CBSE BOARD X, asked by vrindashageela6012, 10 months ago

Sin90×cos0+ sin88×cos 2+sin 86×cos 4+......+sin2×cos88+ sin0×cos90

Answers

Answered by gunjan4321
1

Explanation:

= Sin 1° + sin 2° + sin 3° ... + sin 88° + sin 89°

Multiply both sides with 2 sin 1°.

2 Y sin 1° = 2 sin 1° sin 1° + 2 sin 2° sin 1° + 2 sin 3 sin 1°+ ...

+ 2 sin 88° sin 1° + 2 sin 89 sin 1°

= cos 0° - cos 2° + cos 1° - cos 3° + cos 2° - cos 4° + cos 3° - cos 5...

... + cos 86 - cos 88° + cos 87° - cos 89° + cos 88° - cos 90°

= 1 + cos 1° - cos 89°

Y = [1 + cos 1° - sin 1°] / (2 sin 1°) = 1/2 * [ cot 1/2° - 1]

==============

we can do this by using complex numbers using De Moivre's formula:

Let 1° = 1*π/180 rad = a

\begin{lgathered}Y=Imaginary\ part\ of\ \Sigma_a^{89a}{e^{ia}}\\\\=Im[ \frac{e^{ia}(1-e^{89a})}{1-e^{ia}} ]\\\\=Im[ \frac{e^{ia}-e^{i90a}}{1-e^{ia}}]=IM[ \frac{cos\ a-i\ (1-sin\ a)}{1-cos\ a + i\ sin a} ]\\\\=Im[ \frac{cos\ a + sin\ a -1}{2(1-cos\ a)}*(1 +i) ]\\\\=\frac{1}{2}(cot\ 1^o-1)\end{lgathered}

Y=Imaginary part of Σ

a

89a

e

ia

=Im[

1−e

ia

e

ia

(1−e

89a

)

]

=Im[

1−e

ia

e

ia

−e

i90a

]=IM[

1−cos a+i sina

cos a−i (1−sin a)

]

=Im[

2(1−cos a)

cos a+sin a−1

∗(1+i)]

=

2

1

(cot 1

o

−1)

Similar questions