Is cot((15pi)/8)=cot((7pi)/8) ?

The answer to whether cot((15pi)/8)=cot((7pi)/8) is True

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prove cot((15pi)/8)=cot((7pi)/8)

Solution

prove

cot (\frac{15 π}{8}) = cot (\frac{7 π}{8})

Solution

True

Solution steps

cot (\frac{15 π}{8}) = cot (\frac{7 π}{8})

Manipulating left side

cot (\frac{15 π}{8})

Simplify

cot (\frac{15 π}{8}) : - 2 - 1

cot (\frac{15 π}{8})

cot (\frac{15 π}{8}) = cot (\frac{7 π}{8})

cot (\frac{15 π}{8})

Rewrite

\frac{15 π}{8}

π + \frac{7 π}{8}

= cot (π + \frac{7 π}{8})

Apply the periodicity of

cot

cot (x + π) = cot (x)

cot (π + \frac{7 π}{8}) = cot (\frac{7 π}{8})

= cot (\frac{7 π}{8})

= cot (\frac{7 π}{8})

Rewrite using trig identities:

\frac{1}{tan ( \frac{7 π}{8} )}

cot (\frac{7 π}{8})

Use the basic trigonometric identity:

cot (x) = \frac{1}{tan ( x )}

= \frac{1}{tan ( \frac{7 π}{8} )}

= \frac{1}{tan ( \frac{7 π}{8} )}

Rewrite using trig identities:

tan (\frac{7 π}{8}) = - 3 - 22

tan (\frac{7 π}{8})

Rewrite using trig identities:

- \frac{1 - cos ( \frac{7 π}{4} )}{1 + cos ( \frac{7 π}{4} )}

tan (\frac{7 π}{8})

Write

tan (\frac{7 π}{8})

tan (\frac{\frac{7 π}{4}}{2})

= tan (\frac{\frac{7 π}{4}}{2})

Use the Half Angle identity:

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

Rewrite using trig identities:

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 )}

Substitute

θ

with

\frac{θ}{2}

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

Simplify

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

Square root both sides

Choose the root sign according to the quadrant of

\frac{θ}{2}

range [0, \frac{π}{2}] [\frac{π}{2}, π] quadrant I II tan positive negative

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

= - \frac{1 - cos ( \frac{7 π}{4} )}{1 + cos ( \frac{7 π}{4} )}

= - \frac{1 - cos ( \frac{7 π}{4} )}{1 + cos ( \frac{7 π}{4} )}

Rewrite using trig identities:

cos (\frac{7 π}{4}) = \frac{2}{2}

cos (\frac{7 π}{4})

Rewrite using trig identities:

cos (π) cos (\frac{3 π}{4}) - sin (π) sin (\frac{3 π}{4})

cos (\frac{7 π}{4})

Write

cos (\frac{7 π}{4})

cos (π + \frac{3 π}{4})

= cos (π + \frac{3 π}{4})

Use the Angle Sum identity:

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

= cos (π) cos (\frac{3 π}{4}) - sin (π) sin (\frac{3 π}{4})

= cos (π) cos (\frac{3 π}{4}) - sin (π) sin (\frac{3 π}{4})

Use the following trivial identity:

cos (π) = (- 1)

cos (π)

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= (- 1)

Use the following trivial identity:

cos (\frac{3 π}{4}) = - \frac{2}{2}

cos (\frac{3 π}{4})

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} 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 (π) = 0

sin (π)

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= 0

Use the following trivial identity:

sin (\frac{3 π}{4}) = \frac{2}{2}

sin (\frac{3 π}{4})

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} 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}

= (- 1) (- \frac{2}{2}) - 0 \cdot \frac{2}{2}

Simplify

= \frac{2}{2}

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

Simplify

- \frac{1 - \frac{2}{2}}{1 + \frac{2}{2}} : - 3 - 22

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

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

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

Join

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

1 + \frac{2}{2}

Convert element to fraction:

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

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

Multiply the numbers:

