SOLUTION: If cos(t)=-5/6 where pi < t < 3pi/2 , find the values of the following trigonometric functions. Note: Give exact answers, do not use decimal numbers. The answer should be a fra

Algebra ->  Trigonometry-basics -> SOLUTION: If cos(t)=-5/6 where pi < t < 3pi/2 , find the values of the following trigonometric functions. Note: Give exact answers, do not use decimal numbers. The answer should be a fra      Log On


   

Question 1102370: If cos(t)=-5/6 where pi < t < 3pi/2 , find the values of the following trigonometric functions.

Note: Give exact answers, do not use decimal numbers. The answer should be a fraction or an arithmetic expression. If the answer involves a square root it should be enter as sqrt; e.g. the square root of 2 should be written as sqrt(2).
cos(2t) =
sin(2t)=
cos(t/2)=
sin(t/2)=
Found 2 solutions by Alan3354, ikleyn:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
Look up "Half angle formulas" on Wikipedia.

Answer by ikleyn(52852) About Me  (Show Source):
You can put this solution on YOUR website!
.
If cos(t)=-5/6 where pi < t < 3pi/2 , find the values of the following trigonometric functions.

cos(2t) =
sin(2t)=
cos(t/2)=
sin(t/2)=
~~~~~~~~~~~~~~~~ 

You are given that  cos(t) = -5%2F6  and the angle  "t"  is in  QIII.


Then   sin(t) =  -sqrt%281-cos%5E2%28t%29%29 = -sqrt%281-%28-5%2F6%29%5E2%29 = -sqrt%28%2836-25%29%2F36%29 = -sqrt%2811%2F36%29 = -sqrt%2811%29%2F6.


        Notice that I put the sign "-" (minus) at sqrt, since the function sine is NEGATIVE in QIII.


The next steps are straightforward.  

a)  cos(2t) = cos^2(t) - sin^2(t) = %28-5%2F6%29%5E2+-+%28-sqrt%2811%29%2F6%29%5E2 = 25%2F36+-+11%2F36 = 14%2F36 = 7%2F18.


b)  sin(2t) = 2*sin(t)*cos(t) = 2%2A%28-sqrt%2811%29%2F6%29%2A%28-5%2F6%29 = %285%2Asqrt%2811%29%29%2F18.


c)   cos(t/2) = -sqrt%28%281%2Bcos%28t%29%29%2F2%29 = -sqrt%28%281%2B%28-5%2F6%29%29%2F2%29 = -sqrt%28%286-5%29%2F%286%2A2%29%29 = -sqrt%281%2F12%29 = -sqrt%2812%29%2F12 = %28-4%2Asqrt%283%29%29%2F12%29 = -sqrt%283%29%2F3.


         Again, the sign is "-" at sqrt, since cosine is NEGATIVE function in QII, where the angle t/2 is.


d)  sin(t/2) = sqrt%28%281-cos%28t%29%29%2F2%29 = sqrt%28%281-%28-5%2F6%29%29%2F2%29 = sqrt%28%286%2B5%29%2F%286%2A2%29%29 = sqrt%2811%2F12%29 = sqrt%28132%29%2F12 = %282%2Asqrt%2833%29%29%2F12%29 = sqrt%2833%29%2F6.


         The sign is "+" at sqrt, since the sine function is POSITIVE in QII, where the angle t/2 is.


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There are two useful groups of lessons in this site, relevant to this problem.

First group is the lesson
    - FORMULAS FOR TRIGONOMETRIC FUNCTIONS
and all associated lessons (links) with it.

The second group is these lessons
    - Calculating trigonometric functions of angles
    - Advanced problems on calculating trigonometric functions of angles
    - Evaluating trigonometric expressions
in this site.


Also,  you have this free of charge online textbook in ALGEBRA-II in this site
    - ALGEBRA-II - YOUR ONLINE TEXTBOOK.

The referred lessons are the parts of this online textbook under the topics
"Trigonometry. Formulas for trigonometric functions"   and   "Trigonometry: Solved problems".


Save the link to this textbook together with its description

Free of charge online textbook in ALGEBRA-II
https://www.algebra.com/algebra/homework/complex/ALGEBRA-II-YOUR-ONLINE-TEXTBOOK.lesson

into your archive and use when it is needed.