We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
157
2
avatar+844 

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+102995 
+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+844 
+1

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

YEEEEEET  Feb 3, 2019

47 Online Users

avatar
avatar
avatar