{\circ}) + 7 tan^{4} (15 0^{\circ}) - csc^{2} (24 0^{\circ})

Solution

- \frac{41}{9}

Decimal Notation

- 4.55555 \dots

Solution steps

4 cot^{3} (13 5^{\circ}) + 7 tan^{4} (15 0^{\circ}) - csc^{2} (24 0^{\circ})

Rewrite using trig identities:

csc^{2} (24 0^{\circ}) = 1 + cot^{2} (6 0^{\circ})

csc^{2} (24 0^{\circ})

Use the Pythagorean identity:

csc^{2} (x) = 1 + cot^{2} (x)

= 1 + cot^{2} (24 0^{\circ})

cot (24 0^{\circ}) = cot (6 0^{\circ})

cot (24 0^{\circ})

Rewrite

24 0^{\circ}

18 0^{\circ} + 6 0^{\circ}

= cot (18 0^{\circ} + 6 0^{\circ})

Apply the periodicity of

cot

cot (x + 18 0^{\circ}) = cot (x)

cot (18 0^{\circ} + 6 0^{\circ}) = cot (6 0^{\circ})

= cot (6 0^{\circ})

= 1 + cot^{2} (6 0^{\circ})

= 4 cot^{3} (13 5^{\circ}) + 7 tan^{4} (15 0^{\circ}) - (1 + cot^{2} (6 0^{\circ}))

Simplify

= 4 cot^{3} (13 5^{\circ}) + 7 tan^{4} (15 0^{\circ}) - 1 - cot^{2} (6 0^{\circ})

Use the following trivial identity:

cot (13 5^{\circ}) = - 1

cot (13 5^{\circ})

cot (x)

periodicity table with

18 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} cot (x) \mp \infty 31 \frac{3}{3} 0 - \frac{3}{3} - 1 - 3

= - 1

Rewrite using trig identities:

tan (15 0^{\circ}) = - \frac{3}{3}

tan (15 0^{\circ})

Rewrite using trig identities:

\frac{sin ( 15 0 ^{\circ} )}{cos ( 15 0 ^{\circ} )}

tan (15 0^{\circ})

Use the basic trigonometric identity:

tan (x) = \frac{sin ( x )}{cos ( x )}

= \frac{sin ( 15 0 ^{\circ} )}{cos ( 15 0 ^{\circ} )}

= \frac{sin ( 15 0 ^{\circ} )}{cos ( 15 0 ^{\circ} )}

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:

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}

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

Simplify

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

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

Apply the fraction rule:

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

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

Divide fractions:

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

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

Refine

= - \frac{2}{2 3}

Cancel the common factor:

2

= - \frac{1}{3}

Rationalize

- \frac{1}{3} : - \frac{3}{3}

- \frac{1}{3}

Multiply by the conjugate

\frac{3}{3}

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

1 \cdot 3 = 3

33 = 3

33

Apply radical rule:

a a = a

33 = 3

= 3

= - \frac{3}{3}

= - \frac{3}{3}

= - \frac{3}{3}

Use the following trivial identity:

cot (6 0^{\circ}) = \frac{3}{3}

cot (6 0^{\circ})

cot (x)

periodicity table with

18 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} cot (x) \mp \infty 31 \frac{3}{3} 0 - \frac{3}{3} - 1 - 3

= \frac{3}{3}

= 4 (- 1)^{3} + 7 (- \frac{3}{3})^{4} - 1 - (\frac{3}{3})^{2}

Simplify

4 (- 1)^{3} + 7 (- \frac{3}{3})^{4} - 1 - (\frac{3}{3})^{2} : - \frac{41}{9}

4 (- 1)^{3} + 7 (- \frac{3}{3})^{4} - 1 - (\frac{3}{3})^{2}

4 (- 1)^{3} = - 4

4 (- 1)^{3}

(- 1)^{3} = - 1

(- 1)^{3}

Apply exponent rule:

(- a)^{n} = - a^{n},

n

is odd

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

= - 1^{3}

Apply rule

1^{a} = 1

= - 1

= 4 (- 1)

