Solving Trigonometric Identities: Tips and Examples

  • Thread starter Thread starter enibaraliu
  • Start date Start date
  • Tags Tags
    identities
AI Thread Summary
The discussion revolves around solving specific trigonometric identities, with the original poster expressing difficulty in verifying several equations. Key points include the need for clarity in the notation of the identities, particularly regarding the placement of parentheses in the equations. The first identity is confirmed as valid, while the others require further clarification to determine their validity. Suggestions include using double-angle formulas and substituting known values, such as sin(30°) = 1/2, to assist in solving the identities. Overall, the thread emphasizes the importance of precise notation and systematic approaches in solving trigonometric identities.
enibaraliu
Messages
12
Reaction score
0
Please help me in these identities!

I tried to solve these identities, but I don't think that these are identities exept the first,
1.) sin2α/1+cos2α=tanα
2.) 1-cosα/sinα=tanα/2
3.) tanα+ctanα=csecα
4.) 3cosα+cos3α/3sinα-sin3α=3/2
5.) sin18°+sin30°=sin54

At first i did this:
sin2α=2sinα*cosα
cos2α=cos^2α-sin^2α
1=cos^2α+sin^2α
sin2α/1+cos2α=2sinα*cosα/cos^2α+sin^2α+cos^2α-sin^2α=...=tanα

But I don't know what to do eith others i tried to substitute tanα=sinα/cosα and etc but no solve ,
please help me
 
Physics news on Phys.org


please , I need help
 


You don't have a single parenthesis in what you show, which probably means that what you think you wrote and what people will read are different.

For example, in 1, you have sin2α/1+cos2α=tanα.
Is 1 + cos2a in the denominator? If so, write the equation this way:
sin2α/(1+cos2α)=tanα

In 2, do you mean
(1-cosα)/sinα=(tanα)/2 or
(1-cosα)/sinα=tan(α/2) or something else?

In 4, do you mean (3cosα+cos3α)/(3sinα-sin3α)=3/2?

In the last equation you wrote, you have cos^2a. Do you mean cos^2(a)? You probably do, because cos^(2a) doesn't make any sense.
 


Please clarify what you meant for 1, 2, and 4. By that, I mean, if the numerator or denominator (or both) has more than one term, put parentheses around the terms in the numerator and/or denominator.

Number 5 appears to be true, assuming that what you have on the right side is sin 54°.
 


Try if these are identities

sin18°+sin30°=sin54°

(cos36°)^2 + (sin18°)^2=3/4, please help on these
 


I believe that the first one is an identity. The right side is sin(3*18°), and I would start with it.
sin(3*18°) = sin(2*18° + 18°) = sin(2*18°)cos(18°) + cos(2*18°)sin(18°). Now use the double-angle formulas on that expression and see if that helps. Also, sin(30°) = 1/2.

For the second, cos(36°) = cos(2*18°). What can you do with that?
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
Back
Top