In Exercises 41–43, find the exact value of each of the remaining...

Question

In Exercises 41–43, find the exact value of each of the remaining trigonometric functions of θ.cos θ = 2/5, sin θ < 0

Accepted Answer

Hello everybody today today, we're going to begin this question that states given the following conditions, evaluate the other trigonometric functions of D where the trigonometric function that they give us is cosine of D is equal to four divided by nine. And they also let us know that the sign of D is less than zero. So it's negative. Now, the first thing I want to do is see what quadrant we are working with. So if we have cosine the cosine that they provide us is positive, so they tell us that it's four divided by nine, it's positive four divided by nine. So that we know that cosine theta is positive. And when you think about cosine, you should think about X values. So where are X values positive? Well, they're positive in a graph, they are positive to the right side of the Y axis. And what quadrant is that? Well, this will be in quadrants one and four and then we can look at sine theta because they also give us the sign they told us that it's negative. So if we know sine theta is negative, this will be in two quadrants as well. And when you think about sign, you should think about why values and where they are negative. So in a graph Y values are negative below the X axis. So what quadrants lie below the X axis? Well, those quadrants will be quadrant three and quadrants four. So now we look at what quadrant, the both of these scenarios share in common and we see that it would be quadrant four, which means we are working in quadrant four. Now, another thing we should know what trigonometric functions is a rule that we call so or S O H dash C A H dash T O A. So now with this rule, the first letter of each three letter abbreviation represents either sign cosine or tangent. So we'll work with cosine since that's the value that they gave us. So we'll work with the, the three letters in the middle. So we'll have that cosine data. So we have the first letter equals the second letter divided by the third letter. Now the O could stand for opposite, the H could stand for hypotenuse and the A stands for adjacent. So if we are working with cosine, so C A H, we have that cosine is equal to the a adjacent, divided by the H which is hypotenuse. And they told us that the cosine is four divided by nine. So we know that the adjacent side to our angle, theta will be four units long. And the hypotenuse side will be nine units long. So now we can go ahead and use a graph to help us calculate the rest. So we draw a Y axis and an X axis and I'll leave an emphasis when quad four because that's where we're working with. And then I'll draw a straight line coming out of the origin and into quadrant four and then connect on, use another line to connect that to the x axis to draw a right triangle. And then we'll have theta and we can label what we know. So we know that the adjacent side to theta will be four and the hypotenuse side will be nine. And I'll label the other side as being because if we have a right triangle where we are trying to figure out the lengths of the sides, we use a, a certain theorem that's called the Pythagorean theorem. And the Pythagorean theorem states that A squared plus B squared equals C squared where A and B are the legs of the triangle and C is the hypotenuse. So we will have four squared plus B squared since we don't have the second leg is equal to nine squared because nine is the hypotenuse. So I'll have that 16 plus B squared equals or that B squared equals 65 by subtracting the 16 from the right hand side. And then we take the square root of both sides. We'll have that B is equal to the square roots of 65. And now we have all the information that we need about our right triangle to go ahead and discover and solve for the other trigonometric functions. Now, before we do that, I'm going to make a quick note of quadrant four to know which trigonometric functions are positive. Here, we already said cosine is one of them. So we'll have cosine data is positive and the other other, the only other function that will be positive here will be C and this is because as you see later, Cca's data is just one divided by cosine data. So it's in the same general idea. So in quarter four cos data and second, the R positive, the other trigonometric functions will be negative. So now we can start, so we'll start with sine theta as equal to and we'll have negative. So Sine, we look at S S we have sine is equal to O which is opposite, which is square root of 65 divided by H which is hypotenuse nine. So Sine theta is equal to negative square root of 65 divided by nine. And they gave us co sign. So we'll move on to tangent. So we have that tangent data we look at so we have T is equal to O opposite which is square root of 65. And remember this will be negative divided by a adjacent, which is four. So we have tangent data is equal to negative square root of 65 divided by four. And now we can move on to data. Now the is just one divided by sine theta. So we'll just get the sine data that we already calculated and flip it. So we'll have negative nine divided by the square root of 65. And now to not have a square root of the denominator will multiply both the numerator and the denominator by the screw of 65 to give us that theta is equal to negative nine multiplied by the screw of 65 in the numerator divided by 65 in the denominator. And then we can move on to second the or on our other positive function and sequent theta is one divided by cosine theta as I mentioned earlier. So we just take the cosine that they gave us and flip it and we'll have that second data is equal to nine divided by four. And finally, we move on to cotangent data. Cotangent theta is one divided by tangent data. So we get the tangent data that we obtain and flip it and we'll have negative four divided by the square root of 65. And once again, to not have a square root of 65 on the denominator, we'll multiply both the numerator and denominator by the squared of 65 to give us that co tangent data is equal to negative four multiplying the square root of 65 in the numerator divided by 65 in the denominator. And just like that, we have all six trigonometric functions calculated for our angle D. So in this case, we use data, but let's just make a note that beta equals D. So just like that with the help of some graphing and some trigonometric rules, we were able to calculate every trigonometric function for this specific angle. So I hope that was helpful. And until next time.

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In Exercises 41–43, find the exact value of each of the remaining trigonometric functions of θ.cos θ = 2/5, sin θ < 0

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