Hi everybody, I know, we are on a ASM forum but It's the end of holliday (here in France, in the south) but for Tuesday, I have to do a lot of homework and I 've not so much time to do it. It's in Mathematics : Does someone know how I can solve the following: cos²(x+pi/3)-sin²(2x+pi/4)=0 Thankx very much and sorry to ask this question here Vom-bonjour:-() PS:I did not manage to put the pi symbol.
Posted on 2001-02-16 11:03:00 by Vom-bonjour:-()
what you have there is the difference of two perfect squares. x^2 -y^2 = 0 (x-y)(x+y)=0 x is the cos term and y is the sin term. rewrite cos(a) as sin(90-a) and solve x-y =0 and x+y =0. For x-y=0, you will get sin(c) - sin(d) = 0 c=d ---> solve for x. For x+y=0, you will get sin(c)+sin(d) = 0 therefore, c= 180-d . solve for x. done!
Posted on 2001-02-16 11:22:00 by Hel
Posted on 2001-02-16 12:03:00 by JimmyClif
Thankx very much, i didn't think to a²-b²=... Thankx Vom-bonjour:-()
Posted on 2001-02-16 12:04:00 by Vom-bonjour:-()
Hel, there are actually four solutions. sin(c) = sin(d) implies c = d OR c = 180 - d. Also, sin(c) = - sin(d) implies c = 180 + d (not 180 - d) OR c = -d. Here c = 30 - x, d = 45 + 2x so there are four solutions for x.
Posted on 2001-02-20 16:09:00 by Xmas