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 .                                                                                                                                                                                                                                                                                                                                                                                                               12x + 5y = 0 , x ≥ 0 .

Accepted Answer

Hey, everyone here we are given an angle theta in standard position has its terminal side represented by the given equation. And a restriction on X is also included, we need to provide a sketch of the least positive angle theta and determine its six trigonometric function values. Here we are given the equation three X plus or Y is equal to zero. And we are given the restriction on X where X must be greater than or equal to zero. So to begin this problem, we first want to solve our given equation for Y divided by X. So the first step is to just subtract three X from both sides. So we have four, Y is equal to negative three X. And now getting Y alone, we divide both sides by four. So we have Y is equal to negative quantity of 3/4 multiplied by X. And now we want to get X to be on the left side of the equation. So we just need to divide both sides by X. And we now have Y divided by X is equal to either negative three, divided by four or Y divided by X is equal to positive three divided by negative four. And now we can derive two ordered pairs from both equations here. So from our first equation again, we have Y divided by X is equal to negative three, divided by four, we can get the ordered pair four and negative three. And now, from our second equation, we get the ordered pair negative four and positive three. And now we want to notice our restriction on X, we know that X must be greater than or equal to zero. So we know that our second ordered pair does not fit within the restriction. So we will only focus on using our first ordered pair. And so now we can begin sketching the least positive angle by starting by plotting our ordered pair. So we need to plot the 0.4 and negative three. And now we just want to plot the initial sign or sketch the initial sign by drawing an arrow which starts at the origin and coincides with the positive X axis. And next we draw the terminal side which again is an arrow that starts from the origin and now crosses through or intersects the plotted point we found from earlier. And now we need to note the angle theta which moves counterclockwise from the initial side to the terminal side. And now that we have sketched the least positive angle theta we can now move on to determining its six trigonometric function values. And so the first step is to find our value for R. So we first need to recall that R is equal to the square root of X squared plus Y squared. So now we just need to substitute in our values for X and Y into the formula. So we have R is equal to the square root of four squared plus the quantity of negative three squared. And now simplifying we have the square root of plus nine. Simplifying further, we have the square root of 25. And finally, simplifying we have the square root of 25 which is just five. So we have found again that R is equal to five. And now we can move on to finding our trigonometric function values. And so we can start with finding sine of theta. So we just need to recall that sign of theta is equivalent to the ratio Y divided by R. So now just substituting in our values for Y and R, we have the ratio negative 3/5. And now moving on to cosine of theta, we need to recall that cosine of theta is equivalent to X divided by R substituting in our values, we have 4/5. Next, moving on to tangent of theta, we need to recall that tangent of theta is the same as sine of theta divided by cosine of theta. So now substituting in the values we found for sine and cosine, we have negative 3/5 divided by 4/5 and now just simplifying, we find tangent of theta is negative 3/ and now we can move on to cotangent of theta. And we need to recall that cotangent of theta is equivalent to one divided by tangent of theta. And now just substituting in what we found for tangent, we have one divided by negative 3/4 and just simplifying we have negative four thirds. Now moving further, we can find sin of theta. So just recalling sin of theta is the same as one divided by cosine of theta. So substituting in the ratio we found for cosine, we have one divided by 4/ simplifying we have five fourths. And now moving on to our last function here we have co sequent of theta and cos of theta is equivalent to one divided by sine of theta. So substituting in what we found for sine, we have one divided by negative 3/ simplifying we have negative five thirds. And with this, we have successfully sketched our least positive angle theta as well as determined each of the six trigonometric function values. Thanks for watching. I hope you found this video helpful and I'll see you again next time.

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 . 12x + 5y = 0 , x ≥ 0 .

Key Concepts

Standard Position of an Angle

Trigonometric Functions

Graphing Linear Equations

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