1 \cdot 2 = 2

= \frac{2 + 2}{2}

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

Join

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

1 - \frac{2}{2}

Convert element to fraction:

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

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

Multiply the numbers:

1 \cdot 2 = 2

= \frac{2 - 2}{2}

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

Divide fractions:

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

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

Cancel the common factor:

2

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

Factor

2 - 2 : 2 (2 - 1)

2 - 2

2 = 22

= 22 - 2

Factor out common term

2

= 2 (2 - 1)

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

Factor

2 + 2 : 2 (2 + 1)

2 + 2

2 = 22

= 22 + 2

Factor out common term

2

= 2 (2 + 1)

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

Cancel the common factor:

2

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

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

\frac{2 - 1}{2 + 1} = 3 - 22

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

Multiply by the conjugate

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

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

(2 - 1) (2 - 1) = 3 - 22

(2 - 1) (2 - 1)

Apply exponent rule:

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

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

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

Add the numbers:

1 + 1 = 2

= (2 - 1)^{2}

Apply Perfect Square Formula:

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

a = 2, b = 1

= (2)^{2} - 22 \cdot 1 + 1^{2}

Simplify

(2)^{2} - 22 \cdot 1 + 1^{2} : 3 - 22

(2)^{2} - 22 \cdot 1 + 1^{2}

Apply rule

1^{a} = 1

1^{2} = 1

= (2)^{2} - 2 \cdot 1 \cdot 2 + 1

(2)^{2} = 2

(2)^{2}

Apply radical rule:

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

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

Apply exponent rule:

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

= 2^{\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

= 2

22 \cdot 1 = 22

22 \cdot 1

Multiply the numbers:

2 \cdot 1 = 2

= 22

= 2 - 22 + 1

Add the numbers:

2 + 1 = 3

= 3 - 22

= 3 - 22

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

(2 + 1) (2 - 1)

Apply Difference of Two Squares Formula:

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

a = 2, b = 1

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

Simplify

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

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

Apply rule

1^{a} = 1

1^{2} = 1

= (2)^{2} - 1

(2)^{2} = 2

(2)^{2}

Apply radical rule:

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

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

Apply exponent rule:

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

= 2^{\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

= 2

= 2 - 1

Subtract the numbers:

2 - 1 = 1

= 1

= 1

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

Apply rule

\frac{a}{1} = a

= 3 - 22

= - 3 - 22

= - 3 - 22

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

Simplify

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

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

Apply the fraction rule:

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

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

3 - 22 = 2 - 1

3 - 22

= 2 - 22 + 1

= (2)^{2} - 22 + (1)^{2}

1 = 1

1

Apply rule

1 = 1

= 1

= (2)^{2} - 22 + 1^{2}

22 \cdot 1 = 22

22 \cdot 1

Multiply the numbers:

2 \cdot 1 = 2

= 22

= (2)^{2} - 22 \cdot 1 + 1^{2}

Apply Perfect Square Formula:

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

(2)^{2} - 22 \cdot 1 + 1^{2} = (2 - 1)^{2}

= (2 - 1)^{2}

Apply radical rule:

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

= 2 - 1

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

Rationalize

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

- \frac{1}{2 - 1}

Multiply by the conjugate

\frac{2 + 1}{2 + 1}

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

1 \cdot (2 + 1) = 2 + 1

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

(2 - 1) (2 + 1)

Apply Difference of Two Squares Formula:

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

a = 2, b = 1

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

Simplify

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

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

Apply rule

1^{a} = 1

1^{2} = 1

= (2)^{2} - 1

(2)^{2} = 2

(2)^{2}

Apply radical rule:

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

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

Apply exponent rule:

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

= 2^{\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

= 2

= 2 - 1

Subtract the numbers:

2 - 1 = 1

= 1

= 1

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

Apply rule

\frac{a}{1} = a

= - (1 + 2)

Distribute parentheses

= - (2) - (1)

Apply minus-plus rules

+ (- a) = - a

= - 2 - 1

= - 2 - 1

= - 2 - 1

= - 1 - 2

Manipulating right side

cot (\frac{7 π}{8})

