9.5 Solving trigonometric equations  (Page 6/10)

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Using the pythagorean theorem to model an abstract problem

OSHA safety regulations require that the base of a ladder be placed 1 foot from the wall for every 4 feet of ladder length. Find the angle that a ladder of any length forms with the ground and the height at which the ladder touches the wall.

For any length of ladder, the base needs to be a distance from the wall equal to one fourth of the ladder’s length. Equivalently, if the base of the ladder is “ a” feet from the wall, the length of the ladder will be 4 a feet. See [link] .

The side adjacent to $\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ is a and the hypotenuse is $\text{\hspace{0.17em}}4a.\text{\hspace{0.17em}}$ Thus,

$\begin{array}{ccc}\hfill \mathrm{cos}\text{\hspace{0.17em}}\theta & =& \frac{a}{4a}=\frac{1}{4}\hfill \\ \hfill {\mathrm{cos}}^{-1}\left(\frac{1}{4}\right)& \approx & 75.5°\hfill \end{array}$

The elevation of the ladder forms an angle of $\text{\hspace{0.17em}}75.5°\text{\hspace{0.17em}}$ with the ground. The height at which the ladder touches the wall can be found using the Pythagorean Theorem:

$\begin{array}{ccc}\hfill {a}^{2}+{b}^{2}& =& {\left(4a\right)}^{2}\hfill \\ \hfill {b}^{2}& =& {\left(4a\right)}^{2}-{a}^{2}\hfill \\ \hfill {b}^{2}& =& 16{a}^{2}-{a}^{2}\hfill \\ \hfill {b}^{2}& =& 15{a}^{2}\hfill \\ \hfill b& =& a\sqrt{15}\hfill \end{array}$

Thus, the ladder touches the wall at $\text{\hspace{0.17em}}a\sqrt{15}\text{\hspace{0.17em}}$ feet from the ground.

Key concepts

• When solving linear trigonometric equations, we can use algebraic techniques just as we do solving algebraic equations. Look for patterns, like the difference of squares, quadratic form, or an expression that lends itself well to substitution. See [link] , [link] , and [link] .
• Equations involving a single trigonometric function can be solved or verified using the unit circle. See [link] , [link] , and [link] , and [link] .
• We can also solve trigonometric equations using a graphing calculator. See [link] and [link] .
• Many equations appear quadratic in form. We can use substitution to make the equation appear simpler, and then use the same techniques we use solving an algebraic quadratic: factoring, the quadratic formula, etc. See [link] , [link] , [link] , and [link] .
• We can also use the identities to solve trigonometric equation. See [link] , [link] , and [link] .
• We can use substitution to solve a multiple-angle trigonometric equation, which is a compression of a standard trigonometric function. We will need to take the compression into account and verify that we have found all solutions on the given interval. See [link] .
• Real-world scenarios can be modeled and solved using the Pythagorean Theorem and trigonometric functions. See [link] .

Verbal

Will there always be solutions to trigonometric function equations? If not, describe an equation that would not have a solution. Explain why or why not.

There will not always be solutions to trigonometric function equations. For a basic example, $\text{\hspace{0.17em}}\mathrm{cos}\left(x\right)=-5.$

When solving a trigonometric equation involving more than one trig function, do we always want to try to rewrite the equation so it is expressed in terms of one trigonometric function? Why or why not?

When solving linear trig equations in terms of only sine or cosine, how do we know whether there will be solutions?

If the sine or cosine function has a coefficient of one, isolate the term on one side of the equals sign. If the number it is set equal to has an absolute value less than or equal to one, the equation has solutions, otherwise it does not. If the sine or cosine does not have a coefficient equal to one, still isolate the term but then divide both sides of the equation by the leading coefficient. Then, if the number it is set equal to has an absolute value greater than one, the equation has no solution.

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