Explanation: #sin (pi/3)xxtan (pi/6)# #" "# #=sqrt3/2xxsqrt3/3# #" "# #=sqrt (3^2)/ (2xx3)# #" "# #=3/ (2xx3)# #" "# #=1/2#
Then it decreasing until x = pi where again sin pi = 0 For x<0 the graph is symmetric with respect to the origin, as sin (-x) = -sinx We can also note that in the interval (0,pi) the graph is symmetric with respect to the line y=pi/2 since: sin (pi/2-h) = sin (pi/2)cosh - cos (pi/2)sinh = sin (pi/2)cosh sin (pi/2+h) = sin (pi/2)cosh + cos (pi/2 ...
It is now a simple matter of substituting the values worked out from the list above in place of f^n (x) like so: f (x)=sin (pi/5)+cos (pi/5)x-sin (pi/5)x^2/2-cos (pi/5)x^3/6+...
So, dividing by cos (k), we get tan (k) = sqrt (3)/3 One of the solution of this equation is k=pi/6, then from the first equation of our system c=sin (pi/6)=1/2. f (x) = sin (x-pi/6)+1/2
1/α The key here is to know that: cos (x)=sin (pi/2-x) Given: sec ( (4pi)/9)=α Then: cos ( (4pi)/9)=sin (pi/2- (4pi)/9) cos ( (4pi)/9)=sin ( (9pi-8pi)/18) cos ...
However, #x_1=0# and #x_3=pi# don't provide original problem. Hence #x_2=pi/2# and #x_4= (3pi)/2# are solutions of it. #2)# For #sin4x=sin (pi+2pik)# or #4x=pi+2pik#,
Let x=cos (pi/15)cos ( (2pi)/15)cos ( (4pi)/15)cos ( (7pi)/15) = (2^4sin (pi/15)cos (pi/15)cos ( (2pi)/15)cos ( (4pi)/15)cos (pi- (8pi)/15))/ (2^4sin (pi/15)) =- (2^4sin (pi/15)cos (pi/15)cos ( (2pi)/15)cos ( (4pi)/15)cos ( (8pi)/15))/ (2^4sin (pi/15)) =-sin ( (16pi)/15)/ (2^4sin (pi/15)) =-sin (pi+pi/15)/ (2^4sin (pi/15)) =sin (pi/15)/ (2^4sin ...
The maximum repeats at x = pi/8+npi; n in ZZ The first one is at: 8sin (pi/8)cos (pi/8)+4-8sin^2 (pi/8) = 4sqrt2 Compute the first derivative: (d (8sin (x)cos (x)+4-8sin^2 (x)))/dx = 8 (cos^2 (x)-sin^2 (x))-16sin (x)cos (x) Is see the double angle identities cos (2x) = cos^2 (x)-sin^2 (x) and sin (2x) = 2sin (x)cos (x): (d (8sin (x)cos (x)+4-8sin^2 (x)))/dx = 8 (cos (2x)-sin (2x)) Set the ...