Is cos(-2x)=cosh^2(x)+sinh^2(x) ?

The answer to whether cos(-2x)=cosh^2(x)+sinh^2(x) is False

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prove cos(-2x)=cosh^2(x)+sinh^2(x)

Solution

prove

cos (- 2 x) = cosh^{2} (x) + sinh^{2} (x)

Solution

False

Solution steps

cos (- 2 x) = cosh^{2} (x) + sinh^{2} (x)

Show that the two sides are not equal

For

cos (- 2 x) = cosh^{2} (x) + sinh^{2} (x)

plug in

x = 1

cos (- 2 \cdot 1) = cos (2) (Decimal : - 0.41614 \dots)

cos (- 2 \cdot 1)

Use the following property:

cos (- x) = cos (x)

cos (- 21) = cos (21)

= cos (2 \cdot 1)

= cos (2)

cosh^{2} (1) + sinh^{2} (1) = \frac{2 e ^{4} + 2}{4 e ^{2}} (Decimal : 3.76219 \dots)

cosh^{2} (1) + sinh^{2} (1)

Rewrite using trig identities:

cosh (1) = \frac{e ^{2} + 1}{2 e}

cosh (1)

Use the Hyperbolic identity:

cosh (x) = \frac{e ^{x} + e ^{- x}}{2}

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

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

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

Apply rule

a^{1} = a

e^{1} = e

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

Apply exponent rule:

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

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

Join

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

e + \frac{1}{e}

Convert element to fraction:

e = \frac{ee}{e}

= \frac{ee}{e} + \frac{1}{e}

Since the denominators are equal, combine the fractions:

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

= \frac{ee + 1}{e}

ee + 1 = e^{2} + 1

ee + 1

ee = e^{2}

ee

Apply exponent rule:

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

ee = e^{1 + 1}

= e^{1 + 1}

Add the numbers:

1 + 1 = 2

= e^{2}

= e^{2} + 1

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

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

Apply the fraction rule:

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

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

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

Rewrite using trig identities:

sinh (1) = \frac{e ^{2} - 1}{2 e}

sinh (1)

Use the Hyperbolic identity:

sinh (x) = \frac{e ^{x} - e ^{- x}}{2}

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

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

\frac{e ^{1} - e ^{- 1}}{2}

Apply rule

a^{1} = a

e^{1} = e

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

Apply exponent rule:

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

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

Join

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

e - \frac{1}{e}

Convert element to fraction:

e = \frac{ee}{e}

= \frac{ee}{e} - \frac{1}{e}

Since the denominators are equal, combine the fractions:

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

= \frac{ee - 1}{e}

ee - 1 = e^{2} - 1

ee - 1

ee = e^{2}

ee

Apply exponent rule:

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

ee = e^{1 + 1}

= e^{1 + 1}

Add the numbers:

1 + 1 = 2

= e^{2}

= e^{2} - 1

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

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

Apply the fraction rule:

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

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

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

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

Simplify

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

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

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

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

Apply exponent rule:

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

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

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

(2 e)^{2} = 2^{2} e^{2}

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

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

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

Apply exponent rule:

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

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

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

(2 e)^{2} = 2^{2} e^{2}

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

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

Apply rule

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

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

2^{2} = 4

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

Expand

(e^{2} + 1)^{2} + (e^{2} - 1)^{2} : 2 e^{4} + 2

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

(e^{2} + 1)^{2} : e^{4} + 2 e^{2} + 1

Apply Perfect Square Formula:

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

a = e^{2}, b = 1

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

Simplify

(e^{2})^{2} + 2 e^{2} \cdot 1 + 1^{2} : e^{4} + 2 e^{2} + 1

(e^{2})^{2} + 2 e^{2} \cdot 1 + 1^{2}

Apply rule

1^{a} = 1

1^{2} = 1

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

(e^{2})^{2} = e^{4}

(e^{2})^{2}

Apply exponent rule:

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

= e^{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= e^{4}

2 e^{2} \cdot 1 = 2 e^{2}

2 e^{2} \cdot 1

Multiply the numbers:

2 \cdot 1 = 2

= 2 e^{2}

= e^{4} + 2 e^{2} + 1

= e^{4} + 2 e^{2} + 1

= e^{4} + 2 e^{2} + 1 + (e^{2} - 1)^{2}

(e^{2} - 1)^{2} : e^{4} - 2 e^{2} + 1

Apply Perfect Square Formula:

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

a = e^{2}, b = 1

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

Simplify

(e^{2})^{2} - 2 e^{2} \cdot 1 + 1^{2} : e^{4} - 2 e^{2} + 1

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

Apply rule

1^{a} = 1

1^{2} = 1

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

(e^{2})^{2} = e^{4}

(e^{2})^{2}

Apply exponent rule:

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

= e^{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= e^{4}

2 e^{2} \cdot 1 = 2 e^{2}

2 e^{2} \cdot 1

Multiply the numbers:

2 \cdot 1 = 2

= 2 e^{2}

= e^{4} - 2 e^{2} + 1

= e^{4} - 2 e^{2} + 1

= e^{4} + 2 e^{2} + 1 + e^{4} - 2 e^{2} + 1

Simplify

e^{4} + 2 e^{2} + 1 + e^{4} - 2 e^{2} + 1 : 2 e^{4} + 2

e^{4} + 2 e^{2} + 1 + e^{4} - 2 e^{2} + 1

Group like terms

= e^{4} + e^{4} + 2 e^{2} - 2 e^{2} + 1 + 1

Add similar elements:

2 e^{2} - 2 e^{2} = 0

= e^{4} + e^{4} + 1 + 1

Add similar elements:

e^{4} + e^{4} = 2 e^{4}

= 2 e^{4} + 1 + 1

Add the numbers:

1 + 1 = 2

= 2 e^{4} + 2

= 2 e^{4} + 2

= \frac{2 e ^{4} + 2}{4 e ^{2}}

= \frac{2 e ^{4} + 2}{4 e ^{2}}

The two sides are not equal

\Rightarrow False

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Frequently Asked Questions (FAQ)

Is cos(-2x)=cosh^2(x)+sinh^2(x) ?
The answer to whether cos(-2x)=cosh^2(x)+sinh^2(x) is False