verify the identity 4tan 1/5 = π/4 + arctan 1/239?? heeelp?
回答 (3)
2016-06-18 5:18 am
L.H.S.
= 4 tan(1/5)
≈ 0.8108 (to 4 sig. fig.)
R.H.S.
= (π/4) + arctan(1/239)
≈ 0.7896 (to 4 sig. fig.)
L.H.S. ≠ R.H.S.
This is NOT an identity.
= 4 tan(1/5)
≈ 0.8108 (to 4 sig. fig.)
R.H.S.
= (π/4) + arctan(1/239)
≈ 0.7896 (to 4 sig. fig.)
L.H.S. ≠ R.H.S.
This is NOT an identity.
2016-06-18 4:23 am
Regardless of whether the above 1/5 and 1/239 is in degrees or radians, the statement is false.
You can check it here:
http://www.wolframalpha.com/input/?i=4tan+(1%2F5+degrees)+%3D+%CF%80%2F4+%2B+arctan+(1%2F239+degrees)
http://www.wolframalpha.com/input/?i=4tan+(1%2F5+radian)+%3D+%CF%80%2F4+%2B+arctan+(1%2F239+radian)
You can check it here:
http://www.wolframalpha.com/input/?i=4tan+(1%2F5+degrees)+%3D+%CF%80%2F4+%2B+arctan+(1%2F239+degrees)
http://www.wolframalpha.com/input/?i=4tan+(1%2F5+radian)+%3D+%CF%80%2F4+%2B+arctan+(1%2F239+radian)
2016-06-18 9:46 am
4tan⁻¹(1/5)=π/4+tan⁻¹(1/239) is valid
Let x=tan⁻¹(1/5) so tan(x)=1/5
Need to show that 4x = π/4 + tan⁻¹(1/239) or tan(4x−π/4) = 1/239
tan(2x) = 2tan(x)/(1−tan²(x)) = (2/5)/(1−1/25) = 5/12
tan(4x) = 2tan(2x)/(1−tan²(2x)) = (10/12)/(1−25/144) = 120/119
∴ tan(4x−π/4) = ( 120/119 – 1 ) / ( 1 + 1*120/119 ) = (1/119) / (239/119) = 1/239
Let x=tan⁻¹(1/5) so tan(x)=1/5
Need to show that 4x = π/4 + tan⁻¹(1/239) or tan(4x−π/4) = 1/239
tan(2x) = 2tan(x)/(1−tan²(x)) = (2/5)/(1−1/25) = 5/12
tan(4x) = 2tan(2x)/(1−tan²(2x)) = (10/12)/(1−25/144) = 120/119
∴ tan(4x−π/4) = ( 120/119 – 1 ) / ( 1 + 1*120/119 ) = (1/119) / (239/119) = 1/239
收錄日期: 2021-04-18 15:10:17
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