Is (sin^4(x)-1)/(cos^2(x))=cos^2(x)-2 ?

The answer to whether (sin^4(x)-1)/(cos^2(x))=cos^2(x)-2 is True

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prove (sin^4(x)-1)/(cos^2(x))=cos^2(x)-2

Solution

prove

\frac{sin ^{4} ( x ) - 1}{cos ^{2} ( x )} = cos^{2} (x) - 2

Solution

True

Solution steps

\frac{sin ^{4} ( x ) - 1}{cos ^{2} ( x )} = cos^{2} (x) - 2

Manipulating left side

\frac{sin ^{4} ( x ) - 1}{cos ^{2} ( x )}

Rewrite using trig identities

\frac{sin ^{4} ( x ) - 1}{cos ^{2} ( x )}

Use the Pythagorean identity:

cos^{2} (x) + sin^{2} (x) = 1

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

= \frac{sin ^{4} ( x ) - 1}{1 - sin ^{2} ( x )}

Simplify

\frac{sin ^{4} ( x ) - 1}{1 - sin ^{2} ( x )} : - sin^{2} (x) - 1

\frac{sin ^{4} ( x ) - 1}{1 - sin ^{2} ( x )}

Factor

sin^{4} (x) - 1 : (sin^{2} (x) + 1) (sin (x) + 1) (sin (x) - 1)

sin^{4} (x) - 1

Rewrite

sin^{4} (x) - 1

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

sin^{4} (x) - 1

Rewrite

1

1^{2}

= sin^{4} (x) - 1^{2}

Apply exponent rule:

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

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

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

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

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

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

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

Factor

sin^{2} (x) - 1 : (sin (x) + 1) (sin (x) - 1)

sin^{2} (x) - 1

Rewrite

1

1^{2}

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

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

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

= (sin (x) + 1) (sin (x) - 1)

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

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

Factor

1 - sin^{2} (x) : - (sin (x) + 1) (sin (x) - 1)

1 - sin^{2} (x)

Factor out common term

- 1

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

Factor

sin^{2} (x) - 1 : (sin (x) + 1) (sin (x) - 1)

sin^{2} (x) - 1

Rewrite

1

1^{2}

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

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

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

= (sin (x) + 1) (sin (x) - 1)

= - (sin (x) + 1) (sin (x) - 1)

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

Cancel

- \frac{( sin ^{2} ( x ) + 1 ) ( sin ( x ) + 1 ) ( sin ( x ) - 1 )}{( sin ( x ) + 1 ) ( sin ( x ) - 1 )} : - (sin^{2} (x) + 1)

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

Cancel the common factor:

sin (x) + 1

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

Cancel the common factor:

sin (x) - 1

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

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

Negate

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

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

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

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

Manipulating right side

cos^{2} (x) - 2

Rewrite using trig identities

cos^{2} (x) - 2

Use the Pythagorean identity:

cos^{2} (x) + sin^{2} (x) = 1

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

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

Simplify

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

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

Group like terms

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

Add/Subtract the numbers:

1 - 2 = - 1

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

Is (sin^4(x)-1)/(cos^2(x))=cos^2(x)-2 ?
The answer to whether (sin^4(x)-1)/(cos^2(x))=cos^2(x)-2 is True