Verify Trigonometric Identities

Verifying trigonometric identities through examples with detailed solutions are presented.
No single method works for all identities. However following certain steps might help. To verify an identity, you may start by transforming the more complicated side into the other, using basic trigonometric identities. You may also transform the two sides into one same expression.
More questions with solutions are included.


Examples with Solutions

Example 1
Verify the identity \[ \cos x \cdot \tan x = \sin x \]
Solution to Example 1:
Transform the left side using the identity \( \tan x = \dfrac{\sin x}{\cos x} \). \[ \cos x \cdot \tan x = \cos x \cdot \dfrac{\sin x}{\cos x} \] Simplify to obtain. \[ = \sin x \] The left side was transformed so that it is equal to the right side.



Example 2
Verify the identity \[ \cot x \cdot \sec x \cdot \sin x = 1 \]

Solution to Example 2
Use the identities \( \cot x = \dfrac{\cos x}{\sin x} \) and \( sec x = \dfrac{1}{\cos x} \) to rewrite the left side as \[ \cot x \cdot \sec x \cdot \sin x = \dfrac{\cos x}{\sin x} \cdot \dfrac{1}{\cos x} \cdot \sin x \] Simplify the right side to obtain. \[ = 1 \] The left side is transformed so that it is equal to the right side.



Example 3
Verify the identity \[ \dfrac{\cot x - \tan x }{ \sin x \cdot \cos } = \csc^2 x - \sec^2 x \]

Solution to Example 3
Use the identities \( \cot x = \dfrac{\cos x}{\sin x} \) and \( \tan x = \dfrac{\sin x}{\cos x} \) to transform the left side as follows. \[ \dfrac{\cot x - \tan x }{ \sin x \cdot \cos } = \dfrac {\dfrac{\cos x}{\sin x} - \dfrac{\sin x}{\cos x}}{\sin x \cdot \cos x} \] Rewrite the numerator of the main fraction on the right as the difference of two fractions with a common denominator. \[ = \dfrac {\dfrac{\cos x \cos x }{\cos x \sin x} - \dfrac{\sin x \sin x}{\sin x \cos x}}{\sin x \cdot \cos x} \] Group the fractions in the numerator on the right side. \[ = \dfrac {\dfrac{\cos^2 x - \sin^2 x }{\cos x \cdot \sin x}}{\sin x \cdot \cos x} \] Divide the above and rewrite as. \[ = \dfrac {\cos^2 x - \sin^2 x }{\sin^2 x \cdot \cos^2 x} \] Rewrite as a difference of two fractions. \[ = \dfrac {\cos^2 x}{\sin^2 x \cdot \cos^2 x} - \dfrac {\sin^2 x }{\sin^2 x \cdot \cos^2 x} \] Simplify each fraction. \[ = \dfrac {1}{\sin^2 x} - \dfrac {1}{\cos^2 x} \] We now transform the above side using the identities \( \quad \csc^2 x = \dfrac{1}{\sin^2 x} \) and \( \quad \sec^2 x = \dfrac{1}{\cos^2 x } \). \[ = \csc^2 x - \sec^2 x \] The left side has been transformed so that it is equalt to the right side and hence the identity is verified.




Questions

Verify the identities
  1. \( \quad \sin x + \cos x \cdot \cot x = \csc x \)

  2. \( \quad \dfrac{\csc x}{1 + \csc x} - \dfrac{\csc x}{1 - \csc x} = 2 \sec^2 x \)


Solutions


  1. Use the identity \( \cot x = \dfrac{\cos x}{\sin x} \) to rewrite the left side as follows. \[ \quad \sin x + \cos x \cdot \cot x = \quad \sin x + \cos x \cdot \dfrac{\cos x}{\sin x} \] Rewrite the left side as the sum of fractions with common denominator \[ = \quad \dfrac{\sin^2 x}{\sin x } + \dfrac{\cos^2 x}{\sin x} \] Group the two fractions \[ = \quad \dfrac{\sin^2 x + \cos^2 x}{\sin x } \] Use the identity \( \sin^2 x + \cos^2 x = 1 \) to simplify the above \[ = \quad \dfrac{1}{\sin x } \] Use the identity \( \dfrac{1}{\sin x } = \csc x \) to rewrite the above as \[ = \csc x \] The left side has been transformed so that it is equalt to the right side and hence the identity is verified.


  2. Multiply numerator and denominator of the first rational expression by \( \quad (1 - \csc x) \) and the second rational expression by \( \quad (1 + \csc x) \) in the left side of the given identity \[ \quad \dfrac{\csc x}{1 + \csc x} - \dfrac{\csc x}{1 - \csc x} \\[15pt] = \dfrac{ \csc x (1 - \csc x) }{(1 + \csc x)(1 - \csc x)} - \dfrac{ \csc x (1 + \csc x) }{(1 - \csc x)(1 + \csc x)} \] Expand the numerator of each rational expression \[ = \dfrac{ \csc x - \csc^2 x }{(1 + \csc x)(1 - \csc x)} - \dfrac{ \csc x + \csc^2 x }{(1 - \csc x)(1 + \csc x)} \] The two rational expressions have a common denominator and therefore may be grouped as follows \[ = \dfrac{ \csc x - \csc^2 x - \csc x - \csc^2 x }{(1 + \csc x)(1 - \csc x)} \] Simplify the numeratorc and expand the denominator \[ = \dfrac{ - 2 \csc^2 x }{1 - \csc^2 x} \] Use the identity \( \csc^2 x = \dfrac{1}{\sin^2 x} \) to rewrite the above as \[ = \dfrac{ - 2 \dfrac{1}{\sin^2 x}}{1 - \dfrac{1}{\sin^2 x} } \] Rewrite the terms in the denominator with a common factor \[ = \dfrac{ - 2 \dfrac{1}{\sin^2 x}}{\dfrac{\sin^2 x}{\sin^2 x} - \dfrac{1}{\sin^2 x} } \] Simplify \[ = \dfrac{ - 2 \dfrac{1}{\sin^2 x}}{\dfrac{\sin^2 x}{\sin^2 x} - \dfrac{1}{\sin^2 x} } \] Group the terms in the denominator and simplify \[ = \dfrac{ - 2 }{\sin^2 x - 1 } \] Use the identity \( \sin^2 x + \cos^2 x = 1 \) to write \( \sin^2 x - 1 = - \cos^2 x \) and substitute in the denominator \[ = \dfrac{ - 2 }{ - \cos^2 x } \] Simplify \[ = 2 \dfrac{ 1 }{ \cos^2 x } \] which may be written as \[ = 2 \sec^2 x \] The left side has been transformed so that it is equalt to the right side and hence the identity is verified.



More References

  1. Trigonometric Identities and Their Applications
  2. Trigonometric Formulas and Their Applications

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