Basic Trignometry: f(α).f(β) When α+β=5π/4

In summary, we have an expression f(θ)=cotθ/1+cotθ, where α+β=5π/4. The value of f(α).f(β) can be found by substituting α and β in the expression and simplifying using relevant trigonometric identities. It is recommended to use LaTeX to properly format mathematical expressions for clarity.
  • #1
suresh
2
0

Homework Statement


Let f(θ)=cotθ/1+cotθ and α+β=5π/4
then the value of f(α).f(β) is:

The Attempt at a Solution


f(α)f(β)=
(cotα/1+cotα)* (cotβ/1+cotβ)

= cotα.cotβ/((1+cotα)(1+cotβ))
=cotα.cotβ/(1+cotβ+cotα+cotα.cotβ)
 
Last edited:
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  • #2
Hi suresh. http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif

Was there supposed to be an accompanying question?

When you have an answer, you could always check it by taking any value for α then using your calculator...

Double-check by taking another value. :smile:

Also, your expressions above appear to be invalid, owing to a dearth of essential brackets. To help you fix this, please accept some spares I happen to have lying around: (((((((((())))))))) :wink:
 
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  • #3
You should be looking for relevant trigonometric identities, sum- of-angles identities, and the law of tangents.
 
  • #4
Thanks for all the help guys sorry for not writing clearly if only I knew how to put fractions
Also sorry for using some other symbol which I thought was pi and confusing all of you
 
  • #5
You mean like this? Let f(θ)=cotθ/(1+cotθ)

Whatever the symbol was that you originally used, I read it as Pi. :smile:
 
  • #6
suresh said:
Thanks for all the help guys sorry for not writing clearly if only I knew how to put fractions
Also sorry for using some other symbol which I thought was pi and confusing all of you
You get
$$ f(\theta) = \frac{\cot\theta}{1 + \cot\theta} $$
by typing "f(\theta) = \frac{\cot\theta}{1 + \cot\theta}" in TeX mode. You get TeX mode by either putting your equation inside double dollar signs or inside tex tags.
 
  • #7
suresh said:
Thanks for all the help guys sorry for not writing clearly if only I knew how to put fractions
Also sorry for using some other symbol which I thought was pi and confusing all of you

Doing fractions is easy: if you have just ##\frac{a}{b}## it is sufficient to type a/b. If you have something like ##\frac{a}{b+c}## then you absolutely must use parentheses, but it is easy also: just type a/(b+c).
 
  • #8
suresh said:
Thanks for all the help guys sorry for not writing clearly if only I knew how to put fractions
Also sorry for using some other symbol which I thought was pi and confusing all of you
I recommend you to use latex all the time.

You only have to write \pi for ##\pi##
And \frac{a}{b} for ##\frac{a}{b}##

Use two # signs between the latex code to render it inline.
Use two $ signs between the latex code to render it indented and on a separate line. This can be useful for writing large equations which are rendered stupidly using two #.
 

Related to Basic Trignometry: f(α).f(β) When α+β=5π/4

1. What is the meaning of "f(α).f(β)" in basic trigonometry?

In basic trigonometry, f(α).f(β) refers to the product of two trigonometric functions, one evaluated at an angle α and the other evaluated at an angle β.

2. How do you find the value of f(α).f(β) when given that α + β = 5π/4?

To find the value of f(α).f(β) when α + β = 5π/4, you can use the trigonometric identities and formulas to express the product in terms of the given angles. For example, you can use the double angle formula for cosine to express cos(5π/4) in terms of cos(α) and cos(β).

3. Can you provide an example of finding the value of f(α).f(β) when α + β = 5π/4?

Yes, for example, if α = π/4 and β = 3π/4, then f(α) = sin(π/4) = √2/2 and f(β) = cos(3π/4) = -√2/2. Therefore, f(α).f(β) = (√2/2)(-√2/2) = -1/2.

4. Are there any special cases when finding the value of f(α).f(β) when α + β = 5π/4?

Yes, there are special cases when finding the value of f(α).f(β) when α + β = 5π/4. For example, when α = β = 5π/8, then f(α).f(β) = (sin(5π/8))(cos(5π/8)) = (1/2)(√2/2) = √2/4.

5. Can you use the value of f(α).f(β) to solve for the individual values of α and β?

No, you cannot use the value of f(α).f(β) to solve for the individual values of α and β. This is because there are infinitely many combinations of α and β that can add up to 5π/4 and result in the same value of f(α).f(β).

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