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Use De Moivre's Theorem to calculate cos(t)⁷ in terms of cos(2t),sin(3t) etc.

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Final answer:

To express cos(t)⁷ in terms of functions like cos(2t) and sin(3t), De Moivre's Theorem is utilized by expanding (cos(t) + i sin(t))⁷ and equating the real parts, then applying trigonometric identities to simplify the terms.

Step-by-step explanation:

To calculate cos(t)⁵ using De Moivre's Theorem, we first recognize that De Moivre's Theorem states that (cos(t) + i sin(t))n = cos(nt) + i sin(nt), where i is the imaginary unit and n is an integer. For n=7, this becomes (cos(t) + i sin(t))7 = cos(7t) + i sin(7t). We need to expand the left side using the binomial theorem and equate real parts to find the expression for cos(t)⁵ in terms of other cosines and sines of multiples of t.

The binomial expansion will involve terms like cos(t) raised to various powers multiplied by i sin(t) raised to various powers. After expansion, we can collect all the real terms which will give us the expression for cos(t)⁵. Using known trigonometric identities like cos(2t) = 2cos2(t) - 1 and sin(2t) = 2sin(t)cos(t), we may express higher powers of cosine in terms of these functions.

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