Need help with some identity problems

  • Thread starter thakid87
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  • #1
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I'm having difficulties with a few identity problems and I wanted to make sure I'm doing the ones I believe I did correctly, correctly...

1. (cos^3x)+(sin^2x)(cosx)
(cosx)(cos^2x)+(sin^2x)(cosx)
2cosx

2. (1+cosy)/(1+secy)
(1+cosy)/(1+1/cosy)
(1+cosy)/(1+cosy)
1

3. (tanx)/(secx)
(sinx/cosx)/(1/cosx)
(sinx/cosx)(cosx)
sinx

4.(secx-cosx)/(tanx)
(secx-cosx)(cosx/sinx)
(1/sinx)-(cos^2x/sinx)
(1-cos^2x)/(sinx)
(1/sinx)
cscx

5.(sinx/cscx)+(cosx/secx)
(sinx/1/sinx)+(cosx/1/cosx)
(sinxsinx)+(cosxcosx)
sin^2x+cos^2x
1

6. (1+sinx/cosx)+(cosx/1+sinx)
(1+sinx/1+sinx0(1+sinx/cosx)+(cosx/cosx)(cosx/1+sinx)
(1+2sinx+sin^2x+cos^2x)/(1+sinx)(cosx)
(2+2sinx)/(1+sinx)(cosx)
(2+sinx)/(cosx)
2tanx

7. (tanx)(cosx)(cscx)
(sinx/cosx)(cosx)(1/sinx)
1

The ones that I'm not too sure about are 2 and 8. I feel pretty confident about the other ones, but I wanted to make sure I'm doing them correctly.

Thanks a lot to anybody that helps.
 

Answers and Replies

  • #2
rock.freak667
Homework Helper
6,223
31
For 2
2. (1+cosy)/(1+secy)
(1+cosy)/(1+1/cosy)
(1+cosy)/(1+cosy)
1

From the text in red, multiply both the numerator and denominator by cosy


I don't see a question 8...the only one that looked wrong at a glance was 2.
 
  • #3
First, you did not put a #8 up so I guess you are confident about that one lol.

I didn't look at #6 b/c my eyes are tired, but you will want to check 2 and 4!

for #2 you have that 1+(1/cos(y)) = 1 + cos(y) !! since when is that true. instead, you should multiply the numerator and denominator by cos(y) then see what you get.

#4 the hint is to look at what 1 - (cos(x))^2 is
 
  • #4
HallsofIvy
Science Advisor
Homework Helper
41,847
966
I'm having difficulties with a few identity problems and I wanted to make sure I'm doing the ones I believe I did correctly, correctly...

1. (cos^3x)+(sin^2x)(cosx)
(cosx)(cos^2x)+(sin^2x)(cosx)
2cosx
Where did you get the "2"?

(cosx)(cos^2x)+ (sin^2x)(cosx)= cos(x)(cos^2x+ sin^2x).

2. (1+cosy)/(1+secy)
(1+cosy)/(1+1/cosy)
(1+cosy)/(1+cosy)
How did you make this jump? 1/cosy is NOT equal to cos y.

3. (tanx)/(secx)
(sinx/cosx)/(1/cosx)
(sinx/cosx)(cosx)
sinx
Yes, that's right.

4.(secx-cosx)/(tanx)
(secx-cosx)(cosx/sinx)
(1/sinx)-(cos^2x/sinx)
You've got your parenthes wrong. Also sec x= 1/cosx not 1/sin x
(1/cosx- cosx)(cos(x)/sin(x)= (1/cox)(cosx/sinx)- cos^2x/sinx= 1/sinx- cos^2x/sinx
= (1- cos^2x)/sinx

5.(sinx/cscx)+(cosx/secx)
(sinx/1/sinx)+(cosx/1/cosx)
(sinxsinx)+(cosxcosx)
sin^2x+cos^2x
1
Yes, that's correct.

6. (1+sinx/cosx)+(cosx/1+sinx)
Again, you are not being careful with parentheses. I think you mean
(1+ sinx)/cosx+ cosx/(1+ sin(x)) bit what you wrote is 1+ (sinx/cosx)+ cosx+ sin(x).

(1+sinx/1+sinx0(1+sinx/cosx)+(cosx/cosx)(cosx/1+sinx)[/quuote]
Now this is just much too complicated and without parentheses impossible to understand!

(1+2sinx+sin^2x+cos^2x)/(1+sinx)(cosx)
(2+2sinx)/(1+sinx)(cosx)
(2+sinx)/(cosx)
??What happened to the "1+ sinx" in the denominator? 2+ 2sinx= 2(1+ sinx) in the numerator and then you can cancel with the "1+ sinx" in the numerator:
= 2/cosx


7. (tanx)(cosx)(cscx)
(sinx/cosx)(cosx)(1/sinx)
1
yes, that is correct.

The ones that I'm not too sure about are 2 and 8. I feel pretty confident about the other ones, but I wanted to make sure I'm doing them correctly.

Thanks a lot to anybody that helps.
 

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