How to find the exact value of cos 67.5 ?

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To find the exact value of cos(67.5), recognizing it as the half angle of cos(135) simplifies the process. Initial attempts using cos(45 + 45/2) led to complications, but suggestions to square the formula and simplify proved effective. Multiplying by √(2 - √2) was mentioned as a potential simplification method. The relationship between cos(x) and cos(x + 90) was also discussed, highlighting connections to other trigonometric functions. Ultimately, squaring the initial formula and taking the square root provided a clearer path to the solution.
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One of the questions in the homework for online precalc class I'm taking is to find the exact value for cos (67.5).
At first I didn't realize that I should just take the half angle of 135, so instead I tried to find the value for cos (45 + 45/2), and got stuck at this point.

attachment.php?attachmentid=44118&stc=1&d=1329679233.png


If I calculate the actual value I get the correct answer, but I'm not able to see how I can simplify this further.
I know the solution is as simple as finding the half angle value of

attachment.php?attachmentid=44119&stc=1&d=1329679233.png


Any suggestions of how I can get from my first attempt solution to the simpler (and correct) solution?
 

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How is Cos(x) related to Cos(x+90), how is Cos(x+90) related to other trig functions?
 
Try multiplying both the top and bottom by ##\sqrt{2-\sqrt{2}}##.
 
>>How is Cos(x) related to Cos(x+90), how
yeah, that definitely makes it simpler

attachment.php?attachmentid=44137&stc=1&d=1329699522.png


>>Try multiplying both the top and bottom by
I have tried, but I seem to end up with a square root I can't get rid of, but I will keep trying.


Thanks for your help
Jesper
 

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Hi, jkristia,

Square your formula for cos(45+45/2), simplify, and then take the square root.ehild
 
ehild said:
Hi, jkristia,

Square your formula for cos(45+45/2), simplify, and then take the square root.


ehild

ah - of course, that works
Thank you very much
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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