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secondary 4 | A Maths
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AX
AX

secondary 4 chevron_right A Maths chevron_right Singapore

Help me with this proving question. I start from the left side but can't seem to prove it. Thanks in advance.

Date Posted: 7 months ago
Views: 40
J
J
7 months ago
sin 2A + sin 2B

= sin ((A+B) + (A-B)) + sin ((A+B) - (A-B))

= sin(A+B)cos(A-B) + cos(A+B)sin(A-B)
+ sin(A+B)cos(A-B) - cos(A+B)sin(A-B)

= 2sin(A+B)cos(A-B)


sin 2A - sin 2B

= sin ((A+B) + (A-B)) - sin ((A+B) - (A-B))

= sin(A+B)cos(A-B) + cos(A+B)sin(A-B)
- (sin(A+B)cos(A-B) + cos(A+B)sin(A-B) )

= 2cos(A+B)sin(A-B)


So,

(sin 2A + sin 2B)/(sin 2A - sin 2B)

= (2sin(A+B)cos(A-B))/(2cos(A + B)sin(A-B))
= tan(A+B)cot(A-B)
= tan(A+B)/tan(A-B)

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