https://latex.codecogs.com/legacy/eqneditor/editor.php
Each line must start with:
[math]
and end with a similar /math command.
I'm having trouble getting this bit to display properly so I hope you can read this ok.
If someone has used Latex in a post and you want to see their commands, just click on the Latex. You can copy and paste the commands and then edit as you need for your own post.
Bob
]]>Bob wrote:As you know that
all you need is
But one is the reciprocal of the other so that follows straight away.
Bob
Hello Bob. I want to learn how to write in LaTex form. Do you have a link in this regard?
Thanks.
Hey Mathland,
It's a stickied thread in this section.
]]>As you know that
all you need is
But one is the reciprocal of the other so that follows straight away.
Bob
Hello Bob. I want to learn how to write in LaTex form. Do you have a link in this regard?
Thanks.
]]>all you need is
But one is the reciprocal of the other so that follows straight away.
Bob
]]>20. Verify (sinΘ)^4 + 2(sinΘ)^2(cosΘ)^2 + (cosΘ)^4 = tanΘcotΘ
I have factored it down so that (sinΘ)4 + 2(sinΘ)2(cosΘ)2 + (cosΘ)4 = (sinΘ^2+cosΘ^2)^2, I'm just not sure where to go from here.
]]>The formulas
and
are really just re-arrangements of the same result. Let's go back to the right angled triangle.
If you divide through by H squared you get:
If you divide this formula by cos squared you get
So it's 'in the spirit' of what your teacher has told you to use any of the many versions of the formula to help with these problems.
eg. 12. Express cosecant in terms of tangent.
cosecant is 1/sin so I need a version of the formula that has 1/sin in it.
So I'll start with formula (1) and divide through by sin squared.
A full set of identities is on this page: https://www.mathsisfun.com/algebra/trig … ities.html about 2/3 of the way down the page.
Hope that clears it up for you,
Bob
]]>write cosine in terms of tangent, in a way that avoids using sine. One of the other ways to write cosΘ is to say 1/secΘ
We know that tan2Θ + 1 = sec2Θ
This means that secΘ = √(tan2Θ + 1)
So cosΘ = 1/√(tan2Θ + 1)
Putting this together with what we got before we wind up with an answer:
sinΘ = tanΘ * 1/√(tan2Θ + 1)
Can you help me understand how to figure out my answer in that format?
]]>In a right angled triangle and using A for adjacent, O for opposite and H for hypotenuse:
Pythagoras: O^2 + A^2 = H^2
Divide by H^2: sin^2 (x) + cos^2 (x) = 1 **
This useful identity can be used for many things including all of your tasks.
eg.
Divide by sin^2 (x) : 1 + 1/tan^2(x) = cosec^2 (x)
** For this to work x has to be an angle in a right angled triangle. But trig values beyond the range 0-90, are defined in such a way that the result still holds.
Bob
]]>Welcome to the forum!
See the links in MathsIfFun
and
Hope these solve the problems!
]]>I have been tasked with this question:
12. Express cosecant in terms of tangent.
I think I might have a start, but I'm not sure if I'm correct. Here is what I have so far:
tan = sin/cos
sin= 1/csc
tan= (1/csc)/cos
That's as far as I can think. I also have some other, similar problems that I'm struggling with.
13. Express sine in terms of cotangent.
14. Express cosecant in terms of cosine.
15. Express secant in terms of sine.
Any help would be greatly appreciated
]]>