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Trigonometry 12 Online
OpenStudy (anonymous):

How do i evaluate sopmething like this?It says Use even and odd properties of the trigometric function to find the exact value of the cos(-150 degrees).

OpenStudy (anonymous):

firstly you have to know cos(-150)=cos(150) After that ım rewriting 150 by using two numbers 90+60 It became cos(90+60) when u write a sum as if that (90+x.270+y) you have to change the name of this function e.g cos(90+60) = "sin"60 you can write other number right into paranthesis so ,the result is : \[\sqrt{3}/2 \]

OpenStudy (anonymous):

thanks alot.but how did u get 90 and 60?i am lost from that line straight down

OpenStudy (anonymous):

i know that it is \[\sqrt{3} \div 2 \] from memorization. but -150 is in the third quadrant, where cos is negative, therefor i believe you answer will be negative radical 3 over 2 but maybe vitas can confirm that for me

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