1. **State the problem:** Express $$\frac{3 \left(e^{j400} + e^{-2j00}\right)}{e^{j00}}$$ in terms of cosine only. 2. **Recall Euler's formula:** $$e^{j\theta} = \cos \theta + j \sin \theta$$ and $$e^{-j\theta} = \cos \theta - j \sin \theta$$ Adding these gives: $$e^{j\theta} + e^{-j\theta} = 2 \cos \theta$$ 3. **Simplify the expression:** Rewrite the numerator: $$3 \left(e^{j400} + e^{-2j00}\right) = 3 \left(e^{j400} + e^{-j400}\right)$$ (since $$-2j00 = -j400$$) 4. **Divide by the denominator:** $$\frac{3 \left(e^{j400} + e^{-j400}\right)}{e^{j00}} = 3 \left(e^{j400} + e^{-j400}\right) e^{-j00}$$ 5. **Use Euler's formula to express in cosine:** $$e^{j400} + e^{-j400} = 2 \cos 400$$ 6. **Multiply by $$e^{-j00}$$:** $$3 \times 2 \cos 400 \times e^{-j00} = 6 \cos 400 \times e^{-j00}$$ 7. **Express $$e^{-j00}$$ in terms of cosine and sine:** $$e^{-j00} = \cos 00 - j \sin 00$$ 8. **Final expression:** $$6 \cos 400 \left(\cos 00 - j \sin 00\right) = 6 \cos 400 \cos 00 - 6j \cos 400 \sin 00$$ Since the problem asks for expression in terms of cosine only, the real part is: $$6 \cos 400 \cos 00$$ **Answer:** $$\boxed{6 \cos 400 \cos 00}$$ School of Maths & Science, Singapore Polytechnic.