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avatar+777 

for this i tried squaring both sides which gave me sin^2 x = 2-1 = 1

which didn't work

then i tried going backwards with the cos equation 

cos^2x = (1-sin^2 x) = a + b sqrt2 

rearranged to get sin^2x = -a -b sqrt2 + 1

i didnt know what to do from here 

please help thanks

 Feb 3, 2019
 #1
avatar+98153 
+1

sin x =  sqrt (2)  - 1

 

sin^2 x = [ sqrt (2) - 1 ] ^2  = [ sqrt (2) - 1 ] * [ sqrt (2) - 1 ] =  

 

sqrt (2) *sqrt (2) - 1*sqrt(2) - 1*sqrt(2) + (-1)(-1)  =

 

2 - 2sqrt (2) + 1  =

 

[ 3 - 2sqrt (2) ]

 

So.....using an identity.....

 

cos^2x = 1 - sin^2 x

 

And we have that

 

1 - [ 3 - 2sqrt(2) ] =

 

-2 + 2sqrt (2)   .....so   a = -2   and b = 2 

 

 

cool cool cool

 Feb 3, 2019
 #2
avatar+777 
+1

ah! so you square the entire rhs as a bracket instead, thanks c phill

YEEEEEET  Feb 3, 2019

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