Please help me in these identities

  • Thread starter enibaraliu
  • Start date
  • #1
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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
 

Answers and Replies

  • #2
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please , I need help
 
  • #3
35,237
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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.
 
  • #4
35,237
7,056


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°.
 
  • #5
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Try if these are identities

sin18°+sin30°=sin54°

(cos36°)^2 + (sin18°)^2=3/4, please help on these
 
  • #6
35,237
7,056


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?
 

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