Reduce Trigonometric Series: 2 Problems Solved

[ritwik06]

Homework Statement

Reduce the following series to the most simplified form:

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

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

Homework Equations

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

The Attempt at a Solution

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

Result=
1-cos^{y}\theta




is it correct?
For problem 2:
If I try to expand
sec 2 \theta
I get huge denominators which are difficult to solve. Similarly with tan (\theta/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

 

[ritwik06]

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

 

[HallsofIvy]

Homework Statement

Reduce the following series to the most simplified form:

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

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

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

Homework Equations

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

The Attempt at a Solution

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




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 \theta
I get huge denominators which are difficult to solve. Similarly with tan (\theta/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.

 

[ritwik06]

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


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

 

[ritwik06]

'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

 

[ritwik06]

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!