Question

sin^2 A+sin^2 B =sin^2 c the angle c=

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Answer 1

Answer:

STEP 1 - The given formula looks a lot like the law of cosines - and we can use this quite elegantly.

The given formula looks a lot like the law of cosines - and we can use this quite elegantly.Note that we can interpret A, B, and C as the angles in a triangle with side lengths a, b and c. We can choose a as anything we want, so let a=sin(A), and by the law of sines:

The given formula looks a lot like the law of cosines - and we can use this quite elegantly.Note that we can interpret A, B, and C as the angles in a triangle with side lengths a, b and c. We can choose a as anything we want, so let a=sin(A), and by the law of sines:sin(A)a=sin(B)b=sin(C)c

The given formula looks a lot like the law of cosines - and we can use this quite elegantly.Note that we can interpret A, B, and C as the angles in a triangle with side lengths a, b and c. We can choose a as anything we want, so let a=sin(A), and by the law of sines:sin(A)a=sin(B)b=sin(C)cwe have a triangle with angles A, B, and C such that a=sin(A), b=sin(B), and c=sin(C)

The given formula looks a lot like the law of cosines - and we can use this quite elegantly.Note that we can interpret A, B, and C as the angles in a triangle with side lengths a, b and c. We can choose a as anything we want, so let a=sin(A), and by the law of sines:sin(A)a=sin(B)b=sin(C)cwe have a triangle with angles A, B, and C such that a=sin(A), b=sin(B), and c=sin(C)Now we can apply the law of cosines:

The given formula looks a lot like the law of cosines - and we can use this quite elegantly.Note that we can interpret A, B, and C as the angles in a triangle with side lengths a, b and c. We can choose a as anything we want, so let a=sin(A), and by the law of sines:sin(A)a=sin(B)b=sin(C)cwe have a triangle with angles A, B, and C such that a=sin(A), b=sin(B), and c=sin(C)Now we can apply the law of cosines:c2=a2+b2−2abcos(C)⟹sin2(C)=sin2(A)+sin2(B)−2sin(A)sin(B)cos(C)

The given formula looks a lot like the law of cosines - and we can use this quite elegantly.Note that we can interpret A, B, and C as the angles in a triangle with side lengths a, b and c. We can choose a as anything we want, so let a=sin(A), and by the law of sines:sin(A)a=sin(B)b=sin(C)cwe have a triangle with angles A, B, and C such that a=sin(A), b=sin(B), and c=sin(C)Now we can apply the law of cosines:c2=a2+b2−2abcos(C)⟹sin2(C)=sin2(A)+sin2(B)−2sin(A)sin(B)cos(C)Rearranging: voila, we get the desired equality!

The given formula looks a lot like the law of cosines - and we can use this quite elegantly.Note that we can interpret A, B, and C as the angles in a triangle with side lengths a, b and c. We can choose a as anything we want, so let a=sin(A), and by the law of sines:sin(A)a=sin(B)b=sin(C)cwe have a triangle with angles A, B, and C such that a=sin(A), b=sin(B), and c=sin(C)Now we can apply the law of cosines:c2=a2+b2−2abcos(C)⟹sin2(C)=sin2(A)+sin2(B)−2sin(A)sin(B)cos(C)Rearranging: voila, we get the desired equality!sin2(A)+sin2(B)−sin2(C)=2sin(A)sin(B)cos(C

Step-by-step explanation:

STEP 2 Use Prove sin(A+B)sin(A−B)=sin2A−sin2B,

andsinA=sin[π−(B+C)]=sin(B+C)

sin2A+sin2B−sin2C

=sin2A+sin(B−C)sin(B+C)

=sinA[sinA+sin(B−C)]

=sinA[sin(B+C)+sin(B−C)]

=sinA[2sinBcosC]

=2sinAsinBcosC