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Link to original question: https://web2.0calc.com/questions/simplify-each-expression_1

CPhill answered but he dropped the 2 on the -2cos4x.

 Mar 30, 2018
 #1
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+1

Simplify: Sorry Adam, no LaTex !!
sin(x)^4 - sin(x)^2 cos(x)^2 - 2 cos(x)^4 = 

 

Express sin(x)^2 in terms of cosine via the Pythagorean identity.
sin(x)^2 = 1 - cos(x)^2:
-2 cos(x)^4 - cos(x)^2 1 - cos(x)^2 + sin(x)^4 =

 

Expand -cos(x)^2 (1 - cos(x)^2).
-cos(x)^2 (1 - cos(x)^2) = cos(x)^4 - cos(x)^2:
-2 cos(x)^4 + cos(x)^4 - cos(x)^2 + sin(x)^4 =

 

Express sin(x)^4 in terms of cosine via the Pythagorean identity.
sin(x)^4 = (sin(x)^2)^2 = (1 - cos(x)^2)^2:
-2 cos(x)^4 - cos(x)^2 + cos(x)^4 + (1 - cos(x)^2)^2 =

 

Expand (1 - cos(x)^2)^2.
(1 - cos(x)^2)^2 = 1 - 2 cos(x)^2 + cos(x)^4:
-2 cos(x)^4 - cos(x)^2 + cos(x)^4 + 1 - 2 cos(x)^2 + cos(x)^4 =

 

Evaluate -2 cos(x)^4 - cos(x)^2 + cos(x)^4 + 1 - 2 cos(x)^2 + cos(x)^4.
-2 cos(x)^4 - cos(x)^2 + cos(x)^4 + 1 - 2 cos(x)^2 + cos(x)^4 


= 1 - 3cos(x)^2

 Mar 30, 2018
 #4
avatar+865 
0

Thanks Guest!

AdamTaurus  Mar 30, 2018
 #2
avatar+100571 
+1

Sorry for the earlier error!!!

 

This becomes a little easier if we factor up front

 

sin^4x -sin^2xcos^2x -2cos^4x      factor as

 

(sin^2x  - 2cos^2x) (sin^2x + cos^2x)

 

(sin^2x - 2cos^2x)  (1)

 

sin^2x  - 2cos^2x

 

sin^2x  - 2 (1 - sin^2x)

 

3sin^2x -  2   or

 

3(1 - cos^2x) - 2

 

3 - 3cos^2x - 2

 

1 - 3cos^2x

 

 

cool cool cool

 Mar 30, 2018
 #3
avatar+865 
+1

Thanks CPhill!

AdamTaurus  Mar 30, 2018
 #5
avatar+100571 
0

OK, Adam  !!!

 

 

cool cool cool

CPhill  Mar 30, 2018

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