What is the value of 2tan^2(300)+3sin^2(150)-cos^2(180) ?

The value of 2tan^2(300)+3sin^2(150)-cos^2(180) is 23/4

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2tan^2(300)+3sin^2(150)-cos^2(180)

Solution

2 tan^{2} (30 0^{\circ}) + 3 sin^{2} (15 0^{\circ}) - cos^{2} (18 0^{\circ})

Solution

\frac{23}{4}

Decimal Notation

5.75

Solution steps

2 tan^{2} (30 0^{\circ}) + 3 sin^{2} (15 0^{\circ}) - cos^{2} (18 0^{\circ})

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

tan (30 0^{\circ})

Rewrite

30 0^{\circ}

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

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

Apply the periodicity of

tan

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

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

= tan (12 0^{\circ})

= 2 tan^{2} (12 0^{\circ}) + 3 sin^{2} (15 0^{\circ}) - cos^{2} (18 0^{\circ})

Rewrite using trig identities:

tan^{2} (12 0^{\circ}) = sec^{2} (12 0^{\circ}) - 1

tan^{2} (12 0^{\circ})

Use the Pythagorean identity:

tan^{2} (x) + 1 = sec^{2} (x)

tan^{2} (x) = sec^{2} (x) - 1

= sec^{2} (12 0^{\circ}) - 1

= 2 (sec^{2} (12 0^{\circ}) - 1) + 3 sin^{2} (15 0^{\circ}) - cos^{2} (18 0^{\circ})

Use the following trivial identity:

sec (12 0^{\circ}) = - 2

sec (12 0^{\circ})

sec (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} sec (x) 1 \frac{2 3}{3} 22 Undefined - 2 - 2 - \frac{2 3}{3} 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} sec (x) - 1 - \frac{2 3}{3} - 2 - 2 Undefined 22 \frac{2 3}{3}

= - 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:

cos (18 0^{\circ}) = (- 1)

cos (18 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}

= (- 1)

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

Simplify

2 ((- 2)^{2} - 1) + 3 (\frac{1}{2})^{2} - (- 1)^{2} : \frac{23}{4}

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

2 ((- 2)^{2} - 1) = 6

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

(- 2)^{2} = 4

(- 2)^{2}

Apply exponent rule:

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

n

is even

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

= 2^{2}

2^{2} = 4

= 4

= 2 (4 - 1)

Subtract the numbers:

4 - 1 = 3

= 2 \cdot 3

Multiply the numbers:

2 \cdot 3 = 6

= 6

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

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

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

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

Apply exponent rule:

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

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

Multiply fractions:

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

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

Multiply the numbers:

1 \cdot 3 = 3

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

2^{2} = 4

= \frac{3}{4}

(- 1)^{2} = 1

(- 1)^{2}

Apply exponent rule:

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

n

is even

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

= 1^{2}

Apply rule

1^{a} = 1

= 1

= 6 + \frac{3}{4} - 1

Subtract the numbers:

6 - 1 = 5

= \frac{3}{4} + 5

Convert element to fraction:

5 = \frac{5 \cdot 4}{4}

= \frac{5 \cdot 4}{4} + \frac{3}{4}

Since the denominators are equal, combine the fractions:

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

= \frac{5 \cdot 4 + 3}{4}

5 \cdot 4 + 3 = 23

5 \cdot 4 + 3

Multiply the numbers:

5 \cdot 4 = 20

= 20 + 3

Add the numbers:

20 + 3 = 23

= 23

= \frac{23}{4}

= \frac{23}{4}

Popular Examples

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cos (2 x) = - \frac{3}{2}, - π \leq x \leq \frac{π}{2}

cos(2θ)=cos^2(θ)-1/2

cos (2 θ) = cos^{2} (θ) - \frac{1}{2}

arccos(cos((7pi)/6))

arccos (cos (\frac{7 π}{6}))

cos(60)

cos (6 0^{\circ})

sin^2(2x)=1

sin^{2} (2 x) = 1

Frequently Asked Questions (FAQ)

What is the value of 2tan^2(300)+3sin^2(150)-cos^2(180) ?
The value of 2tan^2(300)+3sin^2(150)-cos^2(180) is 23/4