Give all six trigonometric function values for each angle θ .Rati...

Question

Give all six trigonometric function values for each angle θ .Rationalize denominators when applicable.   sec θ = ―√5 , and θ is in quadrant II

Accepted Answer

Hello, everyone. We are asked to determine the value of the other five trigonometric ratios. We're given that the CCA of theta equals negative radical seven and that theta lies in the second quadrant. So first we call when we're finding values of trig ratios that we have an X A Y and an R that we work with in our ratios. So we're gonna use first the sea can of data to figure out whichever pieces we can of that. So the of the, we know is negative radical seven. So thinking about the ratio that we use for C, can I know that this is R divided by X where our is like our radius, it's our distance from the origin. So R is not gonna be negative R is a positive distance is always positive. So we would do this as the square of seven divided by negative one. So X is gonna be negative one, R is the square to seven. We're gonna use the Pythagorean theorem to solve for why? So X squared plus Y squared equals R squared. So negative one squared plus Y squared equals in parentheses, the squared of seven squared. So negative one squared is a positive one plus Y squared equals the squared of seven squared would be seven subtracting one from both sides. I get that Y squared equals six, taking the square root of both sides. I get Y equals plus or minus radical six. Now knowing we're in quadrant two and in quadrant two, sine and co C can are positive. This means my Y value will be a positive radical six. So Y equals the squared of six. So now that I know these things, I can try to find my other five ratios, I will need to rationalize denominators along the way. So I'm starting with sign. So the sign of the I know was the ratio of Y divided by R. So Y is the square to six divided by R which is the square to seven. I want to rationalize my denominator by multiplying numerator and denominator by the square of seven. And when I do so, my numerator becomes the square root of 42 divided by seven. Sorry, the square to seven multiplied by the square to seven is seven. And since I do not believe the square to 42 has anything I can simplify within it, that's gonna be my answer for the sign of the. So the sin of theta is radical 42 divided by seven. So next I want the cosine of the, so the cosine of the I know is X divided by R I can also think of it since we have the sea can it is the same as one divided by the CCA of theta. So if I do this, I would have one divided by negative radical seven. I again need to rationalize my denominator by multiplying numerator and denominator by the square to seven. And when I do, I will have negative radical seven divided by seven. And that will be my cosign of theta. And this makes sense because in quadrant two cosine is negative. So that would make sense that our cosine is negative. Next, we have the tangent of data and that's the ratio of Y divided by X. So Y is the squared of six divided by X which is negative one. So we can simplify this to be the negative square root of six. And that would be our tangent of theta. Again, in quadrant two tangent would be negative. So this makes perfect sense. We already know the C can of theta. So next, we're gonna go to the co C can of theta. Recall that co C can is the same as one divided by the sign of theta. And sometimes it's looked at as if it's the reciprocal of sign. So this would be R divided by Y. So R is the square root of seven divided by Y which is the square to six. Again, I need to rationalize my denominator by multiplying numerator and denominator. In this case by the square to six square to seven, multiplied by the square to six is the square of divided by the square to six, multiplied by the square to six is six. So the square of 42 divided by six is the cosecant of theta. And our last one should be the cotangent of the. And that is going to be the same as saying one divided by the tangent of the. Also again, sometimes looked at as the reciprocal. So this would be X divided by Y. So X is negative one, Y is radical six. So negative one divided by radical six. I rationalized the denominator by multiplying numerator and denominator by the square to six. And we get negative radical six divided by six. And that is our cotangent of theta. So in the end, I have all five ratios and just thinking through it, the only ones that should be positive if we are in quadrant to our sign and CO C can't and that's what this looks like. So we are on the right track. Have a nice day.

Give all six trigonometric function values for each angle θ .Rationalize denominators when applicable. sec θ = ―√5 , and θ is in quadrant II

Key Concepts

Trigonometric Functions and Their Relationships

Sign of Trigonometric Functions in Quadrants

Rationalizing Denominators

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