Sin 90-cos 0-sin 88-cos 2-sin 86-cos 4- -sin 2-cos 88-sin 0-cos 90-

The correct option is C 23 sin90^∘ . cos0^∘ + sin88^∘ . cos2^∘ + ..........+ sin2^∘ . cos88^∘ + sin0^∘ . cos90^∘ = cos(90^∘ 90^∘). cos0^∘ + cos(90^∘ 88^∘ ...

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Byju's Answer

Standard X

Mathematics

Trigonometric Identities

sin 90∘·cos 0...

Question

$sin 90^{\circ} . cos 0^{\circ} + sin 88^{\circ} . cos 2^{\circ} + sin 86^{\circ} . cos 4^{\circ} + . . . . . . . . . . + sin 2^{\circ} . cos 88^{\circ} + sin 0^{\circ} . cos 90^{\circ}$

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Solution

The correct option is D
23

$sin 90^{\circ} . cos 0^{\circ} + sin 88^{\circ} . cos 2^{\circ} + . . . . . . . . . . + sin 2^{\circ} . cos 88^{\circ} + sin 0^{\circ} . cos 90^{\circ} = cos (90^{\circ} - 90^{\circ}) . cos 0^{\circ} + cos (90^{\circ} - 88^{\circ}) . cos 2^{\circ} + . . . . . . . . + cos (90^{\circ} - 46^{\circ}) . cos (44^{\circ}) + sin (44^{\circ}) . sin (90^{\circ} - 46^{\circ}) + sin (42^{\circ}) . sin (90^{\circ} - 48^{\circ}) + . . . . . . . . . . . . + sin 0^{\circ} . sin (90^{\circ} - 90^{\circ}) = cos 0^{\circ} . cos 0^{\circ} + cos 2^{\circ} . cos 2^{\circ} + . . . . . . . . . . . . + cos 44^{\circ} . cos 44^{\circ} + sin 44^{\circ} . sin 44^{\circ} + . . . . . . . . . . + sin 2^{\circ} . sin 2^{\circ} + sin 0^{\circ} . sin 0^{\circ} = {cos}^{2} 0^{\circ} + {cos}^{2} 2^{\circ} + . . . . . . . . . . . . + {cos}^{2} 44^{\circ} + {sin}^{2} 44^{\circ} + . . . . . . . . . . . . + {sin}^{2} 2^{\circ} + {sin}^{2} 0^{\circ} = 1 + 1 + 1 + . . . . . . . . . + 1 (23 times) = 23$

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Similar questions

s i n 90^{\circ} c o s 0^{\circ} + s i n 88^{\circ} c o s 2^{\circ} + s i n 86^{\circ} c o s 4^{\circ} + . . . . . . . . . . . . . . + s i n 2^{\circ} c o s 88^{\circ} + s i n 0^{\circ} c o s 90^{\circ}

Q. What is the value of

\frac{sin 2^{\circ} sin 4^{\circ} sin 6^{\circ} . . . . . sin 88^{\circ}}{cos 88^{\circ} cos 86^{\circ} cos 84^{\circ} . . . . . cos 2^{\circ}}

(Do not use trigonometric tables)

Evaluate

\begin{matrix} {cos}^{4} \frac{π}{16} + {sin}^{4} \frac{π}{16} - {sin}^{2} \frac{π}{16} {cos}^{2} \frac{π}{16} + {cos}^{4} \frac{3 π}{16} + {sin}^{4} \frac{3 π}{16} - {sin}^{2} \frac{3 π}{16} {cos}^{2} \frac{3 π}{16} \end{matrix}

Q. If

\frac{{cos}^{4} x}{{cos}^{2} y} + \frac{{sin}^{4} x}{{sin}^{2} y} = 1

, then

\frac{{cos}^{4} y}{{cos}^{2} x} + \frac{{sin}^{4} y}{{sin}^{2} x} = ?

Q. If

\frac{{cos}^{4} x}{{cos}^{2} y} + \frac{{sin}^{4} x}{{sin}^{2} y} = 1

then show that

\frac{{cos}^{4} y}{{cos}^{2} x} + \frac{{sin}^{4} y}{{sin}^{2} x} = 1

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sin90∘.cos0∘+sin88∘.cos2∘+sin86∘.cos4∘+..........+sin2∘.cos88∘+sin0∘.cos90∘

$sin 90^{\circ} . cos 0^{\circ} + sin 88^{\circ} . cos 2^{\circ} + sin 86^{\circ} . cos 4^{\circ} + . . . . . . . . . . + sin 2^{\circ} . cos 88^{\circ} + sin 0^{\circ} . cos 90^{\circ}$