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Hint: Rearrange the terms of the equation \[A+B+C+D=2\pi \] and use the properties of cosine and sine functions to prove the given expression. Also use the formula for the sum of two sine and cosine functions to simplify the expression.

Complete step-by-step answer:

We know that \[A+B+C+D=2\pi \]. We have to prove \[\cos A-\cos B+\cos C-\cos D=4\sin \dfrac{A+B}{2}\sin \dfrac{A+D}{2}\cos \dfrac{A+C}{2}\].

We will use a formula for the sum of two cosine or sine functions to get the desired expression.

We will simplify the left side of the expression \[\cos A-\cos B+\cos C-\cos D=4\sin \dfrac{A+B}{2}\sin \dfrac{A+D}{2}\cos \dfrac{A+C}{2}\].

We know that \[\cos x-\cos y=2\sin \left( \dfrac{x+y}{2} \right)\sin \left( \dfrac{y-x}{2} \right)\].

Thus, we have \[\cos A-\cos B=2\sin \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{B-A}{2} \right)\].

Similarly, we have \[\cos C-\cos D=2\sin \left( \dfrac{C+D}{2} \right)\sin \left( \dfrac{D-C}{2} \right)\].

Adding the two equations, we have \[\cos A-\cos B+\cos C-\cos D=2\sin \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{B-A}{2} \right)+2\sin \left( \dfrac{C+D}{2} \right)\sin \left( \dfrac{D-C}{2} \right).....\left( 1 \right)\].

We can write \[C+D=2\pi -(A+B)\] by rearranging the terms of the expression \[A+B+C+D=2\pi \].

Thus, we have \[\dfrac{C+D}{2}=\pi -\dfrac{(A+B)}{2}\].

So, we have \[\sin \left( \dfrac{C+D}{2} \right)=\sin \left( \pi -\dfrac{(A+B)}{2} \right)\].

As we know, \[\sin \left( \pi -\theta \right)=\sin \theta \], we have \[\sin \left( \pi -\dfrac{(A+B)}{2} \right)=\sin \left( \dfrac{A+B}{2} \right)\].

Thus, we have \[\sin \left( \dfrac{C+D}{2} \right)=\sin \left( \pi -\dfrac{(A+B)}{2} \right)=\sin \left( \dfrac{A+B}{2} \right).....\left( 2 \right)\].

Substituting equation \[\left( 2 \right)\] in equation \[\left( 1 \right)\], we have \[\cos A-\cos B+\cos C-\cos D=2\sin \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{B-A}{2} \right)+2\sin \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{D-C}{2} \right)\].

We can rewrite this equation as \[\cos A-\cos B+\cos C-\cos D=2\sin \left( \dfrac{A+B}{2} \right)\left( \sin \left( \dfrac{B-A}{2} \right)+\sin \left( \dfrac{D-C}{2} \right) \right).....\left( 3 \right)\].

We know that \[\sin x+\sin y=2\sin \left( \dfrac{x+y}{2} \right)\cos \left( \dfrac{x-y}{2} \right)\].

Substituting \[x=\dfrac{B-A}{2},y=\dfrac{D-C}{2}\] in the above equation, we have \[\sin \left( \dfrac{B-A}{2} \right)+\sin \left( \dfrac{D-C}{2} \right)=2\sin \left( \dfrac{\dfrac{B-A}{2}+\dfrac{D-C}{2}}{2} \right)\cos \left( \dfrac{\dfrac{B-A}{2}-\dfrac{D-C}{2}}{2} \right)\].

Simplifying the above expression, we have \[\sin \left( \dfrac{B-A}{2} \right)+\sin \left( \dfrac{D-C}{2} \right)=2\sin \left( \dfrac{B+D-\left( A+C \right)}{4} \right)\cos \left( \dfrac{B+C-\left( A+D \right)}{4} \right)\].

We can rewrite \[B+D\] as \[B+D=2\pi -(A+C)\] and \[B+C\] as \[B+C=2\pi -\left( A+D \right)\].

Thus, we have \[\sin \left( \dfrac{B-A}{2} \right)+\sin \left( \dfrac{D-C}{2} \right)=2\sin \left( \dfrac{2\pi -2\left( A+C \right)}{4} \right)\cos \left( \dfrac{2\pi -2\left( A+D \right)}{4} \right)\].

