Trigonometric Equation: Solving tan(\alpha)=\sqrt{2}-1 for \alpha in ]0,90°[

In summary, we can find the value of α by using the formula tan(2α) = 1 and solving for α using the equation tan(2σ) = tan(π/4 + 2kπ), which results in α = π/8.
  • #1
mtayab1994
584
0

Homework Statement


[tex]tan(\alpha)=\sqrt{2}-1[/tex] for every alpha in ]0,90°[

Homework Equations



1-count tan(2α)

2-conclude the value of α

The Attempt at a Solution



1-after using tan(2α)=(2tanα)/1-tan^2α) i got tan(2α)=(√2-1)/2

2- I know that α is pi/8 but i just don't know how to conclude it.
 
Physics news on Phys.org
  • #2
mtayab1994 said:

Homework Statement


[tex]tan(\alpha)=\sqrt{2}-1[/tex] for every alpha in ]0,90°[

Homework Equations



1-count tan(2α)

2-conclude the value of α

The Attempt at a Solution



1-after using tan(2α)=(2tanα)/1-tan^2α) i got tan(2α)=(√2-1)/2

2- I know that α is pi/8 but i just don't know how to conclude it.

Write that as$$\frac {\sqrt 2 - 1}{1}$$Then divide the numerator and denominator by ##\sqrt 2## and see if you recognize the result in form$$
\frac{1-\cos\theta}{\sin\theta}$$ for some ##\theta##, and see if that formula rings any bells.
 
  • #3
mtayab1994 said:

Homework Statement


[tex]tan(\alpha)=\sqrt{2}-1[/tex] for every alpha in ]0,90°[
The above should say, "for alpha in ]0,90°[".
mtayab1994 said:

Homework Equations



1-count tan(2α)

2-conclude the value of α

The Attempt at a Solution



1-after using tan(2α)=(2tanα)/1-tan^2α) i got tan(2α)=(√2-1)/2
Your value for tan(2α) is incorrect. Show us how you got that value, and we'll help you get the right value.
mtayab1994 said:
2- I know that α is pi/8 but i just don't know how to conclude it.
 
  • #4
LCKurtz said:
Write that as$$\frac {\sqrt 2 - 1}{1}$$Then divide the numerator and denominator by ##\sqrt 2## and see if you recognize the result in form$$
\frac{1-\cos\theta}{\sin\theta}$$ for some ##\theta##, and see if that formula rings any bells.

if i write it like you said and i keep solving, it just brings me back to tanα=√2-1
 
  • #5
LCKurtz said:
Write that as$$\frac {\sqrt 2 - 1}{1}$$Then divide the numerator and denominator by ##\sqrt 2## and see if you recognize the result in form$$
\frac{1-\cos\theta}{\sin\theta}$$ for some ##\theta##, and see if that formula rings any bells.

mtayab1994 said:
if i write it like you said and i keep solving, it just brings me back to tanα=√2-1

Show me what you did when you wrote it that way. What ##\theta## works?
 
  • #6
It's much simpler if you follow their hint.
mtayab1994 said:
1-count tan(2α)
I assume this means compute tan(2α). If you do this, you get a very simple value for tan(2α), which you can use to find α.
 
  • #7
yea i solved it tan(2α)=1 and to conclude the value of α i did tan(2σ)=tanpi/4+2kpi and i just solve it out and i get α=pi/8
 

FAQ: Trigonometric Equation: Solving tan(\alpha)=\sqrt{2}-1 for \alpha in ]0,90°[

What is a trigonometric equation?

A trigonometric equation is an equation that involves trigonometric functions, such as sine, cosine, and tangent. These equations often involve angles and are used to solve for unknown values in a triangle or other geometric shape.

How do you solve a trigonometric equation?

To solve a trigonometric equation, you first need to isolate the trigonometric function on one side of the equation. Then, use inverse trigonometric functions, such as arcsine, arccosine, or arctangent, to find the value of the angle. Finally, substitute the angle back into the original equation to solve for the unknown value.

What are the common trigonometric identities used in solving equations?

Some common trigonometric identities used in solving equations include the Pythagorean identities (sin²θ + cos²θ = 1, tan²θ + 1 = sec²θ, cot²θ + 1 = csc²θ), the sum and difference identities (sin(A ± B) = sinAcosB ± cosAsinB, cos(A ± B) = cosAcosB ∓ sinAsinB), and the double angle identities (sin2θ = 2sinθcosθ, cos2θ = cos²θ - sin²θ).

Are there any special cases or restrictions when solving trigonometric equations?

Yes, there are some special cases and restrictions when solving trigonometric equations. For example, the domain of the inverse trigonometric functions is limited, and the range of trigonometric functions is limited to certain values. Additionally, some equations may have multiple solutions or no solution at all.

How are trigonometric equations used in real-life applications?

Trigonometric equations are used in various fields, such as physics, engineering, and navigation. They can help calculate distances, heights, and angles in real-life situations, such as finding the height of a building or the distance between two points. They are also used in designing and building structures, such as bridges and buildings.

Back
Top