Solve tan^2a=tanbtanc: Prove and Simplify

  • Thread starter Thread starter shravan
  • Start date Start date
Click For Summary
SUMMARY

The discussion focuses on proving the equation tan²a = tan b tan c using trigonometric identities. Participants emphasize the importance of manipulating the equation sin(a+b)/sin(a+c) = [sin(2b)/sin(2c)]^(1/2) to derive the desired result. Key strategies include squaring terms and applying fundamental trigonometric identities such as sin(a+b) and cos(a+b). The consensus is that while the proof is complex, it is achievable through systematic algebraic manipulation.

PREREQUISITES
  • Understanding of trigonometric identities, specifically sin(a+b) and cos(a+b).
  • Familiarity with the tangent function and its relationship with sine and cosine.
  • Ability to manipulate algebraic expressions, including squaring and factoring.
  • Knowledge of angle relationships in triangles, particularly the sum of angles equaling 180 degrees.
NEXT STEPS
  • Study the derivation and applications of the sine and cosine addition formulas.
  • Learn advanced techniques for manipulating trigonometric equations, including squaring and factoring methods.
  • Explore the properties of tangent functions and their applications in various trigonometric proofs.
  • Investigate the implications of angle relationships in triangle geometry, particularly in proofs involving sums of angles.
USEFUL FOR

Mathematicians, students studying trigonometry, and educators looking to deepen their understanding of trigonometric proofs and identities.

shravan
Messages
16
Reaction score
0
sin(a+b)/sin(a+c)=[ sin(2b)/sin(2c)]^(1/2)
then prove tan^2a=tanbtanc
I have reached till {tan(a)cos(b)+ sin(b)} * {sin(c)cos(c)}^(1/2)=
{tan(a)cos(c)+sin(c)}* { sin(b) cos(b)}^(1/2)
 
Physics news on Phys.org
shravan said:
sin(a+b)/sin(a+c)=[ sin(2b)/sin(2c)]^(1/2)
then prove tan^2a=tanbtanc
I have reached till {tan(a)cos(b)+ sin(b)} * {sin(c)cos(c)}^(1/2)=
{tan(a)cos(c)+sin(c)}* { sin(b) cos(b)}^(1/2)


Can you assume that a+b+c = 180 degrees ?

I think there is something missing in your question. How elese can you prove the first equality ?

Or is it given ? In that case, you can find a connection between the angles by manipulating this first equality.

marlon
 
you have all the info you need. i squared everything, and then solved for tan^2a. its messy, but it does work out. when you square it all, the tana terms cancel. and then you just have to factor and divide ect.
 
no there is no relation between the angles.the relation given is not going to help much. I think I will be getting a method for that sum.however thank u
 
it works out. i said that already. just keep trucking. like i said, square everything and start gathering terms. its just a lot of math, but it does work.
 
try using some of these identities

Tan=cos/sin

Sin(a+b)=
SinaCosb+CosaSinb

Cos(a+b)
CosaCosb-SinaSinb

ya that's my knowledge of trig
 
the OP already used those identities. the only other identy needed is knowing tanx= sinx/cosx. otherwise its all just rearranging the equation.
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 18 ·
Replies
18
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
4
Views
1K
  • · Replies 8 ·
Replies
8
Views
4K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K