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By the way, you should put braces around your eleven's that are under the square root, so that the entire radican is covered by the bar of the square root. It should look like \sqrt{11} rather than \sqrt11, so that 11, rather than 11, appears. :)
Are you sure this is what we have to prove? Because I just used a calculator and saw that tan(113π)+4tan(112π) is not equal to the square root of 11. Am I somehow misunderstanding your problem?
Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
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The problem actually is as follows, and I request the user to edit it. Iam not giving the solution, and that googling it will give the solution:
Prove that tan113π+4sin112π= 11
Infact, even the following relations hold:
tan113π+4sin112π= 11
tan114π+4sin11π= 11
tan115π−4sin114π= 11
tan112π−4sin115π= −11
tan11π+4sin113π= 11
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By the way, you should put braces around your eleven's that are under the square root, so that the entire radican is covered by the bar of the square root. It should look like \sqrt{11} rather than \sqrt11, so that 11, rather than 11, appears. :)
Are you sure this is what we have to prove? Because I just used a calculator and saw that tan(113π)+4tan(112π) is not equal to the square root of 11. Am I somehow misunderstanding your problem?
But in my book its tan and not sine....but yes it must be sine.....
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maybe some mistake