If A,B and C are the interior angles of triangle ABC, show that sin (a+b/2) = cos c/2
Answers
Answered by
5
as in Δabc
angle (a+b+c)=180
taking lhs from question
sin(a/2+b/2)
sin(180/2-c/2)
sin(90-c/2)
cos c/2
angle (a+b+c)=180
taking lhs from question
sin(a/2+b/2)
sin(180/2-c/2)
sin(90-c/2)
cos c/2
Answered by
53
As we know, for any given triangle, the sum of all its interior angles is equals to 180°.
Thus,
A + B + C = 180° ….(1)
Now we can write the above equation as;
⇒ B + C = 180° – A
Dividing by 2 on both the sides;
⇒ (B + C)/2 = (180° – A)/2
⇒ (B + C)/2 = 90° – A/2
Now, put sin function on both sides.
⇒ sin (B + C)/2 = sin (90° – A/2)
Since,
sin (90° – A/2) = cos A/2
Therefore,
sin (B + C)/2 = cos A/2
Hope it's Helpful....:)
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