Evaluation Using Identities

$\newenvironment {prompt}{}{} \newcommand {\ungraded }[0]{} \newcommand {\log }[0]{\ln } \newcommand {\vector }[1]{\overrightarrow {#1}} \DeclareMathOperator {\arcsec }{arcsec} \newcommand {\ihat }[0]{\hat {\imath }} \newcommand {\jhat }[0]{\hat {\jmath }} \newcommand {\khat }[0]{\hat {k}} \newcommand {\comp }[0]{\operatorname {comp}} \newcommand {\proj }[0]{\operatorname {proj}} \newcommand {\coords }[1]{\left \langle #1 \right \rangle } \newcommand {\magnitude }[1]{\left \lvert {#1}\right \rvert } \newcommand {\HyperFirstAtBeginDocument }[0]{\AtBeginDocument }$

These are problems intended to help you review by using the trigonometric identities.

None of these problems will tell you which identity to use. You just have to be aware of which identity corresponds to each modification of the input (e.g. if there is a negative in the input you will probably want to think about even and odd functions).

1 : If we are told that $\sage {p5dispf}=\sage {p5dispv}$ and that the terminal side of $x$ lies in the first quadrant, evaluate the following trigonometric expressions.

$\cos (-x)=\answer {\sage {p5ans1}}$ $\sin (-x+2\pi )=\answer {\sage {p5ans2}}$ $\tan (\frac {\pi }{2}-x)=\answer {\sage {p5ans3}}$ $\sec (x-2\pi )=\answer {\sage {p5ans4}}$ $\cot (x)=\answer {\sage {p5ans5}}$

2 : If we are told that $\sage {p6dispf}=\sage {p6dispv}$ and that the terminal side of $x$ lies in the first quadrant, evaluate the following trigonometric expressions.

$\cos (-x)=\answer {\sage {p6ans1}}$ $\sin (-x+2\pi )=\answer {\sage {p6ans2}}$ $\tan (\frac {\pi }{2}-x)=\answer {\sage {p6ans3}}$ $\sec (x-2\pi )=\answer {\sage {p6ans4}}$ $\cot (x)=\answer {\sage {p6ans5}}$

3 : If we are told that $\sage {p7dispf}=\sage {p7dispv}$ and that the terminal side of $x$ lies in quadrant three, evaluate the following trigonometric expressions.

$\sin (x)=\answer {\sage {p7ans1}}$ $\cos (x)=\answer {\sage {p7ans2}}$ $2\sec (\frac {\pi }{2}-x)=\answer {\sage {p7ans3}}$ $\sin (x+\pi )^2=\answer {\sage {p7ans4}}$