Proving ( cos^2x sin^2x 1 ) Using Maclaurin Series

In mathematics, particularly in trigonometry, it is well-known that the identity ( cos^2x sin^2x 1 ) holds true. This identity is fundamental and has numerous applications in calculus, physics, and engineering. In this article, we will demonstrate this identity using the Maclaurin series expansions of ( cos x ) and ( sin x ). The Maclaurin series is a special case of the Taylor series that approximates a function around the point ( x 0 ). By using the Maclaurin series expansions, we can explore the identity in a more analytical manner.

Maclaurin Series Expansions

Let's begin by revisiting the Maclaurin series expansions for ( cos x ) and ( sin x ):

Cosine Function

The Maclaurin series for ( cos x ) is given by:

[ cos x 1 - frac{x^2}{2!} frac{x^4}{4!} - frac{x^6}{6!} cdots sum_{n0}^{infty} frac{(-1)^n x^{2n}}{2n!} ]

Sine Function

The Maclaurin series for ( sin x ) is:

[ sin x x - frac{x^3}{3!} frac{x^5}{5!} - frac{x^7}{7!} cdots sum_{n0}^{infty} frac{(-1)^n x^{2n 1}}{(2n 1)!} ]

Squaring the Series

Now, we will compute ( cos^2 x ) and ( sin^2 x ) by squaring the Maclaurin series expansions of ( cos x ) and ( sin x ) respectively:

Cosine Squared

Starting with ( cos^2 x ):

[ cos^2 x left(1 - frac{x^2}{2!} frac{x^4}{4!} - frac{x^6}{6!} cdots right)^2 ]

Using the binomial expansion, we get:

[ cos^2 x 1 - 2 cdot frac{x^2}{2!} left(frac{x^2}{2!}right)^2 text{higher order terms} ]

Calculating the first few terms:

[ cos^2 x 1 - x^2 frac{x^4}{4} cdots ]

Sine Squared

Moving on to ( sin^2 x ):

[ sin^2 x left(x - frac{x^3}{3!} frac{x^5}{5!} - frac{x^7}{7!} cdots right)^2 ]

Calculating the first few terms:

[ sin^2 x x^2 - 2 cdot frac{x^4}{3!} left(frac{x^3}{3!}right)^2 cdots ]

Which simplifies to:

[ sin^2 x x^2 - frac{x^4}{3} cdots ]

Adding the Two Series

Now, adding the series for ( cos^2 x ) and ( sin^2 x ):

[ cos^2 x sin^2 x left(1 - x^2 frac{x^4}{4} cdotsright) left(x^2 - frac{x^4}{3} cdotsright) ]

Combining like terms, the ( x^2 ) terms cancel out:

[ cos^2 x sin^2 x 1 - frac{x^4}{4} frac{x^4}{3} cdots ]

Further simplifying the higher-order terms:

[ cos^2 x sin^2 x 1 - frac{x^4}{12} cdots ]

As ( x ) approaches 0, all higher-order terms vanish, and we find:

[ cos^2 x sin^2 x 1 ]

This confirms that the identity ( cos^2 x sin^2 x 1 ) holds true as shown by the Maclaurin series expansion.

Solving the Equation

Now, let's solve the equation ( cos^2 2x sin 2x 1 ). Using the identity ( cos^2 x 1 - sin^2 x ):

[ cos^2 2x sin 2x 1 ]

Substituting the identity, we get:

[ (1 - sin^2 2x) sin 2x 1 ]

Move all terms to one side:

[ sin^2 2x - sin 2x 0 ]

This can be factored as:

[ sin 2x (sin 2x - 1) 0 ]

The roots of this equation are:

[ sin 2x 0 quad text{and} quad sin 2x 1 ]

Values of ( x )

The solutions are:

[ 2x 0 Rightarrow x 0 ]

And for the second root:

[ 2x n frac{pi}{2} Rightarrow x n frac{pi}{4} quad text{where} quad n in mathbb{N} ]

Where ( n ) is any positive integer.

Conclusion

In conclusion, we have demonstrated that ( cos^2 x sin^2 x 1 ) using the Maclaurin series expansions. Additionally, we have found the values of ( x ) that satisfy the equation ( cos^2 2x sin 2x 1 ).