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Trigonometric Series

  • Thread starter ritwik06
  • Start date
  • #1
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Homework Statement



Reduce the following series to the most simplified form:

1. 2 sin[tex]^{2}[/tex]([tex]\theta[/tex]/2)*(1+cos[tex]\theta[/tex])*(1+cos[tex]^{2}[/tex][tex]\theta[/tex])(1+cos[tex]^{4}[/tex][tex]\theta[/tex])............

2. tan ([tex]\theta[/tex]/2)*(1+sec [tex]\theta[/tex])(1+2[tex]\theta[/tex])(1+sec 4[tex]\theta[/tex])...................................

Homework Equations


For problem 1:
2 sin[tex]^{2}[/tex]([tex]\theta[/tex]/2)=1-cos[tex]\theta[/tex])

The Attempt at a Solution



For problem 1:
the result I got:
y=2[tex]^{(n+1)}[/tex]

Result=
1-cos[tex]^{y}[/tex][tex]\theta[/tex]




is it correct?????
For problem 2:
If I try to expand
sec 2 [tex]\theta[/tex]
I get huge denominators which are difficult to solve. Similarly with tan ([tex]\theta[/tex]/2),I again get a denominator with no pattern following any rule.




Please help me for the second one. and tell me if I am correct in solving the first problem.
regards,
Ritwik
 
Last edited:

Answers and Replies

  • #2
580
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So I have finally got that latex code right. I have been trying the second problem for hours. Please help me.
Thanks in advance.
Regards,
Ritwik
 
  • #3
HallsofIvy
Science Advisor
Homework Helper
41,794
925

Homework Statement



Reduce the following series to the most simplified form:

1. 2 sin[tex]^{2}[/tex]([tex]\theta[/tex]/2)*(1+cos[tex]\theta[/tex])*(1+cos[tex]^{2}[/tex][tex]\theta[/tex])(1+cos[tex]^{4}[/tex][tex]\theta[/tex])............

2. tan ([tex]\theta[/tex]/2)*(1+sec [tex]\theta[/tex])(1+2[tex]\theta[/tex])(1+sec 4[tex]\theta[/tex])...................................
I presume you mean
tan ([tex]\theta[/tex]/2)*(1+sec [tex]\theta[/tex])(1+sec(2[tex]\theta[/tex]))(1+sec 4[tex]\theta[/tex])

Homework Equations


For problem 1:
2 sin[tex]^{2}[/tex]([tex]\theta[/tex]/2)=1-cos[tex]\theta[/tex])

The Attempt at a Solution



For problem 1:
the result I got:
y=2[tex]^{n+1}[/tex]
Result=
1-cos[tex]^{y}[/tex][tex]\theta[/tex]




is it correct?????
There is no "n" in the problem so I have no idea what "y= 2n+1" means!

For problem 2:
If I try to expand
sec 2 [tex]\theta[/tex]
I get huge denominators which are difficult to solve. Similarly with tan ([tex]\theta[/tex]/2),I again get a denominator with no pattern following any rule.




Please help me for the second one. and tell me if I am correct in solving the first problem.
regards,
Ritwik
Since you say "huge denominators" I presume you switched to sine and cosine. That shouldn't be necessary. Use the corresponding identities as in (1) for tangent and secant.
 
  • #4
580
0
I presume you mean
tan ([tex]\theta[/tex]/2)*(1+sec [tex]\theta[/tex])(1+sec(2[tex]\theta[/tex]))(1+sec 4[tex]\theta[/tex])


There is no "n" in the problem so I have no idea what "y= 2n+1" means!
'n' stands for the number of terms
Since you say "huge denominators" I presume you switched to sine and cosine. That shouldn't be necessary. Use the corresponding identities as in (1) for tangent and secant.
Yes I switched to sine and cosine. But I wonder which idntitis to use. I have been working over thse problms for 6 hours now. Could u please help???
Thanks
 
  • #5
580
0
'n' stands for the number of terms

Yes I switched to sine and cosine. But I wonder which idntitis to use. I have been working over thse problms for 6 hours now. Could u please help???
Thanks
can this ever help 1+tan^2=sec^2
this is the only identity i know between sec and tan. Please help me someone
 
  • #6
580
0
I really have been spending a wretched time to solve that
tan (x/2)*(1+sec x)(1+sec 2x)(1+sec 4x)...................................
How can I apply any identity here??
tan (x/2)=2 tan (x/4) / 1- tan^2 (x/4)
Please help!!!!!!
 
Last edited:

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