Solve Trig Substitution Homework with P

Here is a corrected summary:In summary, if sin(54)=p, then cos(18) can be expressed as 2p√(1-p^2) without using a calculator. Similarly, the expression (tan(27) + cot(63))/(1 + tan(207)cot(117)) can be simplified to p/√(1-p^2).
  • #1
DERRAN
34
0

Homework Statement


If sin54=P, express the following in terms of p, without using a calculator.
1] cos18
2] [tex]\frac{tan27+cot63}{1+tan207.cot117}[/tex]


Homework Equations





The Attempt at a Solution



1] cos18=sin(90-18)
=sin(72)

sin72=sin(180-108)
=sin(180-108)
=sin180.cos108-cos180.sin108
=0-(-sin108)
=sin108

sin108=2.sin54.cos54
=2pcos54.[tex]\sqrt{1-p^{2}}[/tex]



2] =[tex]\frac{2tan27}{1-tan27.tan27}[/tex]

=[tex]\frac{2tan27}{1-tan^{2}27}[/tex]

=tan2(27)

=tan54

=[tex]\frac{p}{\sqrt{1-p^{2}}}[/tex]
 
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  • #2
DERRAN said:

Homework Statement


If sin54=P, express the following in terms of p, without using a calculator.

"P" or "p"? You shouldn't use both to mean the same thing.

1] cos18
2] [tex]\frac{tan27+cot63}{1+tan207.cot117}[/tex]


Homework Equations





The Attempt at a Solution



1] cos18=sin(90-18)
=sin(72)

sin72=sin(180-108)
=sin(180-108)
=sin180.cos108-cos180.sin108
=0-(-sin108)
=sin108

sin108=2.sin54.cos54
=2pcos54.[tex]\sqrt{1-p^{2}}[/tex]
Well, you mean, of course,
[tex]sin(108)= 2p\sqrt{1- p^2}[/tex]
Also you haven't finished the problem. Yes, cos(18) is equal to that but you should write that explicitely:
[tex]cos(18)= 2p\sqrt{1- p^2}[/tex]
2] =[tex]\frac{2tan27}{1-tan27.tan27}[/tex]

=[tex]\frac{2tan27}{1-tan^{2}27}[/tex]

=tan2(27)

=tan54

=[tex]\frac{p}{\sqrt{1-p^{2}}}[/tex]
I see nothing wrong with (2).
 
  • #3
Thanks, I made a few typing errors.
 

1. What is trigonometric substitution?

Trigonometric substitution is a technique used in calculus to simplify integrals involving trigonometric functions. It involves replacing the variable in the integral with a trigonometric function, which can help to simplify the integral and make it easier to solve.

2. Why is trigonometric substitution useful?

Trigonometric substitution is useful because it can help to simplify complicated integrals involving trigonometric functions. It also allows us to solve integrals that would otherwise be difficult or impossible to solve using other techniques.

3. How do I know when to use trigonometric substitution?

You should use trigonometric substitution when you have an integral that contains a radical expression (such as √(a^2-x^2) or √(x^2-a^2)) or an expression involving a^2+x^2 or a^2-x^2. These are common forms in which trigonometric substitution can be applied.

4. What are the three main trigonometric substitutions?

The three main trigonometric substitutions are:
▪ u = sinθ
▪ u = cosθ
▪ u = tanθ

5. How do I choose which trigonometric substitution to use?

Generally, you should choose the trigonometric substitution that matches the form of the integral you are trying to solve. For example, if the integral contains √(a^2-x^2), then you should use the substitution u = sinθ. If the integral contains √(x^2-a^2), then you should use the substitution u = cosθ or u = tanθ.

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