" Q."2quad P_(n)=cos^(n)x+sin^(n)x," then "2P_(6)-3P_(4)+1=

If P_(n)=cos^(n)theta+sin^(n)theta then 2P_(6)-3P_(4)+1=

If P_(n) = cos^(n)theta +sin^(n)theta , then P_(2) - P_(1) is equal to :

If P_(n) = cos^(n)theta +sin^(n)theta , then P_(2) - P_(1) is equal to :

If P_n = cos^n theta+ sin^n theta then 2P_6-3P_4 + 1 =

If P_n = cos^n theta+ sin^n theta then 2P_6-3P_4 + 1 =

If P_(n) = cos ^(n)theta + sin^(n)theta then 2. P_(6) - 3. P_(4) + 1 =

If P_n=cos^ntheta+sin^ntheta, show that : 2P_6-3P_4+1=0 .

If p_(n)=cos^(n)theta+sin^(n)theta then p_(n)-p_(n-2)=kp_(n-4) where k

If P_(n)=cos^(n)theta+sin^(n)theta and Q_(n)=cos^(n)theta-sin^(n)theta then show that p_(n-2)=-sin^(2)theta cos^(2) theta p_(n-4) hence show that p_(4)=1-2 sin^(2) theta cos^(2) theta Q_(4)=cos^(2) theta- sin^(2) theta

If p_n =cos^n theta+sin ^ntheta then the value of 2P_6-3P_4+1 will be