NZ Level 7 (NZC) Level 2 (NCEA)
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Solve trigonometric equations in first quadrant
Lesson

Given an angle $\theta^\circ$θ°, we can find the values of $\sin\theta$sinθ, $\cos\theta$cosθ and $\tan\theta$tanθ, and also values of the reciprocal functions $\csc\theta$cscθ, $\sec\theta$secθ and $\cot\theta$cotθ 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 $\theta$θ that produces a required function value. 

Example 1

Assuming $\theta$θ is an angle measured in degrees, find a value of $\theta$θ that makes $\sin\theta=0.4575$sinθ=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 $\theta$θ using a calculator. First, we might try $\sin30^\circ$sin30°. This has the value $0.5$0.5, which is too large. Then, perhaps try $\sin25^\circ=0.4226$sin25°=0.4226. This is too small. So, $\theta$θ must be somewhere between $25^\circ$25° and $30^\circ$30°.

The angle $27.5^\circ$27.5° is half-way between $25^\circ$25° and $30^\circ$30° and we find $\sin27.5^\circ=0.4617$sin27.5°=0.4617. This is too large. So, $\theta$θ must be between $25^\circ$25° and $27.5^\circ$27.5°. Our next try could be $\sin26.5^\circ=0.4462$sin26.5°=0.4462. This is too small. The angle $\theta$θ must be somewhere between $26.5^\circ$26.5° and $27.5^\circ$27.5°. We try $\sin27^\circ=0.454$sin27°=0.454. Slightly too small!

Next, $\sin27.3^\circ=0.4586$sin27.3°=0.4586. Too large! We need to look between $27^\circ$27° and $27.3^\circ$27.3°.

Next, $\sin27.25^\circ=0.4578$sin27.25°=0.4578. Slightly too large! Try a $\theta$θ between $27^\circ$27° and $27.25^\circ$27.25°. We try, $\sin27.2=0.4571$sin27.2=0.4571 . This is close but still slightly too small! We could keep going in this way and we would find increasingly accurate answers.

We conclude that the angle we are looking for is between $27.2^\circ$27.2° and $27.3^\circ$27.3°

 

 

In Example $1$1, we could have solved the equation by looking in the body of a table of sines (these were used before hand-held calculators were invented) to find the function value $0.4575$0.4575, and then we would read off the $\theta$θ-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. On a calculator, they are given the notations $\sin^{-1}$sin1, $\cos^{-1}$cos1 and $\tan^{-1}$tan1. They are also written as $\arcsin$arcsin, $\arccos$arccos and $\arctan$arctan respectively. 

We say, for example, $\arcsin x$arcsinx is the angle whose sine is $x$x.

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)
  $=$= $27.2259^\circ$27.2259°

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 $\theta$θ in the first quadrant such that $\tan\theta=43$tanθ=43.

Since $\tan\theta$tanθ in this case is a positive number, we expect $\theta$θ to be an angle in the first quadrant. That is, an angle between $0^\circ$0° and $90^\circ$90°. We must make sure the calculator is set in degrees. We write

$\tan\theta$tanθ $=$= $43$43
$\theta$θ $=$= $\tan^{-1}(43)$tan1(43)
  $=$= $88.66778$88.66778

This is $\approx88.67^\circ$88.67°.

 

 

 

 

 

Outcomes

M7-7

Form and use linear, quadratic, and simple trigonometric equations

91261

Apply algebraic methods in solving problems

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