Fourier transform of function is:
"\\hat{f}(k) = \\frac{1}{\\sqrt{2\\pi}}\\int_{-\\infty}^{\\infty} dx\\, e^{ikx}f(x)"
and integral representation of Dirac delta-function is
"\\delta(x)=\\frac{1}{2\\pi}\\int_{-\\infty}^{\\infty}e^{ikx}\\, dk"
a)
"f(x)= \\cos(x)\\\\\n\\hat{f}(k) = \\frac{1}{\\sqrt{2\\pi}}\\int_{-\\infty}^{\\infty} dx\\, e^{ikx}\\cos(x) = \\frac{1}{\\sqrt{2\\pi}}\\int_{-\\infty}^{\\infty}dx\\, e^{ikx}\\frac{e^{ix}+e^{-ix}}{2}=\\\\\n\\sqrt{\\frac{\\pi}{2}}\\delta(k-1)+\\sqrt{\\frac{\\pi}{2}}\\delta(k+1)"
b)
"f(x)= \\sin(x)\\\\\n\\hat{f}(k) = \\frac{1}{\\sqrt{2\\pi}}\\int_{-\\infty}^{\\infty} dx\\, e^{ikx}\\sin(x) = \\frac{1}{\\sqrt{2\\pi}}\\int_{-\\infty}^{\\infty}dx\\, e^{ikx}\\frac{e^{ix}-e^{-ix}}{2i}=\\\\\ni\\sqrt{\\frac{\\pi}{2}}\\delta(k-1)-i\\sqrt{\\frac{\\pi}{2}}\\delta(k+1)"
c)
"f(x)= e^{iax}\\\\\n\\hat{f}(k) = \\frac{1}{\\sqrt{2\\pi}}\\int_{-\\infty}^{\\infty} dx\\, e^{ikx}e^{iax} =\\\\\n =\\frac{1}{\\sqrt{2\\pi}}\\int_{-\\infty}^{\\infty} dx\\, e^{i(k+a)x} = \\sqrt{2\\pi}\\delta(k+a)"
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