problems here :P

[hiroya]

i'm answering this book that doesn't have answers at the back(cause usually it does) and i'm having a bit of difficulty and i wish somebody would help me. here it goes:

prove the ff:
1. sin2x+sin2y-sin2(x+y) = 4sinxsinysin(x+y)
i manage to turn the left side to sin2x(sin^2y+cos^2y)+ sin2y(sin^2x+cos^2y) - sin2(x+y)
2. sin2A+cos2A-cos^2A = cscA+cotA
2cosA-sinA-2cos^2A+sinAcosA
i'm a failure here, i can't even start it.

solve for x:
1. 4tan(x/2)sinx=4-sin2xsinx

another one of my failures. i managed to turn the left side to 4+4cossin^4-4sincos^4 but i think i'm not going anywhere if i solved it more.

2. arcsin(sqrt3/2) = arctan(sqrt(3x) - 2 / x) + arcos(-1/2)

i managed to turn this to arctan(sqrt(3x)-2 / x)= -phi/3

btw sqrt is square root of 3x.. and its square root of 3x only then minus 2 all over x



help would be much much appreciated. thanks in advance

 

[soroban]

Hello, hiroya!

All of these are pretty awful . . . but I got #2.

\(\displaystyle \L2)\;\frac{\sin2A\,+\,\cos2A\,-\,\cos^2A}{2\cdot\cos A\,-\,\sin A\,-\,2\cdot\cos^2A\,+\,\sin A\cdot\cos A} \:= \:\csc A\,+\,\cot A\)

The numerator is: \(\displaystyle \L\,(2\cdot\sin A\cdot\cos A)\,+\,(\cos^2A\,-\,\sin^2A)\,-\,\cos^2A\)

. . \(\displaystyle \L= \;2\cdot\sin A\cdot\cos A\,-\,\sin^2A\)

. . \(\displaystyle \L=\; \sin A(2\cdot\cos A \,-\,\sin A)\)


The denominator is: \(\displaystyle \L\,2\cdot\cos A \,-\,2\cdot\cos^2A \,-\,\sin A\,+\,\sin A\cdot\cos A\)

. . \(\displaystyle \L= \;2\cdot\cos A(1\,-\,\cos A)\,-\,\sin A(1\,-\,\cos A)\)

. . \(\displaystyle \L=\;(1\,-\,\cos A)(2\cdot\cos A\,-\,\sin A)\)


The fraction becomes: \(\displaystyle \L\:\frac{\sin A(2\cdot\cos A\,-\,\sin A)}{(1\,-\,\cos A)(2\cdot\cos A\,-\,\sin A)}\;=\;\frac{\sin A}{1\,-\,\cos A}\)


Multiply top and bottom by \(\displaystyle (1\,+\,\cos A):\)

.\(\displaystyle \L\frac{\sin A}{1\,-\,\cos A}\,\cdot\,\frac{1\,+\,\cos A}{1\,+\,\cos A} \;=\;\frac{\sin A(1\,+\,\cos A)}{1\,-\,\cos^2A}\;=\;\frac{\sin A(1\,+\,\cos A)}{\sin^2A} \;=\;\frac{1\,+\,\cos A}{\sin A}\)

Then we have: \(\displaystyle \L\:\frac{1}{\sin A}\,+\,\frac{\cos A}{\sin A} \;=\;\csc A\,+\,\cot A\)


Whew . . . I need a nap!

 

[hiroya]

thanks sir.
actually, our teacher might get some of the items from the book for our exam today. i wish the other one's won't appear. but i answered most of the items that were given. anyway thanks again and wish me luck