Can You Solve These Trigonometric Identity Problems?

Click For Summary

Discussion Overview

The discussion revolves around solving trigonometric identity problems, specifically focusing on proving various identities involving sine, cosine, secant, tangent, and cotangent functions. Participants are seeking assistance with their homework and are exploring different approaches to these problems.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents three trigonometric identities to prove, requesting detailed solutions.
  • Another participant attempts to simplify the second identity, noting that cotangent must not be zero and derives an identity involving secant and tangent.
  • This participant also emphasizes the importance of the Pythagorean identity, stating that it is a basic identity that should be known.
  • For the third identity, a participant expresses doubt about its validity, suggesting that substituting tangent leads to an incorrect statement.
  • Another participant argues that the last identity can be proven true by manipulating the expression and converting everything to sine and cosine.

Areas of Agreement / Disagreement

There is disagreement regarding the validity of the third identity. Some participants believe it does not hold true, while others argue that it can be proven correct through algebraic manipulation.

Contextual Notes

Participants express uncertainty about the correctness of certain identities and the conditions under which they hold. There are also references to specific values (like x = 32 degrees) that may affect the validity of the identities.

Johnny Neutron
Messages
1
Reaction score
0
Trig Identities question for test tomorrow Help!

Need some help with these two problems:

Thanks in advance if you could answer them:

sin (X) / Cos (x) - 1 = show work

Sec ^2/ cot (x) - Tan ^3x = Tan X

show work to prove

last one

Sec x - Cos x/tanx = sinx

show work to prove
 
Mathematics news on Phys.org
Johnny Neutron said:
sin (X) / Cos (x) - 1 = show work
What?
Sec ^2/ cot (x) - Tan ^3x = Tan X
[tex]\frac{\sec ^2 x}{\cot x} - \tan ^3 x = \tan x[/tex]

Note, [itex]\cot x[/itex] must not be zero. Now, multiply by [itex]\cot x[/itex]:

[tex]\sec ^2 x - \tan ^2 x = 1 \ \dots \ (1)[/tex]

Note, if [itex]\cot x[/itex] were zero, then [itex]\cos x[/itex] would have to be zero (since [itex]\cot x = \frac{\cos x}{\sin x}[/itex]), but since it's not, then [itex]\cos x \neq 0[/itex]. So, we can multiply both sides by [itex]\cos ^2 x[/itex]:

[tex]1 - \sin ^2 x = \cos ^2 x[/tex]

[tex]\sin ^2 x + cos ^2 x = 1[/tex]

This is a basic identity you should know. In fact (1) is a commonly used identity too, but I figured I'd get you down at least this far. I assume you won't have to prove this. If you do, then you know that [itex]\sin x[/itex] is the ratio of the side opposite the angle x in a right triangle to the hypoteneuse. You should also know the definition for [itex]\cos x[/itex]. With these two definitions and the Pythagorean Theorem, you should be able to prove those two identities.

Sec x - Cos x/tanx = sinx

As a general approach to any of these kinds of problems, express everything in terms of sine and cosine. Mutiplying both sides by [itex]\sin x \cos x[/itex]:

[tex]\sin x - \cos ^3 x = \sin ^2 x \cos x[/tex]

[tex]\sin x = \cos x(\cos ^2 x + \sin ^2 x)[/tex]

[tex]\sin x = \cos x[/tex]

This is wrong. Try x = 32 degrees. It doesn't work. I guess it's a trick question or you mistyped (or I made a mistake).
 
Yea I'm almost positive that last one doesn't work. If you make that tanx, sinx/cosx you are left with secx = 2sinx, which is not true.
 
But (sec x- cos x)/tan x= sin x is true.

As AKG suggested change everything to sin and cos:

(1/cos x- cos x)/(sin x/cos x)

= ((1- cos<sup>2</sup> x)/cos x)(cos x/sin x)
= (sin<sup>2</sup> x/cos x)(cos x/sin x)
= sin x
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 28 ·
Replies
28
Views
3K
  • · Replies 17 ·
Replies
17
Views
7K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
9
Views
3K