
How do you solve $ \cos \left( {\theta - 2\pi } \right) $ ?
Answer
514.2k+ views
Hint: Trigonometric functions are those functions that tell us the relation between the three sides of a right-angled triangle. Sine, cosine, tangent, cosecant, secant and cotangent are the six types of trigonometric functions. To solve this we use the cosine difference formula. That is \[\cos (a - b) = \cos (a).\cos (b) + \sin (a).\sin (b)\], where \[a = \theta \]and \[b = 2\pi \].
Complete step-by-step answer:
Given,
$ \cos \left( {\theta - 2\pi } \right) $
We know the cosine difference formula,
\[\cos (a - b) = \cos (a).\cos (b) + \sin (a).\sin (b)\]
where \[a = \theta \]and \[b = 2\pi \].
Substituting we have,
$ \Rightarrow \cos \left( {\theta - 2\pi } \right) = \cos \left( \theta \right)\cos \left( {2\pi } \right) + \sin \left( \theta \right)\sin \left( {2\pi } \right) $
We know that $ \cos \left( {n\pi } \right) = {\left( { - 1} \right)^n} $ and $ \sin \left( {n\pi } \right) = 0 $ , knowing this we have $ \cos \left( {2\pi } \right) = 1 $ and $ \sin \left( {2\pi } \right) = 0 $
$ \cos \left( {\theta - 2\pi } \right) = \cos \left( \theta \right) \times 1 + \sin \left( \theta \right) \times 0 $
$ \Rightarrow \cos \left( {\theta - 2\pi } \right) = \cos \left( \theta \right) $
This should make sense. Since $ 2\pi $ is one revolution around the unit circle, the angles $ \theta $ and $ \theta - 2\pi $
Are $ \cos \left( \theta \right) = \cos \left( {\theta - 2\pi } \right) $ .
So, the correct answer is “ $ \cos \left( \theta \right) = \cos \left( {\theta - 2\pi } \right) $ ”.
Note: Remember A graph is divided into four quadrants, all the trigonometric functions are positive in the first quadrant, all the trigonometric functions are negative in the second quadrant except sine and cosine functions, tangent and cotangent are positive in the third quadrant while all others are negative and similarly all the trigonometric functions are negative in the fourth quadrant except cosine and secant. Also sine, cosine and tangent are the main functions while cosecant, secant and cotangent are the reciprocal of sine, cosine and tangent respectively.
We know the cosine sum and difference formula\[\cos (a + b) = \cos (a).\cos (b) - \sin (a).\sin (b)\] and \[\cos (a - b) = \cos (a).\cos (b) + \sin (a).\sin (b)\]. Similarly we have sine sum and difference formula \[\sin (a + b) = \sin (a).\cos (b) + \cos (a).\sin (b)\] and\[\sin (a - b) = \sin (a).\cos (b) - \cos (a).\sin (b)\]. Depending on the angle we use the required formulas.
Complete step-by-step answer:
Given,
$ \cos \left( {\theta - 2\pi } \right) $
We know the cosine difference formula,
\[\cos (a - b) = \cos (a).\cos (b) + \sin (a).\sin (b)\]
where \[a = \theta \]and \[b = 2\pi \].
Substituting we have,
$ \Rightarrow \cos \left( {\theta - 2\pi } \right) = \cos \left( \theta \right)\cos \left( {2\pi } \right) + \sin \left( \theta \right)\sin \left( {2\pi } \right) $
We know that $ \cos \left( {n\pi } \right) = {\left( { - 1} \right)^n} $ and $ \sin \left( {n\pi } \right) = 0 $ , knowing this we have $ \cos \left( {2\pi } \right) = 1 $ and $ \sin \left( {2\pi } \right) = 0 $
$ \cos \left( {\theta - 2\pi } \right) = \cos \left( \theta \right) \times 1 + \sin \left( \theta \right) \times 0 $
$ \Rightarrow \cos \left( {\theta - 2\pi } \right) = \cos \left( \theta \right) $
This should make sense. Since $ 2\pi $ is one revolution around the unit circle, the angles $ \theta $ and $ \theta - 2\pi $
Are $ \cos \left( \theta \right) = \cos \left( {\theta - 2\pi } \right) $ .
So, the correct answer is “ $ \cos \left( \theta \right) = \cos \left( {\theta - 2\pi } \right) $ ”.
Note: Remember A graph is divided into four quadrants, all the trigonometric functions are positive in the first quadrant, all the trigonometric functions are negative in the second quadrant except sine and cosine functions, tangent and cotangent are positive in the third quadrant while all others are negative and similarly all the trigonometric functions are negative in the fourth quadrant except cosine and secant. Also sine, cosine and tangent are the main functions while cosecant, secant and cotangent are the reciprocal of sine, cosine and tangent respectively.
We know the cosine sum and difference formula\[\cos (a + b) = \cos (a).\cos (b) - \sin (a).\sin (b)\] and \[\cos (a - b) = \cos (a).\cos (b) + \sin (a).\sin (b)\]. Similarly we have sine sum and difference formula \[\sin (a + b) = \sin (a).\cos (b) + \cos (a).\sin (b)\] and\[\sin (a - b) = \sin (a).\cos (b) - \cos (a).\sin (b)\]. Depending on the angle we use the required formulas.
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