Math, asked by boby157, 1 year ago

sin1 sin2 sin3 ------sin90; ?​

Answers

Answered by nazeerahmedjan7863
0

Answer is 1/2^49/2 and its correct

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Answered by jassimahi8967
1

Answer:

0.587

Step-by-step explanation:

sin(1) sin(2) sin(3).....sin(89) sin(90)

Using the trigonometric product to sum formula,

[\frac{1}{2} (cos(1-89)-cos(1+89))+\frac{1}{2} (cos(2-88)-cos(2+88))+\frac{1}{2} (cos(3-87)-cos(3+87))+.....+\frac{1}{2} (cos(45-45)-cos(45+45))]sin(90)

We know, sin(90)=1

Putting the value, we get

\frac{1}{2} (cos(1-89)-cos(1+89))+\frac{1}{2} (cos(2-88)-cos(2+88))+\frac{1}{2} (cos(3-87)-cos(3+87))+.....+\frac{1}{2} (cos(45-45)-cos(45+45))

=\frac{1}{2} (cos(-88)-cos(90))+\frac{1}{2} (cos(-86)-cos(90))+\frac{1}{2} (cos(-84)-cos(90))+.....+\frac{1}{2} (cos(0)-cos(90))

Since cos(-x) = cos x,

=\frac{1}{2} (cos(88)-cos(90))+\frac{1}{2} (cos(86)-cos(90))+\frac{1}{2} (cos(84)-cos(90))+.....+\frac{1}{2} (cos(0)-cos(90))

Putting value of cos90 = 0,

We get,

=\frac{1}{2} (cos(88))+\frac{1}{2} (cos(86))+\frac{1}{2} (cos(84))+.....+\frac{1}{2} (cos(0))

Taking \frac{1}{2} common,

=\frac{1}{2} [(cos(88))+(cos(86))+(cos(84))+.....+(cos(0))]

Converting to sin,

=\frac{1}{2} [(cos(2))+(cos(4))+.....+sin(4)+sin(2)+(cos(0))]

Further converting,

we get,

\frac{\sqrt{10-2\sqrt{5} } }{4}

= \frac{2.35}{4} = 0.587

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