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I 28 OPGAVER TIL LØSNING.
5. sin (.r-y) sin (y - z)sin (z- or)4-sin (x-j)sin(y4-s)sin(s4-,r}
+ sin (x+y) sin (y - z) sin (z + x) + sin(x + y)sin (y+ z)sin(z~ x)
= sin (x -y] [sin (_y - s) sin (s - x) + sin (j 4- z) sin (s + .r)
- sin (x+y) sin (s-j-.v) cos (z - x) - sin (x+y) sin (5-x) cos (s 4-.r)]
~ sin (x - 7) sin (s - x) [sin (j - z) - sin ("r + 7) cos (5 4- ,r)
4- sin (s+x)cos(x+y)] = sm(x-t-y)sm(z-x)[sin(y-z) + sin(s-y)] = or
hvormed Formlen er bevist.
6. sin,rsin_ysin 2 (^4"J;)sm (x-y)+sinysmzsin 2(y + z)$m(y-z}
4~ sin z sin .r sin 2 (z + x} sin (z - x)
= | [sin 2 (x-y) sin 2 (x-\-y)-\-(co$(3x-\-y) - cos (^4-3>’))cos(^+7>
4- sin 2 (j - 5) sin 2(y+z) + (cos (3 j + s) - cos (y + 3*)) cos (>. 4- ø)
4- sin 2 (s-.r) sin 2 (z-\- x) + (cos(3^4-x) - cos (^4-3^)) cos (z-\-x)\
- J [COS (4-v4- 2j) 4- COS 2 "r–- COS (2X + 4^) –- COS 2V
4- cos (47 4~ 2z) 4- cos 27 - cos (2_y 4" 4s) - cos 2Z
4- cos (42 + 2A-) 4~ cos 25 - cos (25 4~ 4-r) - cos 2^r]
– - |[s’n (2X+ 2y-\- 2z) sin 2 (.r-s) -f sin (2jr+ 2j’4~ 2s)sin 2(y - A-)
4- sin (2JT4- 2j 4-2s) sin 2 (s - >’)] = - ^[sin (y-z) cos (2x-z–y)
4- sin (^ –_y) cos (z-y}} sin 2S = - sin (x-y) sin (y-z) sin (z-.r) sin 2^
hvormed Formlen er bevist,
7. 2 cos ,vcos y sos z cos ^ - cos (s 4~ .*") cos (s + 7) cos (^ 4- 5)
- cos (s .- x) cos (5- - y) cos (5 - z] = -^ cos s [cos (.r 4" J; + 5)
4- cos (x 4- 7 - s) 4- cos (.r -f s -y) 4- cos (,r - y - z}}
- ^ cos (s-\-z) [cos (2S-\-x-\-y) -j- cos(x-y)] - .^cos(s-z)[cos(2S-x-y)
4- COS (X––_>’)] = 5 [COS 25+ I 4~ COS (2X-\- 2y] 4~ COS 2S-f-COS (2,v4- 25)
4- cos 2_y 4~ cos 2.r -f cos (27 4" 2’8:) - cos 45 - cos (2x 4- 2y]
- COS (2X-{- 2Z]––COS (254- 2y)–-COS 2S - COS25––COS 2.1-COS 2y\
- i[i - cos 45] = l- sin2 25,
hvormed Formlen er bevist.
8. cos2 .r + cos2 j 4- cos2 z -\- 2 cos "vcosjy cos z - i
- (cos x 4~ cos (y 4~ z}) (cos ,r 4- cos (y - z))
., . x - y~z x + y - z x-j
4 cos–––––––cos–––––––cos–––––––cos––––
2222
S
4 cos cos ( -
(s __ \ (j__ \ (
’ (2 *)cos ^ 2 y jcos \
2
hvormed Formlen er bevist. N. Munk.
Opgaver til Løsning.
69. Naar A 4- B + C= TI , skal følgende Størrelser bringes
paa logaritmisk Form:
sin 3A sin ^B sin ^C
~’~~~s^\ i rm""/"/^ /f\ ’
sm(B-C) ’ s’m(C-A) ’ sin (.4 -5)
og
tgj^______tg^s^ r tp: 3C
-) /^\ 1 i. / /^«
tg(^- C) ’ tg(C-^4) ’ tg(
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