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Activity Discussion Math Trigonometry

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• # Trigonometry

Posted by on January 10, 2024 at 6:07 pm

If A, B, and C are interior angles of a triangle ABC, show that sin (B + C/2) = cos A/2.

replied 1 month, 2 weeks ago 2 Members · 1 Reply
• ### Nitesh

Member
January 11, 2024 at 4:32 pm
0

To prove the trigonometric identity sin(B + C/2) = cos(A/2) for the interior angles of a triangle ABC, we can use the following steps:

1. Start with the given triangle ABC and its interior angles A, B, and C.
2. Use the fact that the sum of interior angles in a triangle is 180 degrees: A + B + C = 180Â°.
3. Divide both sides of the equation by 2 to get (A/2) + (B/2) + (C/2) = 90Â°.
4. Subtract (B/2) and (C/2) from both sides of the equation to obtain (A/2) = 90Â° – (B/2) – (C/2).
5. Take the cosine of both sides of the equation to get cos(A/2) = cos(90Â° – (B/2) – (C/2)).
6. Use the cosine identity cos(90Â° – Î¸) = sin(Î¸) to rewrite the equation as cos(A/2) = sin((B/2) + (C/2)).
7. Rearrange the right side of the equation to obtain sin((B/2) + (C/2)) = cos(A/2), which is the desired result.

Therefore, we have shown that sin(B + C/2) = cos(A/2) for the interior angles of triangle ABC.

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