Simplify

cot (\frac{7 π}{8}) : - 2 - 1

cot (\frac{7 π}{8})

Rewrite using trig identities:

\frac{1}{tan ( \frac{7 π}{8} )}

cot (\frac{7 π}{8})

Use the basic trigonometric identity:

cot (x) = \frac{1}{tan ( x )}

= \frac{1}{tan ( \frac{7 π}{8} )}

= \frac{1}{tan ( \frac{7 π}{8} )}

Rewrite using trig identities:

tan (\frac{7 π}{8}) = - 3 - 22

tan (\frac{7 π}{8})

Rewrite using trig identities:

- \frac{1 - cos ( \frac{7 π}{4} )}{1 + cos ( \frac{7 π}{4} )}

tan (\frac{7 π}{8})

Write

tan (\frac{7 π}{8})

tan (\frac{\frac{7 π}{4}}{2})

= tan (\frac{\frac{7 π}{4}}{2})

Use the Half Angle identity:

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

Rewrite using trig identities:

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 )}

Substitute

θ

with

\frac{θ}{2}

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

Simplify

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

Square root both sides

Choose the root sign according to the quadrant of

\frac{θ}{2}

range [0, \frac{π}{2}] [\frac{π}{2}, π] quadrant I II tan positive negative

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

= - \frac{1 - cos ( \frac{7 π}{4} )}{1 + cos ( \frac{7 π}{4} )}

= - \frac{1 - cos ( \frac{7 π}{4} )}{1 + cos ( \frac{7 π}{4} )}

Rewrite using trig identities:

cos (\frac{7 π}{4}) = \frac{2}{2}

cos (\frac{7 π}{4})

Rewrite using trig identities:

cos (π) cos (\frac{3 π}{4}) - sin (π) sin (\frac{3 π}{4})

cos (\frac{7 π}{4})

Write

cos (\frac{7 π}{4})

cos (π + \frac{3 π}{4})

= cos (π + \frac{3 π}{4})

Use the Angle Sum identity:

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

= cos (π) cos (\frac{3 π}{4}) - sin (π) sin (\frac{3 π}{4})

= cos (π) cos (\frac{3 π}{4}) - sin (π) sin (\frac{3 π}{4})

Use the following trivial identity:

cos (π) = (- 1)

cos (π)

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= (- 1)

Use the following trivial identity:

cos (\frac{3 π}{4}) = - \frac{2}{2}

cos (\frac{3 π}{4})

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} 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 (π) = 0

sin (π)

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= 0

Use the following trivial identity:

sin (\frac{3 π}{4}) = \frac{2}{2}

sin (\frac{3 π}{4})

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} 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}

= (- 1) (- \frac{2}{2}) - 0 \cdot \frac{2}{2}

Simplify

= \frac{2}{2}

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

Simplify

- \frac{1 - \frac{2}{2}}{1 + \frac{2}{2}} : - 3 - 22

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

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

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

Join

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

1 + \frac{2}{2}

Convert element to fraction:

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

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

Multiply the numbers:

1 \cdot 2 = 2

= \frac{2 + 2}{2}

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

Join

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

1 - \frac{2}{2}

Convert element to fraction:

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

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

Multiply the numbers:

1 \cdot 2 = 2

= \frac{2 - 2}{2}

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

Divide fractions:

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

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

Cancel the common factor:

2

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

Factor

2 - 2 : 2 (2 - 1)

2 - 2

2 = 22

= 22 - 2

Factor out common term

2

= 2 (2 - 1)

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

Factor

2 + 2 : 2 (2 + 1)

2 + 2

2 = 22

= 22 + 2

Factor out common term

2

= 2 (2 + 1)

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

Cancel the common factor:

2

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

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

\frac{2 - 1}{2 + 1} = 3 - 22

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

Multiply by the conjugate

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

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

(2 - 1) (2 - 1) = 3 - 22

(2 - 1) (2 - 1)

