4 תשובות
ירדן קצת התבלבלה עם הזהויות הטריגונומטריות.
הזהות שהיא השתמשה בה זה לחיבור סינוסים, לא לחיבור סינוס עם קוסינוס. בקרוב אעלה פתרון דומה אך עם המרה של קוסינוס לסינוס כדי להשתמש באותה זהות
הזהות שהיא השתמשה בה זה לחיבור סינוסים, לא לחיבור סינוס עם קוסינוס. בקרוב אעלה פתרון דומה אך עם המרה של קוסינוס לסינוס כדי להשתמש באותה זהות
בקיצור עושים מה שירדן בערך עשתה רק שהופכים קודם את cosx ל- sin (pi/2-x) כדי שיהיה אפשר להשתמש בזהויות ואז זה עובד:)
עדיין לא חברים בעט?
you are wrong. there's no such a trignometric identity which says that
sin(a)+cos(b)=sin((a+b)/2)cos((a-b)/2)
this trigonometric identity is true for
sin(a)+sin(b),
not for sin(a)+cos(b).
there's no identity for sin(a)+cos(b) in trigonometry.
in order to solve this equation,
we have to use the identity of the double angle is sin and cos functions.
we can rewrite sin(4x) as sin(2*2x):
sin(2*2x)+cos(x)=0
and then we can use the following identity:
sin(2a)=2sin(a)cos(a)
for sin(2*2x):
sin(2*2x)=2sin(2x)cos(2x)
||
v
2sin(2x)cos(2x)+cos(x)=0
now we'll use the identity of sin(2a) for sin(2x) in our equation:
i 2*2sin(x)cos(x)cos(2x)+cos(x)=0
now we can factor out cos(x) as a common factor:
cos(x)[2*2sin(x)*cos(2x)+1]=0
cos(x)[4sin(x)cos(2x)+1]=0
and now we'll put each factor from the product as being 0:
cos(x)=0
x=pi/2+pi*k
4sin(x)cos(2x)+1=0
we'll use the identity for the double angle in cos function:
cos(2x)=1-2sin^2(x)
(we'll prefer to take the option with sin(x), for the equation includes sin(x) in it and we want to have a single function - without a mix of sin and cos
4sin(x)[1-2sin^2(x)]+1=0
4sin(x)-8sin^3(x)+1=0
for convenience purposes, we'll set sin(x) as being t:
4t-8t^3+1=0
and now we'll solve a third-degree equation:
8t^3+4t+1=0-
we'll divide both sides by (-1):
8t^3-4t-1=0
as for this equation, we'll use desmos to solve it and we'll get three solutions:
t=-0.5, t=-0.309017, t=0.80017
and now we'll rewrite t as being sin(x) again:
sin(x)=-0.5
x=-pi/6+2pi*k
x=7pi/6+2pi*k
same for the rest 2 solutions.
i hope that it was clear.
you are wrong. there's no such a trignometric identity which says that
sin(a)+cos(b)=sin((a+b)/2)cos((a-b)/2)
this trigonometric identity is true for
sin(a)+sin(b),
not for sin(a)+cos(b).
there's no identity for sin(a)+cos(b) in trigonometry.
in order to solve this equation,
we have to use the identity of the double angle is sin and cos functions.
we can rewrite sin(4x) as sin(2*2x):
sin(2*2x)+cos(x)=0
and then we can use the following identity:
sin(2a)=2sin(a)cos(a)
for sin(2*2x):
sin(2*2x)=2sin(2x)cos(2x)
||
v
2sin(2x)cos(2x)+cos(x)=0
now we'll use the identity of sin(2a) for sin(2x) in our equation:
i 2*2sin(x)cos(x)cos(2x)+cos(x)=0
now we can factor out cos(x) as a common factor:
cos(x)[2*2sin(x)*cos(2x)+1]=0
cos(x)[4sin(x)cos(2x)+1]=0
and now we'll put each factor from the product as being 0:
cos(x)=0
x=pi/2+pi*k
4sin(x)cos(2x)+1=0
we'll use the identity for the double angle in cos function:
cos(2x)=1-2sin^2(x)
(we'll prefer to take the option with sin(x), for the equation includes sin(x) in it and we want to have a single function - without a mix of sin and cos
4sin(x)[1-2sin^2(x)]+1=0
4sin(x)-8sin^3(x)+1=0
for convenience purposes, we'll set sin(x) as being t:
4t-8t^3+1=0
and now we'll solve a third-degree equation:
8t^3+4t+1=0-
we'll divide both sides by (-1):
8t^3-4t-1=0
as for this equation, we'll use desmos to solve it and we'll get three solutions:
t=-0.5, t=-0.309017, t=0.80017
and now we'll rewrite t as being sin(x) again:
sin(x)=-0.5
x=-pi/6+2pi*k
x=7pi/6+2pi*k
same for the rest 2 solutions.
i hope that it was clear.
2 sin ( (4x+2x)/2)*cos ( (4x-2x)/2)=0
2 sin 3x * cos x=0
we'll put each factor from the product as being 0.
sin 3x=0
this is an elementary equation:
3x = (-1)^k*arcsin 0 + k*pi
3x= (-1)^k*0+k*pi
x=k*pi/3, where k is an integer number.
we'll solve the second elementary equation:
cos x=0
x= +/-arccos 0 + 2*k*pi
x=+/- pi/2 + 2*k*pi
the set of solutions:
באותו הנושא: