The discussion revolves around the elimination of the variable theta from two equations involving trigonometric functions, specifically sine and cosine. The original poster presents equations relating cosine and sine to constants m and n, seeking guidance on how to proceed with the elimination of theta.
The discussion is ongoing, with various approaches being considered. Some participants express uncertainty about the methods proposed, while others suggest checking results with specific values for m and n. There is no explicit consensus on the best path forward.
Participants note the requirement to eliminate theta entirely from the final expression, emphasizing that the answer should relate only to m and n. There is a concern about potential loss of information through the operations performed on the equations.
How do we change to ##tan\theta## and ##cot\theta## to eliminate ##\theta##fresh_42 said:Assuming this is right what you have done, then changing to ##\tan ## and ##\cot## seems to be an option.
By dividing the equations. At last you still have ##\sin^2+\cos^2=1##.sahilmm15 said:How do we change to ##tan\theta## and ##cot\theta## to eliminate ##\theta##
By dividing both the equations we get ##cot^3\theta = {\frac{m}{n}}##fresh_42 said:By dividing the equations. At last you still have ##\sin^2+\cos^2=1##.
I can't figure it out. The only way to make use of ##sin^2\theta + cos^2\theta## is by adding both the two equations??fresh_42 said:And? What do you get from the other hint I gave?
I am not sure whether this is necessary. You can already solve for ##\theta## by taking the cubic root. However, the other equation can be used to check the result.sahilmm15 said:I can't figure it out. The only way to make use of ##sin^2\theta + cos^2\theta## is by adding both the two equations??
I think we need to find the independent values of ##sin\theta## and ##cos\theta## (in terms of m and n) and then put in ##sin^2\theta + cos^2\theta ## to get rid of ##\theta##, but that would be tedious.fresh_42 said:I am not sure whether this is necessary. You can already solve for ##\theta## by taking the cubic root. However, the other equation can be used to check the result.
That is not what you wrote. Your text sayssahilmm15 said:I think we need to find the independent values of ##sin\theta## and ##cos\theta## (in terms of m and n) and then put in ##sin^2\theta + cos^2\theta ## to get rid of ##\theta##, but that would be tedious.
And under the assumption you made no mistake, you got ##\theta =\operatorname{arc}\cot \sqrt[3]{\dfrac{m}{n}}##.... then eliminate ##\theta##
The question states under the given conditions eliminate ##\theta##. It means there should be no ##\theta## in the answer. Then it implies answer should be in the terms of m and n( a relation between m and n where there is no ##\theta##).fresh_42 said:That is not what you wrote. Your text says And under the assumption you made no mistake, you got ##\theta =\operatorname{arc}\cot \sqrt[3]{\dfrac{m}{n}}##.
And since we have ##\theta=\theta(n,m)## we can substitute every occurrence of ##\theta ## by an expression which only has ##n,m## in it. This is an elimination. And please stop shouting!sahilmm15 said:The question states under the given conditions eliminate ##\theta##. It means there should be no ##\theta## in the answer. Then it implies answer should be in the terms of m and n( a relation between m and n where there is no ##\theta##).
To be honest, I was not shouting. And I think it is not fair of you to assume anything. There are bold and italics formatting tools to convey ones expressions and what they are thinking, and it doesn't mean one is shouting when they are using them.fresh_42 said:And since we have ##\theta=\theta(n,m)## we can substitute every occurrence of ##\theta ## by an expression which only has ##n,m## in it. This is an elimination. And please stop shouting!
Lol, honestly I didn't knew that it meant shouting. So I misinterpreted you. I am sorry.fresh_42 said:Yes, and boldface means shouting on the internet.
I do not like that solution either, and I am not sure whether we lost information on the way by squaring or dividing. I would check it with some easy values for ##n,m##.