In a triangle ABC `sin (A/2) sin (B/2) sin (C/2) lt= 1/8`

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How to prove ๐‘กโ„Ž๐‘Ž๐‘ก ๐‘ ๐‘–๐‘› ๐ด/2.๐‘ ๐‘–๐‘› ๐ต/2.๐‘ ๐‘–๐‘› ๐ถ/2 โ‰ค 1/8 , ABC is a triangle | Trigonometry inequality | How to prove ๐‘กโ„Ž๐‘Ž๐‘ก ๐‘ ๐‘–๐‘› ๐ด/2.๐‘ ๐‘–๐‘› ๐ต/2.๐‘ ๐‘–๐‘› ๐ถ/2 โ‰ค 1/8 , ABC is a triangle | Trigonometry inequality |
sinA/2 sinB/2 sinC/2 is less 1/8 sinA/2 sinB/2 sinC/2 is less 1/8
Proof of sinA/2.sinB/2.sinC/2 less than or equals 1/8, when A, B, C are the angles of a triangle. Proof of sinA/2.sinB/2.sinC/2 less than or equals 1/8, when A, B, C are the angles of a triangle.
In any triangle `A B C` , prove that following: `\ \ asin(A/2)sin((B-C)/2)+bsin(B/2)sin((C-A)/ In any triangle `A B C` , prove that following: `\ \ asin(A/2)sin((B-C)/2)+bsin(B/2)sin((C-A)/
In any triangle ABC, show that : `2a sin (B/2) sin (C/2)=(b+c-a) sin (A/2)` In any triangle ABC, show that : `2a sin (B/2) sin (C/2)=(b+c-a) sin (A/2)`
Prove (sin2A + sin2B + sin2C )/(sinA + sinB + sinC ) = 8 sin(A/2) sin(B/2) sin(C/2) Prove (sin2A + sin2B + sin2C )/(sinA + sinB + sinC ) = 8 sin(A/2) sin(B/2) sin(C/2)
Prove That : -sinA+sinB+sinC=4cos A/2.sin B/2.sin C/2 Prove That : -sinA+sinB+sinC=4cos A/2.sin B/2.sin C/2
law of sine and law of cosine #trigonometry law of sine and law of cosine #trigonometry