Need some help with a few difficult Trig proofs. (1 Viewer)

[S1M0]

Read the title? Great, i need help with these trig proofs. Damn textbook doesn't have any answer, lazy buggers.

Prove:

Sin 2A. Cos A - Cos 2A . Sin A / Cos 2A. Cos A + Sin 2A. Sin A = Tan A

Tan (45 + A) + (Tan 45 - A) = 2/Cos 2A)

(1 - Cos x) / Sin x = Tan (x/2)

Help with be much appreciated, as always.

Edit: In case i haven't indicated (and i haven't), i got these questions wrong.

 

[lacklustre]

Read the title? Great, i need help with these trig proofs. Damn textbook doesn't have any answer, lazy buggers.

Prove:

Sin 2A. Cos A - Cos 2A . Sin A / Cos 2A. Cos A + Sin 2A. Sin A = Tan A

Tan (45 + A) + (Tan 45 - A) = 2/Cos 2A)

(1 - Cos x) / Sin x = Tan (x/2)

Help with be much appreciated, as always.

Edit: In case i haven't indicated (and i haven't), i got these questions wrong.

What textbook was it by the way?

 

[Evergreen]

For 1st question

make top sine only, bottom cosine only. therefore you will get the result of sinA/cosA = tanA.

2nd question:

use the double angle rule for tan. add the them together by multiplying by (1-tanA) and (1+tanA) to form a common denominator. then simplify the terms then you convert tan squared A into sin squared A/ cos squared A. Then its easy from there.

3rd question:

use t method. i.e. let tan (x/2)= t. then use t results for cosine and sine and then you will get that 1-cosx/sinx= t. .: 1-cosx/sinx= tan (x/2)


The key is that you have RHS to get an idea how to derive it from LHS. So try to think of an expression that would ultimately lead to RHS. I think these questions are from cambridge but i might be wrong.

 

[S1M0]

For 1st question

make top sine only, bottom cosine only. therefore you will get the result of sinA/cosA = tanA.

2nd question:

use the double angle rule for tan. add the them together by multiplying by (1-tanA) and (1+tanA) to form a common denominator. then simplify the terms then you convert tan squared A into sin squared A/ cos squared A. Then its easy from there.

3rd question:

use t method. i.e. let tan (x/2)= t. then use t results for cosine and sine and then you will get that 1-cosx/sinx= t. .: 1-cosx/sinx= tan (x/2)


The key is that you have RHS to get an idea how to derive it from LHS. So try to think of an expression that would ultimately lead to RHS. I think these questions are from cambridge but i might be wrong.

Great, thanks for the help. Its not that i don't know how to do the questions, but rather its that i seem to screw up with the theorems and the actual working.

On, and i just have one more (if thats alright):

By writing 3A in terms of 2A + A. Express:

Sin3A.

 

[webby234]

sin(3A)
= sin(2A + A)
= sin(2A)cos(A) + sin(A)cos(2A)
= 2sinAcos²(A) + sin(A)(1 - 2sin²(A))
= 2sinA(1 - sin²A) + sin(A) - 2sin³(A))
= 2sinA - 2sin³A + sinA - 2sin³A
= 3sinA - 4sin³A

 

[S1M0]

sin(3A)
= sin(2A + A)
= sin(2A)cos(A) + sin(A)cos(2A)
= 2sinAcos²(A) + sin(A)(1 - 2sin²(A))
= 2sinA(1 - sin²A) + sin(A) - 2sin³(A))
= 2sinA - 2sin³A + sinA - 2sin³A
= 3sinA - 4sin³A

Is that in terms of A? Because thats what i got, and according to these answers, i'm wrong.

 

[Evergreen]

Yeah thats the correct general result of sin(3A).

 

[S1M0]

Yeah thats the correct general result of sin(3A).

Great, thanks a lot buddy!