Cos-sin-sin-i cos-4 equalsA- sin 8 -i cos 8 -B- cos 8 -isin 8 -C- sin 8 -i cos 8 -D- cos 8 -isin 8 -

The correct option is D cos8θ + isin8θ (cosθ + isinθ/sinθ+icosθ)^4 =(cosθ + isinθ)^4/i^4(cosθ isinθ)^4=cos4θ + isin4θ/cos4θ isin4θ =(cos4θ + isin4θ)(cos4θ ...

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Byju's Answer

Standard XI

Mathematics

Period of Trigonometric Ratios

cosθ+sinθ/sin...

Question

${(\frac{c o s θ + i s i n θ}{s i n θ + i c o s θ})}^{4}$ equals

sin8 $θ$ - icos8 $θ$

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cos8 $θ$ + isin8 $θ$

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sin8 $θ$ + icos8 $θ$

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cos8 $θ$ + isin8 $θ$

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Solution

The correct option is D
cos8 $θ$ + isin8 $θ$

${(\frac{c o s θ + i s i n θ}{s i n θ + i c o s θ})}^{4}$ = $\frac{{(c o s θ + i s i n θ)}^{4}}{{i^{4} (c o s θ - i s i n θ)}^{4}}$ = $\frac{c o s 4 θ + i s i n 4 θ}{c o s 4 θ - i s i n 4 θ}$

= $\frac{(c o s 4 θ + i s i n 4 θ) (c o s 4 θ + i s i n 4 θ)}{(c o s 4 θ - i s i n 4 θ) (c o s 4 θ + i s i n 4 θ)}$

= $\frac{{(c o s 4 θ + i s i n 4 θ)}^{2}}{c o s^{2} 4 θ + s i n^{2} 4 θ}$ =cos8 $θ$ + isin8 $θ$

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(cosθ+isinθsinθ+icosθ)4 equals

The correct option is D cos8θ + isin8θ (cosθ+isinθsinθ+icosθ)4 =(cosθ+isinθ)4i4(cosθ−isinθ)4=cos4θ+isin4θcos4θ−isin4θ =(cos4θ+isin4θ)(cos4θ+isin4θ)(cos4θ−isin4θ)(cos4θ+isin4θ) =(cos4θ+isin4θ)2cos24θ+sin24θ=cos8θ + isin8θ

${(\frac{c o s θ + i s i n θ}{s i n θ + i c o s θ})}^{4}$ equals