1. **State the problem:** Express $$3 \frac{e^{j400} + e^{-2j100}}{e^{j00}}$$ in terms of cosine only. 2. **Recall Euler's formula:** $$e^{j\theta} = \cos\theta + j\sin\theta$$ This means that expressions like $$e^{j\alpha} + e^{-j\alpha} = 2\cos\alpha$$. 3. **Simplify the denominator:** Since $$e^{j00} = e^{j0} = 1$$, the expression simplifies to: $$3 (e^{j400} + e^{-2j100})$$ 4. **Simplify the exponent in the second term:** Note that $$-2j100 = -j200$$, so the expression is: $$3 (e^{j400} + e^{-j200})$$ 5. **Rewrite the sum using cosine:** We want to express $$e^{j400} + e^{-j200}$$ in terms of cosine. However, the angles are different: 400 and 200. 6. **Check if angles can be related:** Note that $$e^{j400} = e^{j(200 + 200)} = e^{j200} e^{j200}$$, but this does not directly help. 7. **Use Euler's formula separately:** $$e^{j400} = \cos 400 + j \sin 400$$ $$e^{-j200} = \cos 200 - j \sin 200$$ 8. **Sum these:** $$e^{j400} + e^{-j200} = (\cos 400 + \cos 200) + j(\sin 400 - \sin 200)$$ 9. **Since the problem asks for expression in cosine only, the imaginary part must be zero or ignored.** 10. **Rewrite the original expression as:** $$3 (\cos 400 + \cos 200) + j 3 (\sin 400 - \sin 200)$$ 11. **If only the real part is desired (cosine terms), then the expression in terms of cosine only is:** $$3 (\cos 400 + \cos 200)$$ 12. **Final answer:** $$\boxed{3 (\cos 400 + \cos 200)}$$ This is the expression in terms of cosine only.