SOLUTION: Please help me solve these questions pls: 1. cos6x = 32cos^6 (x) -48cos^4 (x) +18cos^2 (x) -1 2. (sin(x+y)/cos(x-y)) +1 = ((1+tany)(1+cotx))/(cotx+tany) 3. 2cos^2 (3x) -sin7xsin

Algebra ->  Trigonometry-basics -> SOLUTION: Please help me solve these questions pls: 1. cos6x = 32cos^6 (x) -48cos^4 (x) +18cos^2 (x) -1 2. (sin(x+y)/cos(x-y)) +1 = ((1+tany)(1+cotx))/(cotx+tany) 3. 2cos^2 (3x) -sin7xsin      Log On


   

Question 1045772: Please help me solve these questions pls:
1. cos6x = 32cos^6 (x) -48cos^4 (x) +18cos^2 (x) -1
2. (sin(x+y)/cos(x-y)) +1 = ((1+tany)(1+cotx))/(cotx+tany)
3. 2cos^2 (3x) -sin7xsinx = 1+cos7xcosx
Thanks for your help :)
Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
1.   cos(3x) = cos(2x+x) = cos(2x)cos(x)-sin(2x)sin(x) =
(2cos˛(x)-1)cos(x)-2sin(x)cos(x)sin(x) =
2cosł(x)-cos(x)-2sin˛(x)cos(x) =
2cosł(x)-cos(x)-2[1-cos˛(x)]cos(x) =
2cosł(x)-cos(x)-2cos(x)+2cosł(x)
4cosł(x)-3cos(x)

So cos(3x)=4cosł(x)-3cos(x) and cos(2x)=2cos˛(x)-1. 
So cos(6x)=2[cos(3x)]˛-1 or 2[4cosł(x)-3cos(x)]˛-1.
We can expand it to become 
2[16cos6(x)-24cos4(x)+9cos˛(x)]-1 

Final result: cos(6x) = 

32cos6(x) - 48cos4(x) + 18cos2(x) - 1

------------------------

2.   

%281%5E%22%22%2Btan%28y%29%29%281%5E%22%22%2Bcot%28x%29%29%22%F7%22%28cot%28x%29%5E%22%22%2Btan%28y%29%29


%22%F7%22%28cos%28x%29%2Fsin%28x%29%2Bsin%28y%29%2Fcos%28y%29%29

%22%F7%22%28cos%28x%29cos%28y%29%2Bsin%28y%29sin%28x%29%29%2F%28sin%28x%29cos%28y%29%29

%22%F7%22%28cos%28x%29cos%28y%29%2Bsin%28x%29sin%28y%29%29%2F%28sin%28x%29cos%28y%29%29

%22%F7%22cos%28x-y%29%2F%28sin%28x%29cos%28y%29%29

%22%22%2A%22%22%28sin%28x%29cos%28y%29%29%2Fcos%28x-y%29

%28sin%28x%2By%29%2Bcos%28x-y%29%29%2F%28cos%28y%29sin%28x%29%29%29%22%22%2A%22%22%28sin%28x%29cos%28y%29%29%2Fcos%28x-y%29

%28sin%28x%2By%29%2Bcos%28x-y%29%29%2F%28cross%28cos%28y%29%29cross%28sin%28x%29%29%29%29%22%22%2A%22%22%28cross%28sin%28x%29%29cross%28cos%28y%29%29%29%2Fcos%28x-y%29

%28sin%28x%2By%29%2Bcos%28x-y%29%29%2F1%22%22%2A%22%221%2Fcos%28x-y%29

sin%28x%2By%29%2Fcos%28x-y%29%2B%22%22%5E1cross%28cos%28x-y%29%29%2Fcross%28cos%28x-y%29%29

sin%28x%2By%29%2Fcos%28x-y%29%2B1

1%2Bsin%28x%2By%29%2Fcos%28x-y%29

----------------------
2cos%5E2%283x%29+-sin%287x%29sin%28x%29+=+1%2Bcos%287x%29cos%28x%29

This one requires some tricks of subtracting and adding the
same quantity to create a use for some double-angle identities. 
We work with the left side:

Subtract 1 and add 1 to create a use for the identity
                cos%282theta%29=2cos%5E2%28theta%29-1

2cos%5E2%283x%29-1%2B1-sin%287x%29sin%28x%29

Using that identity, the first two terms become cos(6x) 

cos%286x%29%2B1-sin%287x%29sin%28x%29

Subtract and add cos(7x)cos(x) to create a
use for the identity cos%28alpha-beta%29=cos%28alpha%29cos%28beta%29%2Bsin%28alpha%29sin%28beta%29
 
cos%286x%29%2B1-sin%287x%29sin%28x%29-cos%287x%29cos%28x%29%2Bcos%287x%29cos%28x%29

Factor a "-" out of 3rd and 4th terms:



Swap the terms in the parentheses to recognize the identity:



Use the identity

cos%286x%29%2B1-cos%287x-x%29%2Bcos%287x%29cos%28x%29

7x-x = 6x

cos%286x%29%2B1-cos%286x%29%2Bcos%287x%29cos%28x%29

Cancel the two opposite terms

cross%28cos%286x%29%29%2B1-cross%28cos%286x%29%29%2Bcos%287x%29cos%28x%29

1%2Bcos%287x%29cos%28x%29

Tricky, huh?

Edwin