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A 1991 hsc quesion (1 Viewer)

Steven12

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I have a question. from 1991 HSC paper. question 7(B)III

I have read the answer but i do not understand it.

It is very hard to explain it cos its a graph question, so if any of you have Sucess One books , can you take a look and explain the answer,


Much appprecited
 

acmilan

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Are you sure your looking at the right year? In my book the 1991 question 7(b)(iii) is a trig question
 

Steven12

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yep. Successone extension 1.

1991 paper Q7(b)III

(III)deduce maximum angle subtended.......
 

withoutaface

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biii) you already have the max is at x=2

:. @= arctan(2)-arctan(1/2)
tan@=tan(arctan2-arctan(1/2))
tan@= [tan(arctan2)-tan(arctan(1/2))]/[1+tan[arctan(2)]tan[arctan(1/2)]]
tan@= [2-1/2]/[1+1]
tan@=(3/2)/(2)
tan@=3/4
@=arctan(3/4)
 
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Steven12

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yea, but i dont get that answer at all

why do you have to divide by (1+arctan(2)arctan(1/2)]

i dont get how they derive this.
 

Xayma

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tan (α-β.)=(tan α-tan β.)/(1+tanαtanβ.)

Where &alpha;=tan<sup>-1</sup> 2
&beta;=tan<sup>-1</sup> 1/2

tan (tan<sup>-1</sup> &gamma;.)=&gamma;
 
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