We can rewrite the above equation as \[\sin \left( \dfrac{B-A}{2} \right)+\sin \left( \dfrac{D-C}{2} \right)=2\sin \left( \dfrac{\pi }{2}-\dfrac{A+C}{2} \right)\cos \left( \dfrac{\pi }{2}-\dfrac{A+D}{2} \right)\].

We know that \[\sin \left( \dfrac{\pi }{2}-x \right)=\cos x,\cos \left( \dfrac{\pi }{2}-x \right)=\sin x\].

Thus, we have \[\sin \left( \dfrac{B-A}{2} \right)+\sin \left( \dfrac{D-C}{2} \right)=2\cos \left( \dfrac{A+C}{2} \right)\sin \left( \dfrac{A+D}{2} \right).....\left( 4 \right)\].

Substituting equation \[\left( 4 \right)\] in equation \[\left( 3 \right)\], we have \[\cos A-\cos B+\cos C-\cos D=2\sin \left( \dfrac{A+B}{2} \right)\left( 2\cos \left( \dfrac{A+C}{2} \right)\sin \left( \dfrac{A+D}{2} \right) \right)\].

Simplifying the above expression, we have \[\cos A-\cos B+\cos C-\cos D=4\sin \dfrac{A+B}{2}\sin \dfrac{A+D}{2}\cos \dfrac{A+C}{2}\].

Hence, we have proved that if \[A+B+C+D=2\pi \] holds, we have \[\cos A-\cos B+\cos C-\cos D=4\sin \dfrac{A+B}{2}\sin \dfrac{A+D}{2}\cos \dfrac{A+C}{2}\].

Note: Trigonometric functions are real functions which relate any angle of a right angled triangle to the ratios of any two of its sides. The widely used trigonometric functions are sine, cosine and tangent. However, we can also use their reciprocals, i.e., cosecant, secant and cotangent. We can use geometric definitions to express the value of these functions on various angles using unit circle (circle with radius \[1\]). We also write these trigonometric functions as infinite series or as solutions to differential equations. Thus, allowing us to expand the domain of these functions from the real line to the complex plane. One should be careful while using the trigonometric identities and rearranging the terms to convert from one trigonometric function to the other one.

Complete step-by-step answer:

We know that \[A+B+C+D=2\pi \]. We have to prove \[\cos A-\cos B+\cos C-\cos D=4\sin \dfrac{A+B}{2}\sin \dfrac{A+D}{2}\cos \dfrac{A+C}{2}\].

We will use a formula for the sum of two cosine or sine functions to get the desired expression.

We will simplify the left side of the expression \[\cos A-\cos B+\cos C-\cos D=4\sin \dfrac{A+B}{2}\sin \dfrac{A+D}{2}\cos \dfrac{A+C}{2}\].

We know that \[\cos x-\cos y=2\sin \left( \dfrac{x+y}{2} \right)\sin \left( \dfrac{y-x}{2} \right)\].

Thus, we have \[\cos A-\cos B=2\sin \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{B-A}{2} \right)\].

Similarly, we have \[\cos C-\cos D=2\sin \left( \dfrac{C+D}{2} \right)\sin \left( \dfrac{D-C}{2} \right)\].

Adding the two equations, we have \[\cos A-\cos B+\cos C-\cos D=2\sin \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{B-A}{2} \right)+2\sin \left( \dfrac{C+D}{2} \right)\sin \left( \dfrac{D-C}{2} \right).....\left( 1 \right)\].

We can write \[C+D=2\pi -(A+B)\] by rearranging the terms of the expression \[A+B+C+D=2\pi \].

Thus, we have \[\dfrac{C+D}{2}=\pi -\dfrac{(A+B)}{2}\].

So, we have \[\sin \left( \dfrac{C+D}{2} \right)=\sin \left( \pi -\dfrac{(A+B)}{2} \right)\].

As we know, \[\sin \left( \pi -\theta \right)=\sin \theta \], we have \[\sin \left( \pi -\dfrac{(A+B)}{2} \right)=\sin \left( \dfrac{A+B}{2} \right)\].

Thus, we have \[\sin \left( \dfrac{C+D}{2} \right)=\sin \left( \pi -\dfrac{(A+B)}{2} \right)=\sin \left( \dfrac{A+B}{2} \right).....\left( 2 \right)\].

