cos(π/4+x)=3/5,17π/12
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cos(π/4+x)=3/5,17π/12cos(π/4+x)=3/5,17π/12cos(π/4+x)=3/5,17π/12cos(π/4+x)=3/5,cos2(π/4+x)=2[cos(π/4+
cos(π/4+x)=3/5,17π/12
cos(π/4+x)=3/5,17π/12
cos(π/4+x)=3/5,17π/12
cos(π/4+x)=3/5,cos 2(π/4+x) = 2 [cos (π/4+x)]^2 -1 = 2* (9/25) -1 = - 7/25 = cos (π/2+ 2x) = - sin 2x
sin 2x= 7/25 ; cos[π/4+(π/4+x)] = cosπ/4* cos(π/4+x)- sin π/4*sin[(π/4+x)]= 1/2 (2^(1/2)* (3/5) - 1/2 (2^(1/2)* (-4/5) = 7* 2^(1/2)/10 = cos (π/2+x) = - sin x,sin x= -7* 2^(1/2)/10,tan x = 1/7,finally,answer is :
(7/25) + 2* (49/625)/ [1- (1/49)]
cos(π/4+x)=3/5,17π/12
cos(π/4+X)=3/5,17π/12
cos(π/4+x)=3/5,17π/12
cos(π/4+x)=3/5,17π/12
cos(π/4+x)=3/5,17/12π
cos(π/4+x)=3/5,17π/12
cos(π/4+x)=3/5,17π/12
cos(π/4+x)=3/5,-17π/12
cos(2x-π)/[根号2cos(x+π/4)]=-1/5,0
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