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.

sin θ = 5/13, θ in quadrant II

Accepted Answer

Hello everybody. Everything right today that we're going to look at this question that states determine the value of the five other trigonometric ratios where the one that we are given is sine theta is equal to 20 divided by where theta is in quadrant two. Now they give us a lot of information here just by telling us that data is in quadrant too. So before we move on to actually plotting this on a plot, let's talk about a very important rule when it comes to trigonometric ratios. And that is so or S O H dash C A H dash T O A. So, so what does this mean? Well, the first letter of each abbreviation of each three letter abbreviation stands for either sign cosine or tangent. The letter O would stand for opposite, the letter H stands for hypotenuse and the letter A stands for adjacent, referring to the side. So in this case, we're going to focus on the sole part where as in my example, because we're dealing with Sine, they give us the sine of data. So if we look at so S O H, we have sine, the sense for sign. So we have sine theta equals. So it would be the first letter equal to the second letter divided by the third letter. So we have S sine theta equals O or opposites divided by H or hypotenuse, which they told us was equal to 20 divided by 29. So we know that the opposite side to theta will be 20 units long. And the hypotenuse with respect to data will be 29 units long. Now, when we're dealing with the hypotenuse, we know we're going to have a right triangle. So I'm going to apply, make a plot, drawing the Y axis and the X axis leaving a lot of open space in quadrant two. Since they told us that's where we will be. And I draw a straight line coming out of the origin and into quadrant two. Then I connect that line to the X axis to create a right triangle and we can draw our theta angle and now we can plug in the numbers that we know. So we said that the opposite side of the was 20 and the hypotenuse of this triangle was 29. Now the side that is left, I will label it as B because if we have a right triangle and we're trying to figure out the measurements of the sides, what do we use? The Pythagorean theorem? And the Python theorem is A squared plus B squared equals C squared where A and B are the lengths of the lengths of the triangle and C is the length of the hypotenuse. So we'll have that A squared is 20 squared plus B squared stays the same as equal to squared. So once we simplify that we'll have 400 plus B squared is equal to 841. B squared is equal to 441. If we move the 400 over from the left hand side to the right hand side, and if we take the square root of both sides, we'll have that B is equal to 21. No, we have all the information that we need to go ahead and solve for the rest of the trigonometric ratios. But before we do that, let's make a note of the quadrant that we are in. So quadrant two, let's make a note of what ratios would be positive here. So in quadrant route two sine theta and data are positive and these are the only two that will be positive, the rest will be negative. So when you're thinking of Sine, think of the Y axis, anything that is above the X ax, any Y value that is above the X axis will be positive. And if we're in quadrant two, those Y values will be positive. Therefore, Sine will be positive when you're thinking of cosine, think of the X axis and X values anything to the left of the Y axis will be negative. X values. So if we're dealing with quadrant two, we're dealing with negative X values and negative cosine. So let's go and start solving cosine as we said, we'll have a negative. And if we look at so we have C H so we have cosine is equal to adjacent, divided by hypotenuse which is negative 21 divided by 29. And then we can move on to tangents and we have tangent part of to which is T O A. So tangent is equal to opposite, divided by adjacent, which is remember we are dealing with a negative here. So it's negative 20 divided by 21. And then we can move on to cos theta. Now theta is one divided by sine theta. So it's just the sign that they gave to us but flipped. So they told us that sine theta is 20 divided by 29. So one divided by sine theta is 29 divided by 20. So we move on to C C the C can theta is one divided by cosine theta. So we look at the cosine data that we calculated and flip it and we'll have negative 29 divided by 21. And finally, we're left with co tangent data which is equal to one divided by tangent theta. So we look at the tangent that we have and flip it and we'll be left with negative 20 divided by and just like that. Now, we have all six trigonometric ratios and all we had to do was graph and use a plot of the design data that they provided us with the data in quarter two. And then use some simple math to solve for the other five trigonometric ratios. 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. sin θ = 5/13, θ in quadrant II

Key Concepts

Trigonometric Functions

Pythagorean Identity

Quadrants and Signs of Trigonometric Functions

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