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Question: How do you solve \( \cos \left( {\theta - 2\pi } \right) \) ?...

How do you solve cos⁡(θ−2π)\cos \left( {\theta - 2\pi } \right) ?

Explanation

Solution

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)\cos (a - b) = \cos (a).\cos (b) + \sin (a).\sin (b), where a=θa = \theta and b=2πb = 2\pi .

Complete step-by-step answer :
Given,
cos⁡(θ−2π)\cos \left( {\theta - 2\pi } \right)
We know the cosine difference formula,
cos⁡(a−b)=cos⁡(a).cos⁡(b)+sin⁡(a).sin⁡(b)\cos (a - b) = \cos (a).\cos (b) + \sin (a).\sin (b)
where a=θa = \theta and b=2πb = 2\pi .
Substituting we have,
⇒cos⁡(θ−2π)=cos⁡(θ)cos⁡(2π)+sin⁡(θ)sin⁡(2π)\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⁡(nπ)=(−1)n\cos \left( {n\pi } \right) = {\left( { - 1} \right)^n} and sin⁡(nπ)=0\sin \left( {n\pi } \right) = 0 , knowing this we have cos⁡(2π)=1\cos \left( {2\pi } \right) = 1 and sin⁡(2π)=0\sin \left( {2\pi } \right) = 0
cos⁡(θ−2π)=cos⁡(θ)×1+sin⁡(θ)×0\cos \left( {\theta - 2\pi } \right) = \cos \left( \theta \right) \times 1 + \sin \left( \theta \right) \times 0
⇒cos⁡(θ−2π)=cos⁡(θ)\Rightarrow \cos \left( {\theta - 2\pi } \right) = \cos \left( \theta \right)
This should make sense. Since 2π2\pi is one revolution around the unit circle, the angles θ\theta and θ−2π\theta - 2\pi
Are cos⁡(θ)=cos⁡(θ−2π)\cos \left( \theta \right) = \cos \left( {\theta - 2\pi } \right) .
So, the correct answer is “ cos⁡(θ)=cos⁡(θ−2π)\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 formulacos⁡(a+b)=cos⁡(a).cos⁡(b)−sin⁡(a).sin⁡(b)\cos (a + b) = \cos (a).\cos (b) - \sin (a).\sin (b) and cos⁡(a−b)=cos⁡(a).cos⁡(b)+sin⁡(a).sin⁡(b)\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)\sin (a + b) = \sin (a).\cos (b) + \cos (a).\sin (b) andsin⁡(a−b)=sin⁡(a).cos⁡(b)−cos⁡(a).sin⁡(b)\sin (a - b) = \sin (a).\cos (b) - \cos (a).\sin (b). Depending on the angle we use the required formulas.