1.3 Trigonometric functions (Page 3/9)

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Solving trigonometric equations

For each of the following equations, use a trigonometric identity to find all solutions.

$1 + cos (2 θ) = cos θ$
$sin (2 θ) = tan θ$

Using the double-angle formula for $cos (2 θ),$ we see that $θ$ is a solution of
$1 + cos (2 θ) = cos θ$

if and only if
$1 + 2 {cos}^{2} θ - 1 = cos θ,$

which is true if and only if
$2 {cos}^{2} θ - cos θ = 0 .$

To solve this equation, it is important to note that we need to factor the left-hand side and not divide both sides of the equation by $cos θ .$ The problem with dividing by $cos θ$ is that it is possible that $cos θ$ is zero. In fact, if we did divide both sides of the equation by $cos θ,$ we would miss some of the solutions of the original equation. Factoring the left-hand side of the equation, we see that $θ$ is a solution of this equation if and only if
$cos θ (2 cos θ - 1) = 0 .$

Since $cos θ = 0$ when
$θ = \frac{π}{2}, \frac{π}{2} \pm π, \frac{π}{2} \pm 2 π,…,$

and $cos θ = 1 / 2$ when
$θ = \frac{π}{3}, \frac{π}{3} \pm 2 π,… or θ = - \frac{π}{3}, - \frac{π}{3} \pm 2 π,…,$

we conclude that the set of solutions to this equation is

$θ = \frac{π}{2} + n π, θ = \frac{π}{3} + 2 n π, and θ = - \frac{π}{3} + 2 n π, n = 0, \pm 1, \pm 2, … .$
Using the double-angle formula for $sin (2 θ)$ and the reciprocal identity for $tan (θ),$ the equation can be written as
$2 sin θ cos θ = \frac{sin θ}{cos θ} .$

To solve this equation, we multiply both sides by $cos θ$ to eliminate the denominator, and say that if $θ$ satisfies this equation, then $θ$ satisfies the equation
$2 sin θ {cos}^{2} θ - sin θ = 0 .$

However, we need to be a little careful here. Even if $θ$ satisfies this new equation, it may not satisfy the original equation because, to satisfy the original equation, we would need to be able to divide both sides of the equation by $cos θ .$ However, if $cos θ = 0,$ we cannot divide both sides of the equation by $cos θ .$ Therefore, it is possible that we may arrive at extraneous solutions. So, at the end, it is important to check for extraneous solutions. Returning to the equation, it is important that we factor $sin θ$ out of both terms on the left-hand side instead of dividing both sides of the equation by $sin θ .$ Factoring the left-hand side of the equation, we can rewrite this equation as
$sin θ (2 {cos}^{2} θ - 1) = 0 .$

Therefore, the solutions are given by the angles $θ$ such that $sin θ = 0$ or ${cos}^{2} θ = 1 / 2 .$ The solutions of the first equation are $θ = 0, \pm π, \pm 2 π,….$ The solutions of the second equation are $θ = π / 4, (π / 4) \pm (π / 2), (π / 4) \pm π,….$ After checking for extraneous solutions, the set of solutions to the equation is

$θ = n π and θ = \frac{π}{4} + \frac{n π}{2}, n = 0, \pm 1, \pm 2, … .$

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Find all solutions to the equation $cos (2 θ) = sin θ .$

$θ = \frac{3 π}{2} + 2 n π, \frac{π}{6} + 2 n π, \frac{5 π}{6} + 2 n π$ for $n = 0, \pm 1, \pm 2,…$

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Proving a trigonometric identity

Prove the trigonometric identity $1 + {tan}^{2} θ = {sec}^{2} θ .$

We start with the identity

{sin}^{2} θ + {cos}^{2} θ = 1 .

Dividing both sides of this equation by ${cos}^{2} θ,$ we obtain

\frac{{sin}^{2} θ}{{cos}^{2} θ} + 1 = \frac{1}{{cos}^{2} θ} .

Since $sin θ / cos θ = tan θ$ and $1 / cos θ = sec θ,$ we conclude that

{tan}^{2} θ + 1 = {sec}^{2} θ .

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Prove the trigonometric identity $1 + {cot}^{2} θ = {csc}^{2} θ .$

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Graphs and periods of the trigonometric functions

We have seen that as we travel around the unit circle, the values of the trigonometric functions repeat. We can see this pattern in the graphs of the functions. Let $P = (x, y)$ be a point on the unit circle and let $θ$ be the corresponding angle $.$ Since the angle $θ$ and $θ + 2 π$ correspond to the same point $P,$ the values of the trigonometric functions at $θ$ and at $θ + 2 π$ are the same. Consequently, the trigonometric functions are periodic functions. The period of a function $f$ is defined to be the smallest positive value $p$ such that $f (x + p) = f (x)$ for all values $x$ in the domain of $f .$ The sine, cosine, secant, and cosecant functions have a period of $2 π .$ Since the tangent and cotangent functions repeat on an interval of length $π,$ their period is $π$ ( [link] ).

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Source: OpenStax, Calculus volume 1. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11964/1.2

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