Difference between revisions of "Tangent"

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{{:Relationship between tangent, Gudermannian, and sinh}}
 
{{:Relationship between tangent, Gudermannian, and sinh}}
 
{{:Relationship between sinh, inverse Gudermannian, and tan}}
 
{{:Relationship between sinh, inverse Gudermannian, and tan}}
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=See Also=
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[[Arctan]] <br />
 +
[[Tanh]] <br />
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[[Arctanh]] <br />
  
 
<center>{{:Trigonometric functions footer}}</center>
 
<center>{{:Trigonometric functions footer}}</center>

Revision as of 18:23, 11 November 2015

The tangent function is defined as the ratio of the sine and cosine functions: $$\tan(z) = \dfrac{\sin(z)}{\cos(z)}.$$

Properties

Theorem

The following formula holds: $$\dfrac{\mathrm{d}}{\mathrm{d}z} \tan(z) = \sec^2(z),$$ where $\tan$ denotes the tangent function and $\sec$ denotes the secant function.

Proof

From the definition, $$\tan(z) = \dfrac{\sin(z)}{\cos(z)},$$ so using the quotient rule, the derivative of sine, the derivative of cosine, the Pythagorean identity for sin and cos, and the definition of secant, $$\dfrac{\mathrm{d}}{\mathrm{d}z} \tan(z) = \dfrac{\mathrm{d}}{\mathrm{d}z} \dfrac{\sin(z)}{\cos(z)} = \dfrac{\cos^2(z) + \sin^2(z)}{\cos^2(z)} = \dfrac{1}{\cos^2(z)} = \sec^2(z),$$ as was to be shown. $\blacksquare$

References

Theorem

The following formula holds: $$\tan(z)=-i\tanh(iz),$$ where $\tan$ is the tangent and $\tanh$ is the hyperbolic tangent.

Proof

References

Theorem

The following formula holds: $$\tanh(z)=-i \tan(iz),$$ where $\tanh$ is the hyperbolic tangent and $\tan$ is the tangent.

Proof

References

Theorem

The following formula holds: $$\tan(\mathrm{gd}(x))=\sinh(x),$$ where $\tan$ denotes tangent, $\mathrm{gd}$ denotes the Gudermannian, and $\sinh$ denotes the hyperbolic sine.

Proof

References

Theorem

The following formula holds: $$\sinh(\mathrm{gd}^{-1}(x))=\tan(x),$$ where $\sinh$ is the hyperbolic sine, $\mathrm{gd}^{-1}$ is the inverse Gudermannian, and $\tan$ is the tangent.

Proof

References

See Also

Arctan
Tanh
Arctanh

<center>Trigonometric functions
</center>