Trigonometry Refresher

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In summary, the conversation is about proving two trigonometry identities involving secant and tangent functions. The first identity involves a sum/difference formula for secant and the second involves proving a complex equation involving multiple trigonometric functions. The person is seeking help with solving the equations and the conversation ends with them deciding to figure it out on their own.
  • #1
The_ArtofScience
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Homework Statement



I'm currently in cal 1 and we are doing trigonometry identities for review. I can't recall ever seeing a sum/difference formula for secant in precal

(1) Prove secant² (4θ) / 2 - secant² (4θ) = secant (8θ)
(2) Prove tan(x)/(1-cot(x)) + cot(x)/(1-tan(x)) = sec(x)csc(x) + 1

The Attempt at a Solution



(1) secant² (4θ) / 2 - secant² (4θ) = -1/2 *(sec²(4θ))

(2) tan(x)/(1-cot(x)) + cot(x)/(1-tan(x))

=[ tan(x)(1-tan(x)) + cot(x)(1-cot(x))] /(1-cot(x))(1-tan(x))

= [tan(x) - tan²(x) + cot(x) - cot²(x)]/ (1-tan(x)-cot(x)+cot(x)tan(x))

= sin(x)/cos(x) - sin²(x)/cos²(x) + cos(x)/sin(x) - cos²(x)/sin²(x) / 2-tan(x)-cot(x)

= [sin3(x)cos(x)-sin4(x) +cos3(x)sin(x)-cos4(x)]/cos²(x)sin²(x)/ 2-tan(x)-cot(x)

= [sin3(x)cos(x) + cos3(x)sin(x) - (sin4(x) + cos4(x))]/cos²(x)sin²(x)/ 2-tan(x)-cot(x)

= [sin(x)cos(x)[sin2(x)+cos2(x)] -1]/cos²(x)sin²(x)/ 2-tan(x)-cot(x)

= sin(x)cos(x) -1 / cos²(x)sin²(x)(2-sin(x)/cos(x)-cos(x)/sin(x))

= sin(x)cos(x) -1/ 2cos²(x)sin²(x) - sin3(x)cos(x) -cos3(x)sin(x)

= sin(x)cos(x) -1/ 2cos²(x)sin²(x) - sin(x)cos(x)(sin²(x) + cos²(x))

= sin(x)cos(x) -1 /sin(x)cos(x)(2sin(x)cos(x)-1)

= 1/2*csc(x)sec(x) This is where I'm stuck I can't find a way for it to look like what is trying to be proven here
 
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  • #2
(1) I think you have to combine secant² (4θ) / 2 - secant² (4θ) into one fraction by making it (secant² (4θ) - 2secant² (4θ) ) / 2 then using double angle or half angle identity.

Your first line for (2) doesn't comply with me. It's like you multiplied by (1-tan(x)) for the first term and your (1-cot(x)) on its denominator disappeared.
 
  • #3
war485 said:
(1) I think you have to combine secant² (4θ) / 2 - secant² (4θ) into one fraction by making it (secant² (4θ) - 2secant² (4θ) ) / 2 then using double angle or half angle identity.

Your first line for (2) doesn't comply with me. It's like you multiplied by (1-tan(x)) for the first term and your (1-cot(x)) on its denominator disappeared.

Hmmm...I think I'll just figure this out on my own. There should have been more brackets in the post to lessen the confusion but I doubt anyone is going to volunteer to solve it
 

What is Trigonometry Refresher?

Trigonometry Refresher is a course or study guide designed to review the concepts and principles of trigonometry, a branch of mathematics that deals with the relationships between angles and sides of triangles.

Why is Trigonometry Refresher important?

Trigonometry is used in various fields such as engineering, physics, and navigation. Trigonometry Refresher helps individuals refresh their knowledge and skills in this subject, which can be useful in their academic or professional pursuits.

Who can benefit from Trigonometry Refresher?

Anyone who needs to use trigonometry in their studies or work can benefit from Trigonometry Refresher. This includes students, professionals, and individuals who are planning to pursue a career in fields that require a strong understanding of trigonometry.

What topics are covered in Trigonometry Refresher?

The topics covered in Trigonometry Refresher may vary, but generally include basic trigonometric functions, trigonometric identities, solving triangles, and applications of trigonometry.

Can I use Trigonometry Refresher as a standalone resource?

Trigonometry Refresher can be used as a standalone resource, but it is recommended to also use it as a supplement to other learning materials such as textbooks or lectures. This will provide a more comprehensive understanding of the subject.

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