What is the value of 4cos^2(150)+3tan(225)+9csc^2(300) ?

The value of 4cos^2(150)+3tan(225)+9csc^2(300) is 18

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4cos^2(150)+3tan(225)+9csc^2(300)

Solution

4 cos^{2} (15 0^{\circ}) + 3 tan (22 5^{\circ}) + 9 csc^{2} (30 0^{\circ})

Solution

18

Solution steps

4 cos^{2} (15 0^{\circ}) + 3 tan (22 5^{\circ}) + 9 csc^{2} (30 0^{\circ})

tan (22 5^{\circ}) = tan (4 5^{\circ})

tan (22 5^{\circ})

Rewrite

22 5^{\circ}

18 0^{\circ} + 4 5^{\circ}

= tan (18 0^{\circ} + 4 5^{\circ})

Apply the periodicity of

tan

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

tan (18 0^{\circ} + 4 5^{\circ}) = tan (4 5^{\circ})

= tan (4 5^{\circ})

= 4 cos^{2} (15 0^{\circ}) + 3 tan (4 5^{\circ}) + 9 csc^{2} (30 0^{\circ})

Rewrite using trig identities:

csc^{2} (30 0^{\circ}) = 1 + cot^{2} (12 0^{\circ})

csc^{2} (30 0^{\circ})

Use the Pythagorean identity:

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

= 1 + cot^{2} (30 0^{\circ})

cot (30 0^{\circ}) = cot (12 0^{\circ})

cot (30 0^{\circ})

Rewrite

30 0^{\circ}

18 0^{\circ} + 12 0^{\circ}

= cot (18 0^{\circ} + 12 0^{\circ})

Apply the periodicity of

cot

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

cot (18 0^{\circ} + 12 0^{\circ}) = cot (12 0^{\circ})

= cot (12 0^{\circ})

= 1 + cot^{2} (12 0^{\circ})

= 4 cos^{2} (15 0^{\circ}) + 3 tan (4 5^{\circ}) + 9 (1 + cot^{2} (12 0^{\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:

tan (4 5^{\circ}) = 1

tan (4 5^{\circ})

tan (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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

= 1

Use the following trivial identity:

cot (12 0^{\circ}) = - \frac{3}{3}

cot (12 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 (- \frac{3}{2})^{2} + 3 \cdot 1 + 9 1 + (- \frac{3}{3})^{2}

Simplify

4 (- \frac{3}{2})^{2} + 3 \cdot 1 + 9 1 + (- \frac{3}{3})^{2} : 18

4 (- \frac{3}{2})^{2} + 3 \cdot 1 + 9 1 + (- \frac{3}{3})^{2}

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

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

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

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

Apply exponent rule:

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

n

is even

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

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

Apply exponent rule:

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

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

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

Multiply fractions:

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

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

Multiply the numbers:

3 \cdot 4 = 12

= \frac{12}{2 ^{2}}

Factor

12 : 2^{2} \cdot 3

Factor

12 = 2^{2} \cdot 3

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

Cancel the common factor:

2^{2}

= 3

3 \cdot 1 = 3

3 \cdot 1

Multiply the numbers:

3 \cdot 1 = 3

= 3

9 1 + (- \frac{3}{3})^{2} = 12

9 1 + (- \frac{3}{3})^{2}

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

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

Apply exponent rule:

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

n

is even

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

= (\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}

= 9 (\frac{1}{3} + 1)

Join

1 + \frac{1}{3} : \frac{4}{3}

1 + \frac{1}{3}

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

1 \cdot 3 + 1 = 4

1 \cdot 3 + 1

Multiply the numbers:

1 \cdot 3 = 3

= 3 + 1

Add the numbers:

3 + 1 = 4

= 4

= \frac{4}{3}

= 9 \cdot \frac{4}{3}

Multiply fractions:

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

= \frac{4 \cdot 9}{3}

Multiply the numbers:

4 \cdot 9 = 36

= \frac{36}{3}

Divide the numbers:

\frac{36}{3} = 12

= 12

= 3 + 3 + 12

Add the numbers:

3 + 3 + 12 = 18

= 18

= 18

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Frequently Asked Questions (FAQ)

What is the value of 4cos^2(150)+3tan(225)+9csc^2(300) ?
The value of 4cos^2(150)+3tan(225)+9csc^2(300) is 18