9.1 Solving trigonometric equations with identities (Page 4/9)

Algebra and trigonometry Page 4 / 9

Verifying an identity using algebra and even/odd identities

Verify the identity:

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

Let’s start with the left side and simplify:

\begin{matrix} \frac{\sin^{2} (- θ) - \cos^{2} (- θ)}{\sin (- θ) - \cos (- θ)} & = & \frac{{[\sin (- θ)]}^{2} - {[\cos (- θ)]}^{2}}{\sin (- θ) - \cos (- θ)} \\ = & \frac{{(- \sin θ)}^{2} - {(\cos θ)}^{2}}{- \sin θ - \cos θ} & \sin (- x) = - \sin x and \cos (- x) = \cos x \\ = & \frac{{(\sin θ)}^{2} - {(\cos θ)}^{2}}{- \sin θ - \cos θ} & Difference of squares \\ = & \frac{(\sin θ - \cos θ) (\sin θ + \cos θ)}{- (\sin θ + \cos θ)} \\ = & \frac{(\sin θ - \cos θ) (\sin θ + \cos θ)}{- (\sin θ + \cos θ)} \\ = & \cos θ - \sin θ \end{matrix}

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Verify the identity $\frac{\sin^{2} θ - 1}{\tan θ \sin θ - \tan θ} = \frac{\sin θ + 1}{\tan θ} .$

$\begin{matrix} \frac{\sin^{2} θ - 1}{\tan θ \sin θ - \tan θ} & = & \frac{(\sin θ + 1) (\sin θ - 1)}{\tan θ (\sin θ - 1)} \\ = & \frac{\sin θ + 1}{\tan θ} \end{matrix}$

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Verifying an identity involving cosines and cotangents

Verify the identity: $(1 - \cos^{2} x) (1 + \cot^{2} x) = 1.$

We will work on the left side of the equation.

\begin{matrix} (1 - \cos^{2} x) (1 + \cot^{2} x) & = & (1 - \cos^{2} x) (1 + \frac{\cos^{2} x}{\sin^{2} x}) \\ = & (1 - \cos^{2} x) (\frac{\sin^{2} x}{\sin^{2} x} + \frac{\cos^{2} x}{\sin^{2} x}) & Find the common denominator . \\ = & (1 - \cos^{2} x) (\frac{\sin^{2} x + \cos^{2} x}{\sin^{2} x}) \\ = & (\sin^{2} x) (\frac{1}{\sin^{2} x}) \\ = & 1 \end{matrix}

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Using algebra to simplify trigonometric expressions

We have seen that algebra is very important in verifying trigonometric identities, but it is just as critical in simplifying trigonometric expressions before solving. Being familiar with the basic properties and formulas of algebra, such as the difference of squares formula, the perfect square formula, or substitution, will simplify the work involved with trigonometric expressions and equations.

For example, the equation $(\sin x + 1) (\sin x - 1) = 0$ resembles the equation $(x + 1) (x - 1) = 0,$ which uses the factored form of the difference of squares. Using algebra makes finding a solution straightforward and familiar. We can set each factor equal to zero and solve. This is one example of recognizing algebraic patterns in trigonometric expressions or equations.

Another example is the difference of squares formula, $a^{2} - b^{2} = (a - b) (a + b),$ which is widely used in many areas other than mathematics, such as engineering, architecture, and physics. We can also create our own identities by continually expanding an expression and making the appropriate substitutions. Using algebraic properties and formulas makes many trigonometric equations easier to understand and solve.

Writing the trigonometric expression as an algebraic expression

Write the following trigonometric expression as an algebraic expression: $2 \cos^{2} θ + \cos θ - 1.$

Notice that the pattern displayed has the same form as a standard quadratic expression, $a x^{2} + b x + c .$ Letting $\cos θ = x,$ we can rewrite the expression as follows:

2 x^{2} + x - 1

This expression can be factored as $(2 x + 1) (x - 1) .$ If it were set equal to zero and we wanted to solve the equation, we would use the zero factor property and solve each factor for $x .$ At this point, we would replace $x$ with $\cos θ$ and solve for $θ .$

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Rewriting a trigonometric expression using the difference of squares

Rewrite the trigonometric expression using the difference of squares: $4 \cos^{2} θ - 1.$

Notice that both the coefficient and the trigonometric expression in the first term are squared, and the square of the number 1 is 1. This is the difference of squares.

\begin{matrix} 4 \cos^{2} θ - 1 & = & {(2 \cos θ)}^{2} - 1 \\ = & (2 \cos θ - 1) (2 \cos θ + 1) \end{matrix}

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Rewrite the trigonometric expression using the difference of squares: $25 - 9 \sin^{2} θ .$

This is a difference of squares formula: $25 - 9 \sin^{2} θ = (5 - 3 \sin θ) (5 + 3 \sin θ) .$

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Source: OpenStax, Algebra and trigonometry. OpenStax CNX. Nov 14, 2016 Download for free at https://legacy.cnx.org/content/col11758/1.6

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