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Mathematics 13 Online
OpenStudy (anonymous):

assume that sin(pi/8) = (sqrt(2-sqrt2))/2 use identities to find the exact function value of: cos(pi/8) sin(17pi/8)

OpenStudy (anonymous):

Ok do you know about the identity where sin(some angle) = sin(some angle + 2(pi))

OpenStudy (anonymous):

yes, periodicity identity right ?

OpenStudy (anonymous):

yup, the sin, cos, and tan functions are all periodic

OpenStudy (anonymous):

sin(pi/8) = sin(17pi/8) it's the same thing

OpenStudy (anonymous):

To find cos(pi/8), u can use the identity cos^2 + sin^2 =1

OpenStudy (anonymous):

Oh okay got that one im kind of struggling with the first one lol.

OpenStudy (anonymous):

Oh okay so it would be √(1 - ((2-√2)/2) ? o-o

OpenStudy (anonymous):

you square your function, plug it in, then find cos^2

OpenStudy (anonymous):

I'm not sure what you're doing. Just square sin(pi/8). You'll get|dw:1398641439819:dw|

OpenStudy (anonymous):

so you have cos^2 + (2-rad2/4) = 1

OpenStudy (anonymous):

solve for cos^2, then rad it

OpenStudy (anonymous):

Oh.

OpenStudy (anonymous):

one sec, doing the problem.

OpenStudy (anonymous):

redoing**

OpenStudy (anonymous):

so it will be|dw:1398641584945:dw|

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