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secondary 3 | A Maths
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Help with tan3A
Date Posted: 3 years ago
Views: 280
tan 2A
= 2tanA / (1 - tan² A)
= 2(½) / (1 - (½)²)
= 1 / (1 - ¼)
= 1 / ¾
= 4/3
= 2tanA / (1 - tan² A)
= 2(½) / (1 - (½)²)
= 1 / (1 - ¼)
= 1 / ¾
= 4/3
tan 3A
= tan (A + 2A)
= (tan A + tan 2A) / (1 - tan A tan 2A)
= (½ + 4/3) / (1 - ½(4/3) )
= (11/6) / (⅓)
= 11/6 x 3/1
= 11/2
= 5½
= tan (A + 2A)
= (tan A + tan 2A) / (1 - tan A tan 2A)
= (½ + 4/3) / (1 - ½(4/3) )
= (11/6) / (⅓)
= 11/6 x 3/1
= 11/2
= 5½
How u get 4/3 for the tan3A
You mean tan 2A. It has already been found to be 4/3 in the first part. So simply substitute it into the second part when you're solving for tan 3A
Right thanks!
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done
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clear
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A rough idea
Date Posted:
3 years ago
Will be better to leave final answer as mixed number 5½
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