An equation of the terminal side of an angle θ in standard positi...

Question

An equation of the terminal side of an angle θ in standard position is given with a restriction on x. Sketch the least positive such angle θ , and find the values of the six trigonometric functions of θ .  See Example 3.   
                                                                                                                                                                                                                                                                                                                                                                                   
  ―5x ― 3y = 0 , x ≤ 0

Accepted Answer

Hello, today we're going to be drawing a sketch of the least positive angle beta, whose equation of the terminal side of the standard position is given below. We also want to use this value to find sine beta cosine beta and tangent beta. So let's go ahead and take a look at the given Standard equation. We are given the equation negative four X minus seven Y is equal to zero. And we only want the portion of the graph where X is less than or equal to zero. Now this is the f the equation of a line in standard form. But we're going to rewrite this in slope intercept form in order to do that, we're going to solve for the variable line to start this process. We'll add four X to both sides of the equation. This is going to leave us with negative seven Y is equal to four X. In order to solve for Y, we'll divide both sides of this equation by negative seven. And that will leave us with Y is equal to negative four divided by seven X. And again, we only want the region of the graph where X is less than or equal to zero. Now, one point that intercepts at this line is going to be the point at negative seven comma four. And this is going to be the left hand side of the negative slope line. Now keep in mind that this line is negative seven units long and has a height of positive four. So the least positive angle beta to this equation is gonna be located inside of this right triangle that we've just created. So what we want to do now is we want to find the values of cosine beta, sine beta and tangent beta. Now, since we've just created a right triangle, we have two sides to the right triangle four and negative seven. But we need to find the value of the hypotenuse of this right triangle. In order to find the value, we can use the Pythagorean identity. The Pythagorean identity is C squared, which is the hypotenuse is equal to the sum of the lengths squared or X squared plus Y squared. In this case, we're going to let X equal to negative seven and we're going to let Y equal to positive four. So that is going to give us C squared is equal to negative seven squared plus four squared. This will simplify to give AC squared is equal to 49 plus 16, 49 plus 16 is going to give us the value of 65. So C squared is equal to 65 And if we take the square root of both sides of this equation, we are left with C is equal to the square root of 65. So that means that the length of this hypotenuse of the right triangle is going to be the square root of 65. So now that we have the values of the sine links of beta, we're going to use these side links to determine sine beta cosine beta and tangent beta. Now recall that the Pythagorean identity for sine beta is equal to the Y value divided by the hypotenuse in which, which this case is C, we know that C is equal to the square root of 65. And we know that the Y value of the right triangle is four. So sine beta is going to equal to four divided by the square root of 65. We cannot leave a square root in the denominator of a fraction. So we're going to have to rationalize this value by multiplying both the numerator and denominator by the square root of 65. This will simplify to leave us with sine beta is equal to four multiplied by the square root of 65 divided by 65. And this is going to be the value of sine beta for the smallest positive angle beta. Now let's go ahead and take a look at the Pythagorean identity for cosine. We know that cosine beta will equal to the X value divided by the hypotenuse C. We know that the X value is equal to negative seven. And we know that the value of the hypotenuse is the square root of 65. So cosign of beta will equal 27 divided by the square root of 65. And just like the value of S we're going to have to rationalize this denominator by multiplying the numerator and denominator by the square root of 65. This is going to leave us with cosine of beta is equal to negative seven, multiplied by the square root of 65 all divided by 65. And this is going to be the value of cosine. Finally, we need to determine the value of tangent of beta. The Pythagorean identity for tangent is tangent of beta is equal to the Y value divided by the X value. Now again, we know that the Y value is equal to positive four and the X value is equal to negative seven. That means that tangent of beta is going to equal to negative four divided by seven. And this is going to be the simplest form of the tangent of beta. So just to reiterate, we went ahead and used the equation to draw the sketch of the terminal side of the angle beta. And we solve for the missing sides of this right triangle to find the values of sine beta cosine beta and tangent beta, which are the sine cosine and tangent values of the smallest positive angle of beta. And this is going to be the solution to this, to this question. So I hope this video helped you and understand you how to approach this type of problem. And I'll go ahead and see you all in the next video.

An equation of the terminal side of an angle θ in standard position is given with a restriction on x. Sketch the least positive such angle θ , and find the values of the six trigonometric functions of θ . See Example 3. ―5x ― 3y = 0 , x ≤ 0

Key Concepts

Standard Position of an Angle

Trigonometric Functions

Graphing Linear Equations

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