In Exercises 23–34, find the exact value of each of the remaining...

Question

In Exercises 23–34, find the exact value of each of the remaining trigonometric functions of θ.sec θ = -3, tan θ > 0

Accepted Answer

Hello everybody. I have everything right today, today we're going to be on this question that states determine the value of the five other trigonometric ratios where the ones they give us are, they tell us that second theta is equal to negative five. And they also let us know that tangent theta is greater than zero. Now let's first work with the information that they give us. So they tell us that second theta is equal to negative five. Now, what does that mean? Well, what is second the the is equal to one divided by cosine theta. Therefore, when looking at what region of a graph this is in, we know that we have to be negative because they tell us that sin theta is negative five. So we have to look for an area in the graph where the sequent will be negative. And if sequent is one divided by cosine, we have to look for an area where cosine is negative and we know that cosine is negative in quadrants two and three. So second data will be negative in quadrants two and three. So those are the possible quadrants for a second theta of negative five. Now we look at tangent, now tangent theta is equal to sine theta divided by cosine theta. So we'll have to look for a quadrant where sine theta and cosine data are both positive or are both negative and this occurs and two quadrants. So we will be positive and quadrants one where both sine data and cosine data will be positive and quadrant three where they're both negative. So we would have a tangent of that is positive. So 10 quadrant three is the quadrant that both of these scenarios share in common. So we know our graph will be located in quadrant three. No, before we get to graphene to continue to work through a problem. Let's remember a rule that we know and that is the rule of soul. So this is an abbreviation A Pneumonic of swords where so the S stands for sign and sign is equal to opposite over Hypo K stands for cosine and cosine is equal to adjacent over hypotenuse and to a stands for tangent is equal to opposite, divided by adjacent. Now let's apply that to the second data that we know. So we have that second theta is equal to one divided by cosine theta. And by looking at our Pneumonic, we know that cosine data is adjacent, divided by hypos. So we have that one divided by cosine data is equal to one divided by adjacent, divided by hypo. So that means that second and theta is equal to hypo divided by adjacent. Now they tell us that second theta is equal to negative five. Now I'm going to drop the negative because we're dealing with the measurements here. So I'll have that hypo divided by adjacent is equal to five. This is the same thing as saying, sequent data is equal to five. So this will tell us that the hypotenuse equals five and the adjacent side to the data angle will equal one because you have a hypo of five and an adjacent side of one, you have five divided by one, which is equals to five. So this is a measurement we can work with, it could be any, it could be another number as long as the ratio stays the same 5 to 1. So now I'm going to start graphing so we can decipher the other sides and the other trigonometric ratios. So I'm going to draw a small graph here and we know, as we said, our theta angle will be in quadrant three. So we draw a line the hypotenuse stemming from the origin into quadrant three. And then we draw a line connecting that to the X axis making a right triangle. And we'll label theta as the angle quadrant three. And then we already calculated the hypotenuse is five and the adjacent side to the. So on the X axis that will have a unit of one. And then we're left solving for the opposite side of theta which are labeled as side B and how, when we have a right triangle, what do we use to calculate the measurement of the angles we used by? And the, and that is A squared plus B squared equals C squared. Now A squared, in this case will be one. So what we have, we'll have one squared plus B squared because we don't have that opposite side equals C squared C being the hypotenuse. So it will be equal to five squared, which means we'll be left with one plus B squared equals 25 which means that B squared equals 24 or that B is equal to two multiplying the square root of six. Now, we must remember that we are in quadrant three. And before we move on to the trigonometric ratios, we should know that quadrant three only has tangent and col tangent as positive. So we know that whatever ratio we calculate, we know that the tangent and the co tenure will be positive. So let's start calculating, let's do sine the first. Now we said that sine theta, if we look at so sine theta is equal to opposite, divided by hypos. The opposite side of the beta angle is the B and we calculated B as being two multiplying the squared of six. And this will be negative because we are in a negative in a, in a quadrant which sine is negative. So we have negative and then a fraction with two multiplying the squared of six on the numerator divided by what? So says that it's opposite, divided by hypo and the hypothesis is five. So will be divided by five. So sine theta is equal to negative two, multiplying the squared of six divided by five. And then we can move on to cosign data which by looking at. So we know it's adjacent, divided by high partners where the adjacent is. And once again, this will be negative because we are in a quadrant where cosine is negative. So we will have negative adjacent which is one divided by the hypo which is five. So cosine theta equals negative 1/5. And then we move on to tangent data 10 in the, if we look at so is opposite, divided by adjacent. So we'll have that tangent theta is equal to opposite, which is two multiplied by the squared of six divided by adjacent, which is one. So tangent data is equal to two multiplying the square roots of six. And now we can move on to co sequences. Now remember that they gave us the, so that would be one of the trigonometric ratios, we're calculating the other five. So cos theta is equal to one divided by sine theta. And we already calculated sine theta. So we'll have one divided by, if we have one divided by sine theta, we would just look at the reciprocal of sine theta because we had a fraction. So we'll have negative five divided by two, multiplying the squared of six. And to get rid of the square root in the denominator, we multiply every deum and the denominator by the squared of six, which means we'll have cos theta is equal to negative five, multiplying the squared of six divided by 12. And that 12 comes from, if we divide, if we multiply the denominator by the squared of six, we'll have a squared of six multiplying the squared of six, which is six and six multiplied by two is 12. So that would be code sequent. And we can now move on to code tangents, which will be the last trigonometric ratio for this case. So we have that cotangent is equal to one divided by tangent theta. And we already calculated tangent data as being two multiplied by a square of six. So we take the reciprocal of that. So we'll have one divided by two multiply and the squared of six. And once again to get rid of the squared of six, and the denominator will multiply both the top and the bottom by the squared of six leaving us with code tangents, theta is equal to the squared of six divided by 12. So just like that with the help of the small graph that we drew, we were able to calculate the six trigonometric ratios. So they gave us one of them and we calculated the other five. So I hope that was helpful and until next time.

In Exercises 23–34, find the exact value of each of the remaining trigonometric functions of θ.sec θ = -3, tan θ > 0

Key Concepts

Trigonometric Functions

Secant and Tangent Relationships

Quadrants and Signs of Trigonometric Functions

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