Respuesta :
Answer:
[tex]a+b-c[/tex]
*Note c could be written as a/b
Step-by-step explanation:
-sin(-t - 8 π) + cos(-t - 2 π) + tan(-t - 5 π)
The identities I'm about to apply:
[tex]\sin(a-b)=\sin(a)\cos(b)-\sin(b)\cos(a)[/tex]
[tex]\cos(a-b)=\cos(a)\cos(b)+\sin(a)\sin(b)[/tex]
[tex]\tan(a-b)=\frac{\tan(a)-\tan(b)}{1+\tan(a)\tan(b)}[/tex]
Let's apply the difference identities to all three terms:
[tex]-[\sin(-t)\cos(8\pi)+\cos(-t)\sin(8\pi)]+[\cos(-t)\cos(2\pi)+\sin(-t)\sin(2\pi)]+\frac{\tan(-t)-\tan(5\pi)}{1+\tan(-t)\tan(5\pi)}[/tex]
We are about to use that cos(even*pi) is 1 and sin(even*pi) is 0 so tan(odd*pi)=0:
[tex]-[\sin(-t)(1)+\cos(-t)(0)]+[\cos(-t)(1)+\sin(-t)(0)]+\frac{\tan(-t)-0}{1+\tan(-t)(0)[/tex]
Cleaning up the algebra:
[tex]-[\sin(-t)]+[\cos(-t)]+\frac{\tan(-t)}{1}[/tex]
Cleaning up more algebra:
[tex]-\sin(-t)+\cos(-t)+\tan(-t)[/tex]
Applying that sine and tangent is odd while cosine is even. That is,
sin(-x)=-sin(x) and tan(-x)=-tan(x) while cos(-x)=cos(x):
[tex]\sin(t)+\cos(t)-\tan(t)[/tex]
Making the substitution the problem wanted us to:
[tex]a+b-c[/tex]
Just for fun you could have wrote c as a/b too since tangent=sine/cosine.
