in any triangle, show that sin(A + B + C) = sinAc

Question

in any triangle, show that sin(A + B + C) = sinAcosBcosC + sinBcosAcosC + sinCcosAcosB, where A, B, C : are angles.

Arun · Accepted Answer

Dear Student

Use the formula
sin(a + b) = sina cosb + sinb cos a

Then sin(a + b + c)
= sin((a + b) + c)
= sin(a + b) cosc + cos(a + b) sin c

Then use the formula
cos(a + b) = cosa cosb - sina sinb to get

sin(a + b) cosc + cos(a + b) sin c
= (sina cosb + sinb cosa) cosc + (cosa cosb - sina sinb) sinc
= sina cosb cosc + sinb cosa cosc + sinc cosa cosb - sina sinb sinc

Regards

Arun (askIITians forum expert)

Trigonometry> in any triangle, show that sin(A + B + C)...

in any triangle, show that sin(A + B + C) = sinAcosBcosC + sinBcosAcosC + sinCcosAcosB, where A, B, C : are angles.

mmtefy , 8 Years ago

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Trigonometry> in any triangle, show that sin(A + B + C)...

in any triangle, show that sin(A + B + C) = sinAcosBcosC + sinBcosAcosC + sinCcosAcosB, where A, B, C : are angles.

mmtefy , 8 Years ago

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