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Grade 12

Solve trigonometric equations in first quadrant (rad)

Lesson

Given a number $x$x, we can find the values of $\sin x$sinx, $\cos x$cosx and $\tan x$tanx, and also $\csc x$cscx, $\sec x$secx and $\cot x$cotx by pressing the right buttons on a calculator, by looking up a table or by reading the result off a graph.

Solving a trigonometric equation is the reverse of this process. We find a value of $x$x that produces a required function value. 

Example 1

Find a value of $x$x that makes $\sin x=0.4575$sinx=0.4575.

Although it is not the most efficient way of solving the equation, as we will see, we could try out different values of $x$x using a calculator. First, we might try $\sin\frac{\pi}{6}$sinπ6. This has the value $0.5$0.5, which is too large. Then, perhaps try $\sin\frac{\pi}{7}=0.4339$sinπ7=0.4339. This is too small. So, $x$x must be somewhere between $\frac{\pi}{7}$π7 and $\frac{\pi}{6}$π6.

(We are going to use the fact that given fractions $\frac{a}{b}$ab and $\frac{c}{d}$cd, the fraction $\frac{a+c}{b+d}$a+cb+d is always in-between.)

The number $\frac{2\pi}{13}$2π13 is between $\frac{\pi}{7}$π7 and $\frac{\pi}{6}$π6 and we find $\sin\frac{2\pi}{13}=0.4647$sin2π13=0.4647. This is still too large. So, $x$x must be between $\frac{\pi}{7}$π7 and $\frac{2\pi}{13}$2π13. Our next try is $\sin\frac{3\pi}{20}=0.454$sin3π20=0.454. This is close but too small. The number $x$x must be somewhere between $\frac{3\pi}{20}$3π20 and $\frac{2\pi}{13}$2π13. I try $\sin\frac{5\pi}{33}=0.4582$sin5π33=0.4582. Too large!

Next, $\sin\frac{8\pi}{53}=0.4566$sin8π53=0.4566. Too small! We need to look between $\frac{8\pi}{53}$8π53 and $\frac{5\pi}{33}$5π33.

Next, $\sin\frac{13\pi}{86}=0.4572$sin13π86=0.4572. Slightly too small! Try an $x$x between $\frac{13\pi}{86}$13π86 and $\frac{5\pi}{33}$5π33. We try, $\sin\frac{18\pi}{119}=0.457525...$sin18π119=0.457525... . This is near enough!

After trying $8$8 possible values for $x$x, We can conclude that $x=\frac{18\pi}{119}\approx0.4752$x=18π1190.4752 is very close to the solution.

 

 

In Example $1$1, we could have solved the equation by looking in the body of a table of sines to find the function value $0.4575$0.4575, and then read off the $x$x-value that produces that result. This is equivalent to looking up a table of the inverse sine function.

An inverse trigonometric function takes a function value as its input and returns a value of the variable that produces that function value.

The inverse functions of sine, cosine and tangent respectively are called arcsine, arccosine and arctangent. They are given the notations $\sin^{-1}$sin1, $\cos^{-1}$cos1 and $\tan^{-1}$tan1

In the example above, we had $\sin x=0.4575$sinx=0.4575. To solve this using the inverse sine function, we write

$\sin x$sinx $=$= $0.4575$0.4575
$x$x $=$= $\sin^{-1}(0.4575)$sin1(0.4575)
  $=$= $0.4752$0.4752

The values of the inverse trigonometric functions are built into most calculators so that one only needs to press the right buttons to obtain the desired result.

Note that the values returned by the inverse trigonometric functions are always angles with the smallest possible absolute value. For positive values of sine, cosine and tangent the angle returned by the inverse function is always in the first quadrant of the unit circle diagram.

If further solutions are required, we use the periodic properties of the functions to find other values of the variable that also lead to the same function value.  

 

Example 2

Find $x$x in the first quadrant such that $\tan x=43$tanx=43.

We normally understand $x$x to be a number or an angle in radian measure. We must make sure the calculator is set in radians. We write

$\tan x$tanx $=$= $43$43
$x$x $=$= $\tan^{-1}(43)$tan1(43)
  $=$= $1.5475$1.5475

If the same operation is done with the calculator set in degrees, this is $\approx88.67^\circ$88.67°.

 

 

 

 

 

Outcomes

12F.B.3.4

Solve linear and quadratic trigonometric equations, with and without graphing technology, for the domain of real values from 0 to 2 p, and solve related problems

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