Substituting equation \[\left( 2 \right)\] in equation \[\left( 1 \right)\], we have \[\cos A-\cos B+\cos C-\cos D=2\sin \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{B-A}{2} \right)+2\sin \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{D-C}{2} \right)\].

We can rewrite this equation as \[\cos A-\cos B+\cos C-\cos D=2\sin \left( \dfrac{A+B}{2} \right)\left( \sin \left( \dfrac{B-A}{2} \right)+\sin \left( \dfrac{D-C}{2} \right) \right).....\left( 3 \right)\].

We know that \[\sin x+\sin y=2\sin \left( \dfrac{x+y}{2} \right)\cos \left( \dfrac{x-y}{2} \right)\].

Substituting \[x=\dfrac{B-A}{2},y=\dfrac{D-C}{2}\] in the above equation, we have \[\sin \left( \dfrac{B-A}{2} \right)+\sin \left( \dfrac{D-C}{2} \right)=2\sin \left( \dfrac{\dfrac{B-A}{2}+\dfrac{D-C}{2}}{2} \right)\cos \left( \dfrac{\dfrac{B-A}{2}-\dfrac{D-C}{2}}{2} \right)\].

Simplifying the above expression, we have \[\sin \left( \dfrac{B-A}{2} \right)+\sin \left( \dfrac{D-C}{2} \right)=2\sin \left( \dfrac{B+D-\left( A+C \right)}{4} \right)\cos \left( \dfrac{B+C-\left( A+D \right)}{4} \right)\].

We can rewrite \[B+D\] as \[B+D=2\pi -(A+C)\] and \[B+C\] as \[B+C=2\pi -\left( A+D \right)\].

Thus, we have \[\sin \left( \dfrac{B-A}{2} \right)+\sin \left( \dfrac{D-C}{2} \right)=2\sin \left( \dfrac{2\pi -2\left( A+C \right)}{4} \right)\cos \left( \dfrac{2\pi -2\left( A+D \right)}{4} \right)\].

We can rewrite the above equation as \[\sin \left( \dfrac{B-A}{2} \right)+\sin \left( \dfrac{D-C}{2} \right)=2\sin \left( \dfrac{\pi }{2}-\dfrac{A+C}{2} \right)\cos \left( \dfrac{\pi }{2}-\dfrac{A+D}{2} \right)\].

We know that \[\sin \left( \dfrac{\pi }{2}-x \right)=\cos x,\cos \left( \dfrac{\pi }{2}-x \right)=\sin x\].

Thus, we have \[\sin \left( \dfrac{B-A}{2} \right)+\sin \left( \dfrac{D-C}{2} \right)=2\cos \left( \dfrac{A+C}{2} \right)\sin \left( \dfrac{A+D}{2} \right).....\left( 4 \right)\].

Substituting equation \[\left( 4 \right)\] in equation \[\left( 3 \right)\], we have \[\cos A-\cos B+\cos C-\cos D=2\sin \left( \dfrac{A+B}{2} \right)\left( 2\cos \left( \dfrac{A+C}{2} \right)\sin \left( \dfrac{A+D}{2} \right) \right)\].

Simplifying the above expression, we have \[\cos A-\cos B+\cos C-\cos D=4\sin \dfrac{A+B}{2}\sin \dfrac{A+D}{2}\cos \dfrac{A+C}{2}\].

Hence, we have proved that if \[A+B+C+D=2\pi \] holds, we have \[\cos A-\cos B+\cos C-\cos D=4\sin \dfrac{A+B}{2}\sin \dfrac{A+D}{2}\cos \dfrac{A+C}{2}\].

Note: Trigonometric functions are real functions which relate any angle of a right angled triangle to the ratios of any two of its sides. The widely used trigonometric functions are sine, cosine and tangent. However, we can also use their reciprocals, i.e., cosecant, secant and cotangent. We can use geometric definitions to express the value of these functions on various angles using unit circle (circle with radius \[1\]). We also write these trigonometric functions as infinite series or as solutions to differential equations. Thus, allowing us to expand the domain of these functions from the real line to the complex plane. One should be careful while using the trigonometric identities and rearranging the terms to convert from one trigonometric function to the other one.

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