What is the value of tan^2(15) ?

The value of tan^2(15) is 7-4sqrt(3)

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Popular >

tan^2(15)

Solution

tan^{2} (1 5^{\circ})

Solution

7 - 43

Decimal Notation

0.07179 \dots

Solution steps

tan^{2} (1 5^{\circ})

Rewrite using trig identities:

\frac{1 - cos ( 3 0 ^{\circ} )}{1 + cos ( 3 0 ^{\circ} )}

tan^{2} (1 5^{\circ})

Use the following identity:

tan^{2} (θ) = \frac{1 - cos ( 2 θ )}{1 + cos ( 2 θ )}

Use the following identity

tan (θ) = \frac{sin ( θ )}{cos ( θ )}

Square both sides

tan^{2} (θ) = \frac{sin ^{2} ( θ )}{cos ^{2} ( θ )}

Rewrite using trig identities:

sin^{2} (θ) = \frac{1 - cos ( 2 θ )}{2}

Use the Double Angle identity

cos (2 θ) = 1 - 2 sin^{2} (θ)

Switch sides

2 sin^{2} (θ) - 1 = - cos (2 θ)

Add

1

to both sides

2 sin^{2} (θ) = 1 - cos (2 θ)

Divide both sides by

2

sin^{2} (θ) = \frac{1 - cos ( 2 θ )}{2}

Rewrite using trig identities:

cos^{2} (θ) = \frac{1 + cos ( 2 θ )}{2}

Use the Double Angle identity

cos (2 θ) = 2 cos^{2} (θ) - 1

Switch sides

2 cos^{2} (θ) - 1 = cos (2 θ)

Add

1

to both sides

2 sin^{2} (θ) = 1 + cos (2 θ)

Divide both sides by

2

cos^{2} (θ) = \frac{1 + cos ( 2 θ )}{2}

tan^{2} (θ) = \frac{\frac{1 - c o s ( 2 θ )}{2}}{\frac{1 + c o s ( 2 θ )}{2}}

Simplify

tan^{2} (θ) = \frac{1 - cos ( 2 θ )}{1 + cos ( 2 θ )}

= \frac{1 - cos ( 2 \cdot 1 5 ^{\circ} )}{1 + cos ( 2 \cdot 1 5 ^{\circ} )}

Simplify

= \frac{1 - cos ( 3 0 ^{\circ} )}{1 + cos ( 3 0 ^{\circ} )}

= \frac{1 - cos ( 3 0 ^{\circ} )}{1 + cos ( 3 0 ^{\circ} )}

Use the following trivial identity:

cos (3 0^{\circ}) = \frac{3}{2}

cos (3 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{1 - \frac{3}{2}}{1 + \frac{3}{2}}

Simplify

\frac{1 - \frac{3}{2}}{1 + \frac{3}{2}} : 7 - 43

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

Join

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

1 + \frac{3}{2}

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

Multiply the numbers:

1 \cdot 2 = 2

= \frac{2 + 3}{2}

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

Join

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

1 - \frac{3}{2}

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

Multiply the numbers:

1 \cdot 2 = 2

= \frac{2 - 3}{2}

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

Divide fractions:

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

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

Cancel the common factor:

2

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

Rationalize

\frac{2 - 3}{2 + 3} : 7 - 43

\frac{2 - 3}{2 + 3}

Multiply by the conjugate

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

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

(2 - 3) (2 - 3) = 7 - 43

(2 - 3) (2 - 3)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

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

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

Add the numbers:

1 + 1 = 2

= (2 - 3)^{2}

Apply Perfect Square Formula:

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

a = 2, b = 3

= 2^{2} - 2 \cdot 23 + (3)^{2}

Simplify

2^{2} - 2 \cdot 23 + (3)^{2} : 7 - 43

2^{2} - 2 \cdot 23 + (3)^{2}

2^{2} = 4

2^{2}

2^{2} = 4

= 4

2 \cdot 23 = 43

2 \cdot 23

Multiply the numbers:

2 \cdot 2 = 4

= 43

(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

= 4 - 43 + 3

Add the numbers:

4 + 3 = 7

= 7 - 43

= 7 - 43

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

(2 + 3) (2 - 3)

Apply Difference of Two Squares Formula:

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

a = 2, b = 3

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

Simplify

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

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

2^{2} = 4

2^{2}

2^{2} = 4

= 4

(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

= 4 - 3

Subtract the numbers:

4 - 3 = 1

= 1

= 1

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

Apply rule

\frac{a}{1} = a

= 7 - 43

= 7 - 43

= 7 - 43

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Frequently Asked Questions (FAQ)

What is the value of tan^2(15) ?
The value of tan^2(15) is 7-4sqrt(3)