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

Question

Give all six trigonometric function values for each angle θ . Rationalize denominators when applicable. 
 
 sec θ = 5/4 , and θ is in quadrant IV

Accepted Answer

Hello, everyone. We are asked to determine the value of the other five trigonometric ratios. We are given that the CCA of theta equals seven divided by two and theta lies in quadrant four. So first rewriting the sea can of theta as a fraction seven halves, I recall that the ratio that goes with this would be R divided by X. I also recall that I need to find X why and are to be able to solve these trick ratios. So we know X is two, R is seven, we need to find out what Y is. So we can use the Pythagorean theorem to do that. So X squared plus Y squared equals R squared, X is two. So two squared, yep, two squared plus Y squared equals seven squared. Two squared is four plus Y squared equals 49. Subtracting four from both sides. Y squared would equal 45. Taking the square root of both sides. We get Y equals plus or minus the square of 45. I'm gonna keep the plus or minus for this second and I'm gonna break the square root of 45 down into the squared of nine multiplied by the square to five, the squared of nine would give us plus or minus three radical five. And since we are in quadrant four, we want why to be, we want why to be negative? So our Y value will be negative three radical five. I know it's negative because in quadrant four, only cosine and co C can would be positive. And those are the XSYS are more associated with sine. So now that we know our X value, our Y value and our R value, we can find all five trade ratios. So I'm gonna start with the sign of data and I recall that sine is the ratio Y divided by R. So why is negative three radical five divided by R which is seven? Nothing to simplify there. This is just the sin of theta is negative three radical five divided by seven. The cosine of the cosine is can be looked at as one divided by the C can of theta. Since we have the C can of theta, I can look this also as the reciprocal of CCA. So X divided by R which would be two divided by seven. So our cos can of theta is 2/7. Next thinking of the tangent of data tangent is the ratio Y divided by X. So why is negative three radical five divided by X which is two I'm liking these ones where I don't have to rationalize the denominator. But I don't think they're gonna last much longer. We already have the sea can of theta. So we need to find the co C can of theta and recall that CO CK is the same as one divided by the sign of theta. So we can look at that as the reciprocal of the ratio. So it would be R divided by Y R is seven divided by Y which is negative three radical five. This one I do need to rationalize my denominator. So I multiply numerator and denominator by the square root of five. And in the numerator that gives me seven radical five in my denominator, I have my negative three multiplied by the square of five, multiplied by the square of five, the square to five multiplied by itself would be five. So five multiplied by negative three would be negative 15. So we're gonna rewrite this with the negative in the top. So we'd have negative seven radical five divided by 15. And that's our of the and last one, we need our cotangent of the recall cotangent is the same as one divided by the tangent of theta. And its ratio can be looked at as the reciprocal. So it'd be X divided by Y so X is two. Why is negative three radical five? So we need to rationalize that denominator again, just like we did with the previous ratio, we're multiplying numerator and denominator by the square of five. And that'll give us two radical five divided by negative 15. And I wanna put the negative in my numerator. So I'll have negative two radical five divided by 15. And that is the cotangent of the so quickly double checking because I'm in quadrant four and I know only cosine and C can should be positive in quadrant four. I'm just doing a quick check that everything else is negative and they are so we should be all set. Have a nice day.

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Give all six trigonometric function values for each angle θ . Rationalize denominators when applicable. sec θ = 5/4 , and θ is in quadrant IV