Apply exponent rule:

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

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

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

Add the numbers:

1 + 1 = 2

= (2 - 1)^{2}

Apply Perfect Square Formula:

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

a = 2, b = 1

= (2)^{2} - 22 \cdot 1 + 1^{2}

Simplify

(2)^{2} - 22 \cdot 1 + 1^{2} : 3 - 22

(2)^{2} - 22 \cdot 1 + 1^{2}

Apply rule

1^{a} = 1

1^{2} = 1

= (2)^{2} - 2 \cdot 1 \cdot 2 + 1

(2)^{2} = 2

(2)^{2}

Apply radical rule:

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

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

Apply exponent rule:

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

= 2^{\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

= 2

22 \cdot 1 = 22

22 \cdot 1

Multiply the numbers:

2 \cdot 1 = 2

= 22

= 2 - 22 + 1

Add the numbers:

2 + 1 = 3

= 3 - 22

= 3 - 22

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

(2 + 1) (2 - 1)

Apply Difference of Two Squares Formula:

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

a = 2, b = 1

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

Simplify

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

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

Apply rule

1^{a} = 1

1^{2} = 1

= (2)^{2} - 1

(2)^{2} = 2

(2)^{2}

Apply radical rule:

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

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

Apply exponent rule:

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

= 2^{\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

= 2

= 2 - 1

Subtract the numbers:

2 - 1 = 1

= 1

= 1

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

Apply rule

\frac{a}{1} = a

= 3 - 22

= - 3 - 22

= - 3 - 22

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

Simplify

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

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

Apply the fraction rule:

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

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

3 - 22 = 2 - 1

3 - 22

= 2 - 22 + 1

= (2)^{2} - 22 + (1)^{2}

1 = 1

1

Apply rule

1 = 1

= 1

= (2)^{2} - 22 + 1^{2}

22 \cdot 1 = 22

22 \cdot 1

Multiply the numbers:

2 \cdot 1 = 2

= 22

= (2)^{2} - 22 \cdot 1 + 1^{2}

Apply Perfect Square Formula:

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

(2)^{2} - 22 \cdot 1 + 1^{2} = (2 - 1)^{2}

= (2 - 1)^{2}

Apply radical rule:

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

= 2 - 1

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

Rationalize

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

- \frac{1}{2 - 1}

Multiply by the conjugate

\frac{2 + 1}{2 + 1}

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

1 \cdot (2 + 1) = 2 + 1

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

(2 - 1) (2 + 1)

Apply Difference of Two Squares Formula:

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

a = 2, b = 1

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

Simplify

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

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

Apply rule

1^{a} = 1

1^{2} = 1

= (2)^{2} - 1

(2)^{2} = 2

(2)^{2}

Apply radical rule:

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

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

Apply exponent rule:

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

= 2^{\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

= 2

= 2 - 1

Subtract the numbers:

2 - 1 = 1

= 1

= 1

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

Apply rule

\frac{a}{1} = a

= - (1 + 2)

Distribute parentheses

= - (2) - (1)

Apply minus-plus rules

+ (- a) = - a

= - 2 - 1

= - 2 - 1

= - 2 - 1

= - 1 - 2

We showed that the two sides could take the same form

\Rightarrow True

Popular Examples

prove 7-sin(x)=1-(cos^2(x))/(1-sin(x))prove

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

prove 1/(csc^2(x))=1-sin^2(x)prove

\frac{1}{csc ^{2} ( x )} = 1 - sin^{2} (x)

prove sin^4(x)=(sin^2(x))^2 prove

sin^{4} (x) = (sin^{2} (x))^{2}

prove sin(2x)-cos(2x)= 1/2 prove

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

prove (cot(α))/(cos(α))=csc(α)prove

\frac{cot ( α )}{cos ( α )} = csc (α)

Frequently Asked Questions (FAQ)

Is cot((15pi)/8)=cot((7pi)/8) ?
The answer to whether cot((15pi)/8)=cot((7pi)/8) is True