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Question: How do I prove \(\cos \left( 2\pi -\theta \right)=\cos \theta \) ?...

How do I prove cos⁡(2π−θ)=cos⁡θ\cos \left( 2\pi -\theta \right)=\cos \theta ?

Explanation

Solution

To prove that cos⁡(2π−θ)=cos⁡θ\cos \left( 2\pi -\theta \right)=\cos \theta in the given question, we will take the Left Hand Side of the equation and try to get the value of Right hand side of the equation. So, we will start from the Left Hand Side of the equation by expanding the equation cos⁡(2π−θ)\cos \left( 2\pi -\theta \right) with use of formula cos⁡(a−b)=cos⁡a.cos⁡b+sin⁡a.sin⁡b\cos \left( a-b \right)=\cos a.\cos b+\sin a.\sin b . After that will use the value of applying the value of cos⁡2π\cos 2\pi in the expression of the equation that will help us to get the final result that is equal to the right hand side of the equation.

Complete step by step solution:
Since, the given question is cos⁡(2π−θ)=cos⁡θ\cos \left( 2\pi -\theta \right)=\cos \theta , where we need to prove that Left Hand Side is equal to Right Hand Side that means cos⁡(2π−θ)\cos \left( 2\pi -\theta \right) is equal to cos⁡θ\cos \theta . So, we take Left Hand Side of the equation as:
⇒cos⁡(2π−θ)\Rightarrow \cos \left( 2\pi -\theta \right)
Now, we can write the above equation as cos⁡2π.cos⁡θ+sin⁡2π.sin⁡θ\cos 2\pi .\cos \theta +\sin 2\pi .\sin \theta by using the formula cos⁡(a−b)=cos⁡a.cos⁡b+sin⁡a.sin⁡b\cos \left( a-b \right)=\cos a.\cos b+\sin a.\sin b as:
⇒cos⁡2π.cos⁡θ+sin⁡2π.sin⁡θ\Rightarrow \cos 2\pi .\cos \theta +\sin 2\pi .\sin \theta … (i)\left( i \right)
As we know that the value of π\pi is 180∘180{}^\circ . So, by using it we can get the value of 2π2\pi that must be equal to 360∘360{}^\circ as:
⇒π=180∘\Rightarrow \pi =180{}^\circ
Now, we will multiply by 2 in the above equation as:
⇒2×π=2×180∘\Rightarrow 2\times \pi =2\times 180{}^\circ
So, we will get the value of 2π2\pi as:
⇒2π=360∘\Rightarrow 2\pi =360{}^\circ
Here, we will put this value in the equation (i)\left( i \right) that will give us the equation as:
⇒cos⁡360∘.cos⁡θ+sin⁡360∘.sin⁡θ\Rightarrow \cos 360{}^\circ .\cos \theta +\sin 360{}^\circ .\sin \theta
Now, we will put the value of cos⁡360∘\cos 360{}^\circ and sin⁡360∘\sin 360{}^\circ that is 11and 00 respectively as:
⇒1×cos⁡θ+0×sin⁡θ\Rightarrow 1\times \cos \theta +0\times \sin \theta
Since, we know that the multiplication of 00 with any number gives always 00 . So, we will have cos⁡θ\cos \theta from the above equation as:
⇒cos⁡θ+0\Rightarrow \cos \theta +0
⇒cos⁡θ\Rightarrow \cos \theta
And this is equal to the Right Hand Side of the equation. Hence, the equation has been proved.

Note: Here, we need to remember that the values of cos⁡θ\cos \theta and sin θ\theta at different quadrants. Here is diagram related to the trigonometric function with respect to quadrant as: