In a triangle ABC, sin 2A=?a)2sin(B+C)cos( B-C)b)2sin(B-C)cos( B-C)c)2...
To find the value of sin 2A in triangle ABC, we can start by using the double angle formula for sine.
Using the double angle formula for sine, we have sin 2A = 2sin A cos A.
Now, let's analyze triangle ABC to find the values of sin A and cos A.
We know that the sum of the angles in a triangle is 180 degrees. So, we have A + B + C = 180.
Rearranging this equation, we get A = 180 - B - C.
Now, let's use the sine rule in triangle ABC, which states that sin A / a = sin B / b = sin C / c, where a, b, and c are the lengths of the sides opposite to angles A, B, and C, respectively.
From the sine rule, we have sin A / a = sin B / b.
Rearranging this equation, we get sin A = a sin B / b.
Substituting the value of A from the earlier equation, we have sin (180 - B - C) = a sin B / b.
Using the property of sine function, sin (180 - x) = sin x, we have sin (B + C) = a sin B / b.
Rearranging this equation, we get sin (B + C) = (a / b) sin B.
Comparing this equation with the double angle formula for sine, we can conclude that sin 2A = sin (B + C).
Now, using the property of sine function, sin (B + C) = sin (B - C).
So, sin 2A = sin (B - C).
Finally, using the double angle formula for sine, we have sin 2A = 2sin(B - C)cos(B - C).
Hence, the correct answer is option 'D': 2sin(B - C)cos(B - C).
In a triangle ABC, sin 2A=?a)2sin(B+C)cos( B-C)b)2sin(B-C)cos( B-C)c)2...
Please solve it
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