How do you find the remaining trigonometric functions of #theta# given #costheta=sqrt2/2# and #theta# terminates in QI?

1 Answer
Nov 8, 2017

Answer:

#"see explanation"#

Explanation:

#"using "sin^2theta+cos^2theta=1#

#rArrsintheta=+-sqrt(1-cos^2theta)#

#"since "theta" is in first quadrant then all ratios are positive"#

#costheta=sqrt2/2#

#•color(white)(x)rArrsintheta=+sqrt(1-((sqrt2)/2)^2)#

#color(white)(rArrsinthetaxxx)=sqrt(1-1/2)=sqrt(1/2)=1/sqrt2=sqrt2/2#

#•color(white)(x)tantheta=sintheta/costheta=(sqrt2/2)/(sqrt2/2)=1#

#•color(white)(x)cottheta=1/tantheta=1#

#•color(white)(x)sectheta=1/costheta=1/(sqrt2/2)=2/sqrt2=sqrt2#

#•color(white)(x)csctheta=1/sintheta=1/(sqrt2/2)=sqrt2#