Trig identities double angle help: (cos 3 @ - cos 7 @)/(sin 7 @ + sin 3 @) = tan 2 @

[Barry123]

Hi

I have a question which I am really struggling if anybody can help

prove the identity

cos 3 theta - cos 7 theta divided by sin 7 theta + sin 3 theta = tan 2 theta

If anyone can point me in the right direction it would be appreciated

sorry for for the lack of symbols in the message

Thanks

 

[Deleted member 4993]

Hi

I have a question which I am really struggling if anybody can help

prove the identity

cos 3 theta - cos 7 theta divided by sin 7 theta + sin 3 theta = tan 2 theta

If anyone can point me in the right direction it would be appreciated

sorry for for the lack of symbols in the message

Thanks

Utilize the fact that:

3 = 5 - 2 → cos(3Θ) = cos(5Θ)*cos(2Θ) + sin(5Θ) * sin(2Θ)

7 = 5 + 2 and continue...

What are your thoughts?

Please share your work with us ...even if you know it is wrong

If you are stuck at the beginning tell us and we'll start with the definitions.

You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:

http://www.freemathhelp.com/forum/th...Before-Posting!!

 

[Barry1234]

Utilize the fact that:

3 = 5 - 2 → cos(3Θ) = cos(5Θ)*cos(2Θ) + sin(5Θ) * sin(2Θ)

7 = 5 + 2 and continue...

What are your thoughts?

Please share your work with us ...even if you know it is wrong

If you are stuck at the beginning tell us and we'll start with the definitions.

You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:

http://www.freemathhelp.com/forum/th...Before-Posting!!

Hi

Sorry, i have ha sme issues with my forum login this morning, so i have created a new one.

I have followed on from your advice, and finished off the rest of the idenites to;

3 = 5 - 2 → cos(3Θ) = cos(5Θ)*cos(2Θ) + sin(5Θ) * sin(2Θ)


7 = 5 + 2 → cos(7Θ) = cos(5Θ)*cos(2Θ) - sin(5Θ) * sin(2Θ)


7 = 5 + 2 → sin(7Θ) = sin(5Θ)*cos(2Θ) + cos(5Θ) * sin(2Θ)

3 = 5 - 2 → sin(3Θ) = sin(5Θ)*cos(2Θ) - cos(5Θ) * sin(2Θ)

Which equate to the original

(cos(5Θ)*cos(2Θ) + sin(5Θ) * sin(2Θ) ) - (cos(5Θ)*cos(2Θ) - sin(5Θ) * sin(2Θ))
____________________________________________________________________
(sin(5Θ)*cos(2Θ) + cos(5Θ) * sin(2Θ) ) +(sin(5Θ)*cos(2Θ) - cos(5Θ) * sin(2Θ)

From this dos it cancel out to pretty much 1 , or am i going in the wrong direction??

Thanks again for your assistance.

 

[Deleted member 4993]

Hi

Sorry, i have ha sme issues with my forum login this morning, so i have created a new one.

I have followed on from your advice, and finished off the rest of the idenites to;

3 = 5 - 2 → cos(3Θ) = cos(5Θ)*cos(2Θ) + sin(5Θ) * sin(2Θ)


7 = 5 + 2 → cos(7Θ) = cos(5Θ)*cos(2Θ) - sin(5Θ) * sin(2Θ)


7 = 5 + 2 → sin(7Θ) = sin(5Θ)*cos(2Θ) + cos(5Θ) * sin(2Θ)

3 = 5 - 2 → sin(3Θ) = sin(5Θ)*cos(2Θ) - cos(5Θ) * sin(2Θ)

Which equate to the original

(cos(5Θ)*cos(2Θ) + sin(5Θ) * sin(2Θ) ) - (cos(5Θ)*cos(2Θ) - sin(5Θ) * sin(2Θ))
____________________________________________________________________
(sin(5Θ)*cos(2Θ) + cos(5Θ) * sin(2Θ) ) +(sin(5Θ)*cos(2Θ) - cos(5Θ) * sin(2Θ)

From this dos it cancel out to pretty much 1 , or am i going in the wrong direction??

Thanks again for your assistance.

No it does NOT cancel out to "1".

(sin(5Θ)*cos(2Θ) + cos(5Θ) * sin(2Θ) ) +(sin(5Θ)*cos(2Θ) - cos(5Θ) * sin(2Θ)

= sin(5Θ)*cos(2Θ) + sin(5Θ)*cos(2Θ)

= 2 * sin(5Θ)*cos(2Θ)

Now continue....

 

[Barry1234]

No it does NOT cancel out to "1".

(sin(5Θ)*cos(2Θ) + cos(5Θ) * sin(2Θ) ) +(sin(5Θ)*cos(2Θ) - cos(5Θ) * sin(2Θ)

= sin(5Θ)*cos(2Θ) + sin(5Θ)*cos(2Θ)

= 2 * sin(5Θ)*cos(2Θ)

Now continue....

Hi Subhotosh Khan

Thank you for your assistance with this.

Carrying on from your example then does this go to

2 cos (5Θ) * Cos (2Θ) (from a identity sheet)

If this is not correct then, does this cancel to 1 this time, as you are adding them and then taking them away?

Sorry for my incompetence on this, i am finding it hard to grasp

Thanks again

 

[Barry1234]

I am really struggling with this!!

I have done some work and i think this should cancel to

(cos(5Θ)*cos(2Θ) + sin(5Θ) * sin(2Θ) ) - (cos(5Θ)*cos(2Θ) - sin(5Θ) * sin(2Θ))

I need to get this to

→ (1 - tan² ) But i just cant work out how!!

Sorry for me not graspng this. Any help would be appreciated

 

[Barry1234]

Am i correct in thinking that the top of the fraction goes to

2sin(5Θ)*sin(2Θ)

and the fraction is then

2 * sin(5Θ)*sin(2Θ)
________________
2 * sin(5Θ)*cos(2Θ)

Thanks for your patience with me

 

[Harry_the_cat]

Am i correct in thinking that the top of the fraction goes to

2sin(5Θ)*sin(2Θ)

and the fraction is then

2 * sin(5Θ)*sin(2Θ)
________________
2 * sin(5Θ)*cos(2Θ)

Thanks for your patience with me

Yes so do some cancelling and you have \(\displaystyle tan (2\theta)\)

 

[Steven G]

Am i correct in thinking that the top of the fraction goes to

2sin(5Θ)*sin(2Θ)

and the fraction is then

2 * sin(5Θ)*sin(2Θ)
________________
2 * sin(5Θ)*cos(2Θ)

Thanks for your patience with me

Hint: Sin(273)/cos(273) = tan (273). To prove trig identities you MUST know the basic identities!!

 

[Barry1234]

Thanks for your assistance.

The fraction should be

-2* Sin 5Θ *Sin 2Θ
__________________
2 * sin(5Θ)*cos(2Θ)

On my notes, it says that the -2* Sin 5Θ *Sin 2Θ can be turned into 2* Sin 5Θ *Sin 2Θ by multiplying by -1, does this have to be included in the working out, and will it affect the equation in any way?

I am getting the concepts very slowly :/