value of 6p10-15p8+10p6+7
Answers
Answer: 8
Concept: Use of trigonometric formula
Given: Pn= cos^nx+sin^nx
To Find: Value of 6p10-15p8+10p6+7
Step-by-step explanation:
Let's first find the value of Pn-Pn-2
Pn-Pn-2= cos^(n)x+sin^(n)x- cos^(n-2)x- sin^(n-2)x
=cos^(n-2)x[cos^2x-1]+sin^(n-2)x[sin^2x-1] ...................(1)
Now as per trigonometry formula
sin^2x+cos^2x=1
⇒sin^2x-1= -cos^2x ..................... (a)
⇒cos^2x-1= -sin^2x .......................(b)
Now place (a) and (b) in (1)
Pn- Pn-2= cos^(n-2)x[ -sin^2x] + sin^(n-2)x [-cos^2x]
= -sin^2x.cos^2x [ cos^(n-4)x+ sin^(n-4)x ]
= -sin^2x.cos^2x [ Pn-4 ] ........................( 2)
Now we have to find value of
6P10-15P8+10P6+7
= 6P10-6P8-9P8+9P6+P6-P4+P4-P2+P2+7
= 6[P10-P8]-9[P8-P6]+[P6-P4]+[P4-P2]+P2+7
Now using (2) in above equation
= -6[ sin^2x.cos^2x.P6]+9[sin^2xcos^2x.P4]+P2+P0+sin^2x+cos^2x+7
= -sin^2x.cos^2x[6P6-9P4+9P2+P0]+1+7
= -sin^2x.cos^2x[6P6-6P4-3P4+3P2-2P2+P0]+8
= -sin^2x.cos^2x[6P2-3P0]+8 .......................(3)
Now P0=cos^0.x + sin^0.x= 1+1= 2
Placing value of P0 in (3)
= -sin^2x.cos^2x [6-6]+8
= 0+8
= 8
The correct answer is 8.
Project code:#SPJ2