Trigonometry & equations

Cannot solve this:

tg0 - tg 80 (the answer written in the textbook is 0.. why?)

Equations :
-------------------
| 2^X*9^Y=162
| 3^X*4^Y=48
|------------------


cos(a + b ) = cos a*cos b - sin a*sin b
cos(a - b ) = cos a*cos b - sin a*sin b
cos(p/2 - a ) = sin a
sin(p/2 - a ) = cos a
cos(p/2 - b ) = sin b

How do I write sin( a + b ) and sin(a - b ) using the formulae above?
How do I form the sine formulae out of these ones, in short?

Please help !

There are various ways you can go at it, but I think a good visualization of this problem can be done with a factor tree:

162
/ \
18 9 you can try 81 and 2 also
/ \
2 9 Just count the number of 9's and 2's to get their exponents: (9^2)(2^1)=162

You can also try traditional division and multiplication, just remember that sqrtX is X^1/2, cbrtX is x^1/3, etc. and X^a^b = x^(ab)

I'll let you find 3^X*4^Y=48 on your own, now that you know two methods.

Also:
sin(A+ = sinA(cosB) + cosA(sinB)
sin(A- = sinA(cosB) - cosA(sinB)
Note: These 2 can be derived from the Pythagorean theorem a^2+b^2=c^2, although it does take a while.

Sorry, forgot to sign in. FYI, those are B ) after the auto html feature gets through them. And I'm not sure what you mean by tg in tg0-tg80. Maybe it's this heat wave that's causing my confusion...

Hmmm, I tried to solve both of the equations by multiplicating...I got this :
3^5-2y*4^y=48. And I got stuck.. x=5-2y.

You haven't understood. I do know these formulae
The fact is that... well, let's presume that I have only

cos(a + b ) = cos a*cos b - sin a*sin b
cos(a - b ) = cos a*cos b - sin a*sin b
cos(p/2 - a ) = sin a
sin(p/2 - a ) = cos a
cos(p/2 - b ) = sin b

(i.e. I dunno the original sin( a + b ) and sin ( a - b ) formulae).

How do I get sin( a + b ) and sin(a - b ) out of what I wrote above?

I think the easiest way to do the first problem is as follows:

(2^x)*(9^y)=162 & (3^x)*(4^y)=48

search for a simple solution and check

first if you divide 162 by 9 twice you can rewrite as:

(2^x)*(9^(y-2))=2 now divide each side by 2


(2^(x-1))*(9^(y-2))=1 and anything raised to the zero power = 1 with 1*1=1

this is not the only solution but rather a simple solution. Much the same as the checking method of roots explained above. So we set..

x-1=0 and y-2=0 x=1 y=2 if you check it it works in both eqns

take log both sides

x log(2) + y log(9) = log(162)
x log(3) + y log(4) = log(48)

solve simultaneously from there onwards

I'm guessing that in your case you did it this way (I'm going to use 90=pi/2 for simplicity):
sin(a+=cos[90-(a+]=cos(90)*cos(a++sin(90)*sin(a+.
Since cos(90)=0 we ignore that term and sin(90)=1 so we can leave that out
=> sin(a+=sin(a+
circular reasoning

Well, the idea was right, just:
sin(x)=cos[90-(a+]=cos[(90-a)-b]
And we use the cosine of difference of angle identiy and...:
=cos(90-a)*cos(+sin(90-a)*sin(
=sin(a)*cos(+cos(a)*sin(

I'm guessing that in your case you did it this way (I'm going to use 90=pi/2 for simplicity):
sin(a+=cos(90-(a+)=cos(90)*cos(a++sin(90)*sin(a+.
Since cos(90)=0 we ignore that term and sin(90)=1 so we can leave that out
=> sin(a+=sin(a+
circular reasoning

Well, the idea was right, just:
sin(x)=cos(90-(a+)=cos((90-a)-
And we use the cosine of difference of angle identiy and...:
=cos(90-a)*cos(+sin(90-a)*sin(
=sin(a)*cos(+cos(a)*sin(

...darn, well, you get the idea.