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\frac{\sin \alpha \sin 11 \alpha+\sin 3 \alpha \sin 7 \alpha}{\sin \alpha \cos 11 \alpha+\sin 3 \alpha \cos 7 \alpha}=\tan 8 \alpha |
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Answer» Sin11AsinA+sin7Asin3A/cos11AsinA+cos7Asin3A=2sin11AsinA+2sin7Asin3A/2cos11AsinA+2cos7Asin3A=[{cos(11A-A)-cos(11A+A)}+{cos(7A-3A)-cos(7A+3A)}]/[{sin(11A+A)-sin(11A-A)}+{sin(7A+3A)-sin(7A-3A)}]=(cos10A-cos12A+cos4A-cos10A)/(sin12A-sin10A+sin10A-sin4A)=(cos4A-cos12A)/(sin12A-sin4A)=[2sin(4A+12A)/2sin(12A-4A)/2]/[2cos(12A+4A)/2sin(12A-4A)/2]=sin8Asin4A/cos8Asin4A=sin8A/cos8A=tan8A (Proved) |
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