HW2sol - MAC1114 Homework 2 Due Thursday January 20 1 Show...

Info icon This preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: MAC1114 - Homework 2 Due Thursday, January 20 1. Show that sin(45 ◦ ) = cos(45 ◦ ) = √ 2 2 (Hint: Draw an appropriate triangle and use the Pythagorean theorem). The key thing here is to note that since you have an isosceles triangle, the two short sides are the same. So you may have set up something like x 2 + x 2 = 1 ⇒ 2 x 2 = 1 ⇒ x 2 = 1 2 ⇒ x = q 1 2 = √ 2 2 . 2. Evaluate each of the six trigonometric functions at the following values of t : (a) t =- 315 ◦ It helps to nd a coterminal angle,- 315 ◦ + 360 ◦ = 45 ◦ . You showed in (1) that sin(45 ◦ ) = √ 2 2 and cos(45 ◦ ) = √ 2 2 , so tan(45 ◦ ) = 1 , sec(45 ◦ ) = √ 2 , csc(45 ◦ ) = √ 2 , and cot(45 ◦ ) = 1 . (b) t = 3 π 2 This is straight o the unit circle. sin 3 π 2 =- 1 , cos 3 π 2 = 0 , tan 3 π 2 = unde ned, csc 3 π 2 =- 1 , sec 3 π 2 = unde ned, cot 3 π 2 = 0 . 3. In class we showed that sin( t + 2 πn ) = sin t and cos( t + 2 πn ) = cos t whenever n is an integer, and we learned that...
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern