Asked by Anonymous
                Prove that each equation is an identity.
I tried to do the problems, but I am stuck.
1. cos^4 t-sin^4 t=1-2sin^2 t
2. 1/cos s= csc^2 s - csc s cot s
3. (cos x/ sec x -1)- (cos x/ tan^2x)=cot^2 x
4. sin^3 z cos^2 z= sin^3 z - sin^5 z
            
        I tried to do the problems, but I am stuck.
1. cos^4 t-sin^4 t=1-2sin^2 t
2. 1/cos s= csc^2 s - csc s cot s
3. (cos x/ sec x -1)- (cos x/ tan^2x)=cot^2 x
4. sin^3 z cos^2 z= sin^3 z - sin^5 z
Answers
                    Answered by
            Steve
            
    cos^4 - sin^4 = (cos^2+sin^2)(cos^2-sin^2)
Remember algebra I and your double-angle formulas?
working with the right side, we have
csc^2 - csc cot
1/sin^2 (1 - cos)
1/(1-cos^2) (1-cos)
(1-cos) / (1+cos)(1-cos)
1/(1+cos)
Hmmm. I don't get 1/cos
(cos x/ (sec x -1))- (cos x/ tan^2x)
(cos/(1/cos-1))-cos^3/sin^2
cos(cos-1) - cos^3/sin^2
cos^2-1 - cos^3/sin^2
-1/sin^2 (sin^4 - cos^3)
Hmmm. I don't see how to make that
cos^2/sin^2
excuse me?
cos^2 = 1-sin^2
and it drops right out
    
Remember algebra I and your double-angle formulas?
working with the right side, we have
csc^2 - csc cot
1/sin^2 (1 - cos)
1/(1-cos^2) (1-cos)
(1-cos) / (1+cos)(1-cos)
1/(1+cos)
Hmmm. I don't get 1/cos
(cos x/ (sec x -1))- (cos x/ tan^2x)
(cos/(1/cos-1))-cos^3/sin^2
cos(cos-1) - cos^3/sin^2
cos^2-1 - cos^3/sin^2
-1/sin^2 (sin^4 - cos^3)
Hmmm. I don't see how to make that
cos^2/sin^2
excuse me?
cos^2 = 1-sin^2
and it drops right out
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