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What is the minimum positive period of y = SiNx (1 + TaNx * Tan (x / 2))?
Y = SiNx (1 + TaNx * Tan (x / 2)) = SiNx {1 + (SiNx / cosx) * [(1-cosx) / SiNx]} = SiNx [1 + 1 / cosx-1] = SiNx / cosx = TGX. So the minimum period is the minimum period of TGX!
What are the periods of y = Tan π X and y = Cotx TaNx?
Y = Tan π x period T = π / w = π / π = 1
y=cotx-tanx=cosx/sinx-sinx/cosx
=(cos^2x-sin^2x)/sinxcosx
=2cos2x / sin2x = 2cot2x period T = π / w = π / 2
How to prove that the minimum positive period of y = 1-tanx * TaNx / 1 + Tan * Tan is pie?
The title should be as follows: y = [1 - (TaNx) ^ 2] / [1 + (TaNx) ^ 2] then multiply the numerator denominator of the right formula by (cosx) ^ 2 at the same time to get y = [(cosx) ^ 2 - (SiNx) ^ 2] / [(cosx) ^ 2 + (SiNx) ^ 2] = cos2x / 1 = cos2x, and the period of the function y = cosx is t = 2K π + 2 π, K is an integer, then y = cos2x
Function y = TaNx ÷ (1-tan) ² 10) The minimum positive period of is
y=tanx÷（1-tan ² x）=2tanx÷（1-tan ² x）÷2=tan2x/2
The period of y = TaNx is π
So the minimum positive period should be π / 2
Function y = TaNx ÷ (1-tan) ² 10) Minimum positive period of
Because y = TaNx ÷ (1-tan) ² x)=tan(2x)/2
So the minimum positive period is t = π / 2
Find the derivative of y = ln (a ^ 2-x ^ 2)
y'=(a^2-x^2)'/(a^2-x^2)
=-2x/(a^2-x^2)
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