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how do I find out how wolfram turned my trig equation into sqrt fractions?

Posted 9 years ago

Fairly simple question, just looking for some help as I work my way through basic physics.

((xcos60)/cos30)sin30 +(xsin60) = 98 turns into (sqrt(3) x)/2+x/(2 sqrt(3)) = 98

so my question is how do i get it to show me the conversion?

Also any links to guides for this would be great, I get the idea but I just have trouble converting it between the 2.

Thank you for any help, it is much appreciated.

POSTED BY: Kyle wright
4 Replies
Posted 9 years ago

basically I am trying to compare steps in this, which i copied off a handout , brushing up on my math skills I would like to see all the steps, which i cant seem to replicate, mainly from the first equation to the 2nd one, where did that other t2 go? haha i am using X instead of the t2 and i dont use the N but having not a heck of a lot of luck right now getting the step by step to show on wolfram

(T2cos60/cos30) sin30 + T2sin60 = 98N

The rest is just algebra...

T2 (cos60 sin30/cos30 + sin60) = 98N

T2 = 98N/(cos60 sin30/cos30 + sin60)

T2 = 98N / 1.155 = 84.9N

POSTED BY: Kyle wright

If i wanted it to be a simplified version of what I said solving for X value, with cos and sin still in there, how would I go about that?

One way is not to use numbers for the angles, but symbols. something like

Clear[x, y];
((x Cos[y])/Cos[y/2]) Sin[y/2] + (x Sin[y]) == 98

Now Mathematica can't replace the trig functions with numerical values, since it does not know what the angle y is.

POSTED BY: Nasser M. Abbasi
Posted 9 years ago

Oh ,I hadn't realized that it simply was a replacement process at all. Thank you very much for the information. If i wanted it to be a simplified version of what I said solving for X value, with cos and sin still in there, how would I go about that?

POSTED BY: Kyle wright
((xcos60)/cos30)sin30 +(xsin60) = 98 turns into (sqrt(3) x)/2+x/(2 sqrt(3)) = 98

You must have typed

 ((x Cos[60 Degree])/Cos[30 Degree]) Sin[30 Degree] + (x Sin[60 Degree]) == 98

Mathematica simply evaluated the trig functions in there, since they have known values and replaced them with their values. The result is what comes out

enter image description here

POSTED BY: Nasser M. Abbasi
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