Remove parentheses:

(- a) = - a

= - 4 \cdot 1

Multiply the numbers:

4 \cdot 1 = 4

= - 4

= - 4 + 7 (- \frac{3}{3})^{4} - 1 - (\frac{3}{3})^{2}

7 (- \frac{3}{3})^{4} = \frac{7}{9}

7 (- \frac{3}{3})^{4}

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

(- \frac{3}{3})^{4}

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- \frac{3}{3})^{4} = (\frac{3}{3})^{4}

= (\frac{3}{3})^{4}

Apply exponent rule:

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

= \frac{( 3 ) ^{4}}{3 ^{4}}

(3)^{4} : 3^{2}

Apply radical rule:

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

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

Apply exponent rule:

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

= 3^{\frac{1}{2} \cdot 4}

\frac{1}{2} \cdot 4 = 2

\frac{1}{2} \cdot 4

Multiply fractions:

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

= \frac{1 \cdot 4}{2}

Multiply the numbers:

1 \cdot 4 = 4

= \frac{4}{2}

Divide the numbers:

\frac{4}{2} = 2

= 2

= 3^{2}

= \frac{3 ^{2}}{3 ^{4}}

Apply exponent rule:

\frac{x ^{a}}{x ^{b}} = \frac{1}{x ^{b - a}}

\frac{3 ^{2}}{3 ^{4}} = \frac{1}{3 ^{4 - 2}}

= \frac{1}{3 ^{4 - 2}}

Subtract the numbers:

4 - 2 = 2

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

= 7 \cdot \frac{1}{3 ^{2}}

Multiply fractions:

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

= \frac{1 \cdot 7}{3 ^{2}}

Multiply the numbers:

1 \cdot 7 = 7

= \frac{7}{3 ^{2}}

3^{2} = 9

= \frac{7}{9}

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

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

Apply exponent rule:

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

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

(3)^{2} : 3

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

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

Cancel the common factor:

3

= \frac{1}{3}

= - 4 + \frac{7}{9} - 1 - \frac{1}{3}

Group like terms

= \frac{7}{9} - \frac{1}{3} - 4 - 1

Subtract the numbers:

- 4 - 1 = - 5

= \frac{7}{9} - 5 - \frac{1}{3}

Convert element to fraction:

5 = \frac{5}{1}

= - \frac{5}{1} + \frac{7}{9} - \frac{1}{3}

Least Common Multiplier of

1, 9, 3 : 9

1, 9, 3

Least Common Multiplier (LCM)

Prime factorization of

1

Prime factorization of

9 : 3 \cdot 3

9

9

divides by

39 = 3 \cdot 3

= 3 \cdot 3

Prime factorization of

3 : 3

3

3

is a prime number, therefore no factorization is possible

= 3

Compute a number comprised of factors that appear in at least one of the following:

1, 9, 3

= 3 \cdot 3

Multiply the numbers:

3 \cdot 3 = 9

= 9

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

9

For

\frac{5}{1} :

multiply the denominator and numerator by

9

\frac{5}{1} = \frac{5 \cdot 9}{1 \cdot 9} = \frac{45}{9}

For

\frac{1}{3} :

multiply the denominator and numerator by

3

\frac{1}{3} = \frac{1 \cdot 3}{3 \cdot 3} = \frac{3}{9}

= - \frac{45}{9} + \frac{7}{9} - \frac{3}{9}

Since the denominators are equal, combine the fractions:

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

= \frac{- 45 + 7 - 3}{9}

Add/Subtract the numbers:

- 45 + 7 - 3 = - 41

= \frac{- 41}{9}

Apply the fraction rule:

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

= - \frac{41}{9}

= - \frac{41}{9}

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Frequently Asked Questions (FAQ)

What is the value of 4cot^3(135)+7tan^4(150)-csc^2(240) ?
The value of 4cot^3(135)+7tan^4(150)-csc^2(240) is -41/9