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Sum of Trigonometric Ratios in Terms of Their Product

If T n = s...

Question

If $T_{n} = {sin}^{n} θ + {cos}^{n} θ$ , prove that
(i) $2 T_{6} - 3 T_{4} + 1 = 0$
(ii) $6 T_{10} - 15 T_{8} + 10 T_{6} - 1 = 0$

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Solution

$L H S = 2 (s i n^{6} θ + c o s^{6} θ) - 3 (s i n^{4} θ + c o s^{4} θ) + 1 = 2 [(s i n^{2} θ)^{3} + (c o s^{2} θ)^{3}] - 3 [(s i n^{2} θ)^{2} + (c o s^{2} θ)^{2}] + 1 = 2 [(s i n^{2} θ + c o s^{2} θ)^{3} - 3 s i n^{2} θ {cos}^{2} θ (s i n^{2} θ + c o s^{2} θ)] - 3 [(s i n^{2} θ + c o s^{2} θ)^{2} - 2 s i n^{2} θ {cos}^{2} θ] + 1 = 2 (1 - 3 s i n^{2} θ c o s^{2} θ) - 3 (1 - 2 s i n^{θ} c o s^{2} θ) + 1 = 2 - 6 s i n^{2} θ c o s^{2} θ - 3 + 6 s i n^{2} θ c o s^{2} θ + 1 = 0 = R H S$

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Similar questions

If $T_{n} = s i n^{n} θ + c o s^{n} θ$ , prove that

$(i) \frac{T_{3} - T_{5}}{T_{1}} = \frac{T_{5} - T_{7}}{T_{3}}$

$(i i) 2 T_{6} - 3 T_{4} + 1 = 0$

$(i i i) 6 T_{10} - 15 T_{8} + 10 T_{6} - 1 = 0$

T_{n} = \sin^{n} θ + \cos^{n} θ

, prove that

(i)

\frac{T_{3} - T_{5}}{T_{1}} = \frac{T_{5} - T_{7}}{T_{3}}

2 T_{6} - 3 T_{4} + 1 = 0

6 T_{10} - 15 T_{8} + 10 T_{6} - 1 = 0

T_{n} = \sin^{n} x + \cos^{n} x

, prove that

(i)

\frac{T_{3} - T_{5}}{T_{1}} = \frac{T_{5} - T_{7}}{T_{3}}

2 T_{6} - 3 T_{4} + 1 = 0

6 T_{10} - 15 T_{8} + 10 T_{6} - 1 = 0

T_{n} = s i n^{n} θ + c o s^{n} θ

(i) \frac{T_{3} - T_{5}}{T_{1}} (i i) 2 T_{6} - 3 T_{4} + 1 = 0

T_{n} = {sin}^{n} x + {cos}^{n} x

, then find the value of

6 T_{10} - 15 T_{8} + 10 T_{6}

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Sum of Trigonometric Ratios in Terms of Their Product

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