time scaling and shifting of cosine in Fourier transform I've met some problems when calculating the Fourier transform of $\cos(at+b)$. I want to use the shifting and scaling properties to solve this problem. First, when I look up in the book and some online resources, I saw a post about the Fourier transform of the signal $x(3t-6)$, where it tells me that $x(3t-6)$ is scaling $x(t)$ by $3$ and shift it by $2$, which I will get the answer that
$$F\{x(3t-6)\} = \frac13 X(j\omega/3)exp(-j2\omega)$$
This indicates that the Fourier transform of $x(at+b)$ is $$\frac{1}{|a|}X(j\omega/a)\cdot \exp(j\omega b/a)$$

Thus, I apply the conclusion to the Fourier transform of $cos(at+b)$ is 

$$\frac{1}{|a|}\exp(j\omega b/a)\,\pi\, [\delta(\omega/a−1)+\delta(\omega/a+1)]$$

How do I proceed to the right answer which is

$$\pi [\exp(jb)\delta(\omega-b)+\exp(-jb)\delta(\omega+b)]$$ ? I appreciate your help.

(I'm new here so I haven't figure out how to type the fractions or exponentials, sorry for the bad typing)
Using the shifting and scaling properties of the Fourier transform is a rather complicated way of computing the Fourier transform of $\cos(at+b)$. A more straightforward way is to realize that

$$\cos(at+b)=\frac12\left[e^{jat}e^{jb}+e^{-jat}e^{-jb}\right]\tag{1}$$

From the Fourier transform pair

$$e^{jat}\Longleftrightarrow 2\pi\delta(\omega-a)\tag{2}$$

we immediately obtain

$$\cos(at+b)\Longleftrightarrow \pi\left[e^{jb}\delta(\omega-a)+e^{-jb}\delta(\omega+a)\right]\tag{3}$$

But having said that, your result is correct. You just need to simplify it appropriately. In order to do so, you need to know two things. First,

$$\delta\left(\frac{\omega}{a}\right)=|a|\delta(\omega)\tag{4}$$

and, second,

$$f(\omega)\delta(\omega-\omega_0)=f(\omega_0)\delta(\omega-\omega_0)\tag{5}$$

for any $f(\omega)$ that is continuous at $\omega_0$.

From $(4)$ you get

$$\delta(\omega/a-1)=|a|\delta(\omega-a)\tag{6}$$

And from $(5)$ you get

$$e^{j\omega b/a}\delta(\omega-a)=e^{jb}\delta(\omega-a)\tag{7}$$

Applying similar transformations to the second term in your result leads to the form given in $(3)$.