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Question: How do you express \(\cos \left( {\pi + x} \right){\text{ and sin}}\left( {2\pi - x} \right)\) in te...

How do you express cos⁡(π+x) and sin(2π−x)\cos \left( {\pi + x} \right){\text{ and sin}}\left( {2\pi - x} \right) in terms of sin⁡x and cos⁡x ?\sin x{\text{ and }}\cos x{\text{ ?}}

Explanation

Solution

Hint : Here, we have been given two functions i.e. cos⁡(π+x) and sin(2π−x)\cos \left( {\pi + x} \right){\text{ and sin}}\left( {2\pi - x} \right) and we asked to express these functions in the form of cosine and sine functions respectively. To solve these types of questions, we should remember standard trigonometric identities and formulae. For example here, (1)cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B\left( 1 \right)\cos \left( {A + B} \right) = \cos A\cos B - \sin A\sin B and (2)sin⁡(A−B)=sin⁡Acos⁡B−cos⁡Asin⁡B\left( 2 \right)\sin \left( {A - B} \right) = \sin A\cos B - \cos A\sin B. Using these two standard identities, we will solve the given functions and try to simplify them.

Complete step-by-step answer :
Let us solve the given functions one by one.
(1)cos⁡(π+x)=?\left( 1 \right)\cos \left( {\pi + x} \right) = ?
The given trigonometric function is ;
=cos⁡(π+x) ......(1)= \cos \left( {\pi + x} \right){\text{ }}......\left( 1 \right)
By the standard trigonometric identity for cosine function, we know that;
⇒cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B ......(2)\Rightarrow \cos \left( {A + B} \right) = \cos A\cos B - \sin A\sin B{\text{ }}......\left( 2 \right)
Comparing equation 11 with equation 22 , let ;
⇒A=π and B=x\Rightarrow A = \pi {\text{ and }}B = x
We will try to expand equation 11 by using the property mentioned in equation 22 for cosine function and try to get the resultant function in terms of cosine function;
⇒cos⁡(π+x)=cos⁡πcos⁡x−sin⁡πsin⁡x ......(3)\Rightarrow \cos \left( {\pi + x} \right) = \cos \pi \cos x - \sin \pi \sin x{\text{ }}......\left( 3 \right)
By the standard values for sine and cosine functions, we know that;
∵sin⁡π=0\because \sin \pi = 0 and ∵cos⁡π=−1\because \cos \pi = - 1
Putting the above values of sin⁡π and cos⁡π\sin \pi {\text{ and }}\cos \pi in equation 33 , we get;
\Rightarrow \cos \left( {\pi + x} \right) = \left( { - 1} \right) \times \left\\{ {\cos \left( x \right)} \right\\} - \left\\{ {0 \times \sin \left( x \right)} \right\\}
On further simplification, we get;
\Rightarrow \cos \left( {\pi + x} \right) = \left\\{ { - \cos \left( x \right)} \right\\} - 0
⇒cos⁡(π+x)=−cos⁡x\Rightarrow \cos \left( {\pi + x} \right) = - \cos x
By the above trigonometric equality, we can say that cos⁡(π+x)=−cos⁡x\cos \left( {\pi + x} \right) = - \cos x .

(2)sin⁡(2π−x)=?\left( 2 \right)\sin \left( {2\pi - x} \right) = ?
The given trigonometric function is ;
=sin⁡(2π−x) ......(1)= \sin \left( {2\pi - x} \right){\text{ }}......\left( 1 \right)
By the standard trigonometric identity for cosine function, we know that;
⇒sin⁡(A−B)=sin⁡Acos⁡B−cos⁡Asin⁡B ......(2)\Rightarrow \sin \left( {A - B} \right) = \sin A\cos B - \cos A\sin B{\text{ }}......\left( 2 \right)
Comparing equation 11 with equation 22 , let ;
⇒A=2π and B=x\Rightarrow A = 2\pi {\text{ and }}B = x
We will try to expand equation 11 by using the property mentioned in equation 22 for sine function and try to get the resultant function in terms of sine function;
⇒sin⁡(2π−x)=sin⁡2πcos⁡x−cos⁡2πsin⁡x ......(3)\Rightarrow \sin \left( {2\pi - x} \right) = \sin 2\pi \cos x - \cos 2\pi \sin x{\text{ }}......\left( 3 \right)
By the standard values for sine and cosine functions, we know that;
∵sin⁡2π=0\because \sin 2\pi = 0 and ∵cos⁡2π=1\because \cos 2\pi = 1
Putting the above values of sin⁡2π and cos⁡2π\sin 2\pi {\text{ and }}\cos 2\pi in equation 33 , we get;
\Rightarrow \sin \left( {2\pi - x} \right) = 0 \times \left\\{ {\cos \left( x \right)} \right\\} - \left\\{ {1 \times \sin \left( x \right)} \right\\}
On further simplification, we get;
⇒sin⁡(2π−x)=0−sin⁡(x)\Rightarrow \sin \left( {2\pi - x} \right) = 0 - \sin \left( x \right)
⇒sin⁡(2π−x)=−sin⁡x\Rightarrow \sin \left( {2\pi - x} \right) = - \sin x
By the above trigonometric equality, we can say that sin⁡(2π−x)=−sin⁡x\sin \left( {2\pi - x} \right) = - \sin x.

Note : We need to know some rules of conversion to answer this question.
As the angle (θ)\left( \theta \right) of a trigonometric function changes, then we have to take care of two things in the resultant function :
(1)\left( 1 \right) Conversion to another function:
Rules of conversion: Conversion of one trigonometric function to another as the angle varies from quadrant to quadrant. The important point to note here is that the conversion will only be done when the reference angle involves the vertical axis or y axis, if the angle of reference is with respect to horizontal axis or x axis there will be no conversion and function will remain the same.
Conversion rules are:
(i)sin⁡↔cos⁡\left( {\text{i}} \right)\sin \leftrightarrow \cos
(ii)sec⁡↔cos⁡ec\left( {{\text{ii}}} \right)\sec \leftrightarrow \cos ec
(iii)tan⁡↔cot⁡\left( {{\text{iii}}} \right)\tan \leftrightarrow \cot