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Question: If \[A,\,B\] and \(C\) are interior angles of a triangle \(ABC\) ,then show that \(\sin \left( {\dfr...

If A, BA,\,B and CC are interior angles of a triangle ABCABC ,then show that sin⁡(A+B2)=cos⁡A2\sin \left( {\dfrac{{A + B}}{2}} \right) = \cos \dfrac{A}{2}

Explanation

Solution

To solve this type of question we can use the property which is angle sum property of a triangle which gives sum of all sides of a triangle gives 180180 degrees. Then by doing proper rearrangement and using trigonometric identity we can prove the given question.

Complete step-by-step answer:
Given A, BA,\,B and CC are interior angles of a triangle ABCABC ,
Consider a triangle ABCABC ,
Triangle is the smallest polygon which has three sides and three interior angles.

By using the interior angle sum property of a triangle,
Interior angle sum property of a triangle states that the sum of interior angles gives 180180 degrees.
So we get,
Sum of interior angles of a triangle =180∘= 180^\circ
A+B+C=180∘   A + B + C = 180^\circ \\\ \\\
By rearranging the above result we get,
B+C=180∘−AB + C = 180^\circ - A
By multiplying on both left hand side and right hand side with 12\dfrac{1}{2} , the above result changes to,
B+C2=180∘−A2\dfrac{{B + C}}{2} = \dfrac{{180^\circ - A}}{2}
Separating the terms we get,
B2+C2=180∘2−A2\dfrac{B}{2} + \dfrac{C}{2} = \dfrac{{180^\circ }}{2} - \dfrac{A}{2}
Perform the arithmetic operation on above result,
B+C2=90∘−A2\dfrac{{B + C}}{2} = 90^\circ - \dfrac{A}{2}
Taking the left hand side from the expression,
L.H.S =sin⁡(B+C2) = \sin \left( {\dfrac{{B + C}}{2}} \right)
Substitute the value of (B+C2)\left( {\dfrac{{B + C}}{2}} \right) with 90∘−A290^\circ - \dfrac{A}{2}
Hence,
x=sin⁡(90∘−A2)x = \sin \left( {90^\circ - \dfrac{A}{2}} \right)
We can perform the trigonometric identities in the above result to further step,
We know that,
cos⁡(90−θ)=sin⁡θ sin⁡(90−θ)=cos⁡θ  \cos (90 - \theta ) = \sin \theta \\\ \sin (90 - \theta ) = \cos \theta \\\
Applying the above result in our expansion,
x=sin⁡(90∘−A2)x = \sin \left( {90^\circ - \dfrac{A}{2}} \right)
To get,
x=cos⁡A2x = \cos \dfrac{A}{2}
== R.H.S
Which gives the right hand side,
We get the left hand side and right hand side as equal.
sin⁡(A+B2)=cos⁡A2\sin \left( {\dfrac{{A + B}}{2}} \right) = \cos \dfrac{A}{2}
Hence proved.

Note: We use the property of interior angle sum property of a triangle to prove the above question which gives sum of interior angles of triangle is 180180 degrees and also we have to perform some of the trigonometric identities. Students should remember the important trigonometric formulas, identities and standard trigonometric angles for solving these types of problems.