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Question: How do you verify \[\sin \left( {2\pi - \theta } \right) = - sin{\text{ }}\theta \] ?...

How do you verify sin⁡(2π−θ)=−sin θ\sin \left( {2\pi - \theta } \right) = - sin{\text{ }}\theta ?

Explanation

Solution

In this question, we need to prove a trigonometric identity that is sin⁡(2π−θ)=−sin θ\sin \left( {2\pi - \theta } \right) = - sin{\text{ }}\theta . We will do this by using the properties of trigonometric functions and its identity. First, we will use this sin⁡(x−y)=sin⁡xcos⁡ y−sin⁡ycos⁡x\sin \left( {x - y} \right) = \sin x\cos {\text{ }}y - \sin y\cos x identity and then, find the values and prove the required identity.

Formula used: sin⁡(x−y)=sin⁡xcos⁡ y−sin⁡ycos⁡x\sin \left( {x - y} \right) = \sin x\cos {\text{ }}y - \sin y\cos x
cos⁡360∘=1\cos 360^\circ = 1
sin⁡360∘=0\sin 360^\circ = 0

Complete step-by-step solution:
We need to prove a trigonometric identity that is sin⁡(2π−θ)=−sin θ\sin \left( {2\pi - \theta } \right) = - sin{\text{ }}\theta .
First of all, these identities are always proved or derived from the basic trigonometric identity or formula. Trigonometric functions are functions of an angle. They are used to relate the angles of a triangle to the lengths of the sides of a triangle. Therefore all the trigonometric functions and the properties are interlinked with each other.
Now to prove this identity, we can write it like, sin⁡(2π−θ)=sin⁡(360−θ)\sin (2\pi - \theta ) = \sin (360 - \theta )
We know that sin⁡(x−y)=sin⁡xcos⁡ y−sin⁡ycos⁡x\sin \left( {x - y} \right) = \sin x\cos {\text{ }}y - \sin y\cos x
Using this identity, in the previous one, that is sin⁡(360−θ)\sin (360 - \theta ) ,
⇒\Rightarrow sin⁡(360−θ)=sin⁡360cos⁡θ−sin⁡θcos⁡360\sin (360 - \theta ) = \sin 360\cos \theta - \sin \theta \cos 360
Since we know that, sin⁡360∘=0\sin 360^\circ = 0 , and cos⁡360∘=1\cos 360^\circ = 1 , applying this in the above
⇒\Rightarrow sin⁡(360−θ)=sin⁡360cos⁡θ−sin⁡θcos⁡360\sin (360 - \theta ) = \sin 360\cos \theta - \sin \theta \cos 360
⇒\Rightarrow sin⁡(360−θ)=(0)cos⁡θ−sin⁡θ(1)\sin (360 - \theta ) = (0)\cos \theta - \sin \theta (1)
And it becomes,
⇒\Rightarrow sin⁡(360−θ)=0−sin⁡θ\sin (360 - \theta ) = 0 - \sin \theta
Ultimately, we get,
⇒\Rightarrow sin⁡(360−θ)=−sin⁡θ\sin (360 - \theta ) = - \sin \theta
Left hand side = right hand side
Hence the given identity is proved.

Note: Trigonometry is an important branch of Mathematics and the trigonometric functions and its identities are very important identities to learn as we can’t solve most of the trigonometric problems without using it.
Here are some of the important identities or formulas which will be frequently used and asked for.
sin⁡(x−y)=sin⁡xcos⁡ y−sin⁡ycos⁡x\sin \left( {x - y} \right) = \sin x\cos {\text{ }}y - \sin y\cos x
sin⁡(x+y)=sin⁡xcos⁡ y+sin⁡ycos⁡x\sin \left( {x + y} \right) = \sin x\cos {\text{ }}y + \sin y\cos x
cos (x − y) = cos xcos⁡ y + sin⁡ysin⁡x{\text{cos }}\left( {x{\text{ }} - {\text{ }}y} \right){\text{ }} = {\text{ cos }}x\cos {\text{ }}y{\text{ + }}\sin y\sin x
cos (x + y) = cos xcos⁡ y - sin⁡ysin⁡x{\text{cos }}\left( {x{\text{ + }}y} \right){\text{ }} = {\text{ cos }}x\cos {\text{ }}y{\text{ - }}\sin y\sin x
cos⁡(π−θ)=−cos⁡θ\cos (\pi - \theta ) = - \cos \theta
cos⁡(π+θ)=−cos⁡θ\cos (\pi + \theta ) = - \cos \theta
cos⁡(2π−θ)=cos⁡θ\cos (2\pi - \theta ) = \cos \theta
sin⁡(π−θ)=sin⁡θ\sin (\pi - \theta ) = \sin \theta
sin⁡(π+θ)=−sin⁡θ\sin (\pi + \theta ) = - \sin \theta
sin⁡(2π−θ)=−sin⁡θ\sin (2\pi - \theta ) = - \sin \theta
These identities alone aren’t enough for a better preparation. You need to study the proof and memorise the identities to solve all kinds of trigonometric functions.