Question

Let sin t = a​, cos t = b​, and tan t = c. Write the expression in terms of​ a, b, and c.

Answer 1

Answer:

$a+b-c$

*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:

$\sin(a-b)=\sin(a)\cos(b)-\sin(b)\cos(a)$

$\cos(a-b)=\cos(a)\cos(b)+\sin(a)\sin(b)$

$\tan(a-b)=\frac{\tan(a)-\tan(b)}{1+\tan(a)\tan(b)}$

Let's apply the difference identities to all three terms:

$-[\sin(-t)\cos(8\pi)+\cos(-t)\sin(8\pi)]+[\cos(-t)\cos(2\pi)+\sin(-t)\sin(2\pi)]+\frac{\tan(-t)-\tan(5\pi)}{1+\tan(-t)\tan(5\pi)}$

We are about to use that cos(even*pi) is 1 and sin(even*pi) is 0 so tan(odd*pi)=0:

$-[\sin(-t)(1)+\cos(-t)(0)]+[\cos(-t)(1)+\sin(-t)(0)]+\frac{\tan(-t)-0}{1+\tan(-t)(0)$

Cleaning up the algebra:

$-[\sin(-t)]+[\cos(-t)]+\frac{\tan(-t)}{1}$

Cleaning up more algebra:

$-\sin(-t)+\cos(-t)+\tan(-t)$

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):

$\sin(t)+\cos(t)-\tan(t)$

Making the substitution the problem wanted us to:

$a+b-c$

Just for fun you could have wrote c as a/b too since tangent=sine/cosine.