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(reminder: all quotes here are fiddled, probably.)

cos (x-A) = a


cos (x-A) = a
sin (x-B) = b
prove cos^2 (A-B) = a^2 + b^2 - 2ab sin (A-B)


Solution:



M1

Letting p = x - A , and q = x - B ,
the statement in question becomes
cos^2 (q - p) = a^2 + b^2 - 2ab sin (q - p)

LHS
= (cos p cos q + sin p sin q)^2
= a^2 cos^2 q + b^2 sin ^2 p + 2ab sin p cos q

Note that the 1st term = a^2 (1 - b^2)
and the 2nd term = b^2 (1 - a^2) .
So, LHS becomes a^2 + b^2 - 2abH ,
where H = ab - sin p cos q = sin (q - p) .
QED.



M2

Basement cat's solution (see comments).
Besides the brute force, another apporach:

Let A, B be fixed.
Consider a, b to be a function of x .
f(x) = a^2 + b^2 - 2ab sin (A - B)
f'(x) = 2aa' + 2bb' - 2(ab' + a'b) sin (A - B)

And note that
sin (A - B) = sin (A - x + x - B) = a'b' + ab
(a')^2 + a^2 = 1
(b')^2 + b^2 = 1

Thus, f'(x) = 0 .
i.e. f(x) is a constant. In particular,
f(B) = cos^2 (A - B)

Labels:

9 Comments:

:: Blogger Derek (11.04.09, 10:16   ) sagt...

LHS不是cos^2(A-B)嗎﹖



:: Blogger sf (12.04.09, 01:47   ) sagt...

yes, you are right.
revised now. thank you.

i found this question very difficult.
do you think so?



:: Anonymous basement cat (12.04.09, 02:58   ) sagt...

Besides the brute force, another apporach:

Let A, B be fixed
consider a, b to be a function of x

f(x)=a^2 + b^2 - 2ab sin (A-B)

consider sin(A-B) = sin(A-x+x-B) =a'b'+ab

It can be shown that f'(x) = 0

f(x) is a constant. In particular,
f(x) = f(B) = cos^2 (A-B)



:: Anonymous basement cat (12.04.09, 09:02   ) sagt...

I messed up whe I typed. The steps should be

f(x)=a^2 + b^2 - 2ab sin (A-B)
f'(x)=2aa'+2bb'-2(ab'+a'b)sin(A-B)

and becuase
sin(A-B) = sin(A-x+x-B) =a'b'+ab
(a')^2+a^2=1
(b')^2+b^2=1

Thus, f'(x)=0.



:: Blogger Derek (12.04.09, 21:27   ) sagt...

sf:

I spent quite a while to figure out how to represent sin(x-A) and cos(x-B)=.=

basement:

Wa! I dun even think about solving it by differentiation. You teach me new thing.



:: Blogger sf (14.04.09, 01:17   ) sagt...

basement cat, 精采精采. 你從LHS跟x無關而想到做微分, 精采.

derek, 前幾天發現了幾張自己用過的草稿紙, 上面寫了這道題目和三種解法. 記下問題, 順手把草稿紙掉了, 誰知計唔番, 苦也.



:: Blogger Derek (14.04.09, 21:25   ) sagt...

三種咁多﹖我淨係諗到一種



:: Blogger seikomatic (16.04.09, 16:38   ) sagt...

耶穌典解要用人喜歡見到的方法去講佢既道理呢?例如行神蹟?

定系行神蹟系為左證明人既罪可以被赦免?

證明佢系【人子】?

如果人要見到神蹟先信上帝,咁俺要見到上帝先信耶穌,得唔得呢?



:: Blogger sf (16.04.09, 23:01   ) sagt...

是的是的, 不該譯做「證明」, 已改.



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