Question

The value of (sin square 1° + sin square 3°+ sin square 5°+...+ sin square 85 °+ sin square 87°+ sin square 89° )is

Answer 1

Answer:

Step-by-step explanation:

Value=N/2

N=last term-first term/difference of the terms+1

=89-1/3-1+1

=88/2+1

=45

(Sin square 1°+sin square 89°)+(sin square 3°+sin square 87°)+...........

1+1+1..........45 terms

Ans=N/2

=45/2

Answer 2

Answer:

The value of the given sin series expansion is 45/2.  

Step-by-step explanation:

Given the sin series expansion

$sin^2 1^o + sin^2 3^o + sin^2 5^o +...+ sin ^2 85^o + sin ^2 87^o + sin ^2 89^o$

This can be written as

$sin^2 1^o + sin^2 3^o + sin^2 5^o +...+ sin ^2 (90^o-5^o) + sin ^2 (90^o-3^o) + sin ^2 (90^o-1^o)$

Since $sin(90-\theta)=cos\theta$, the above expansion can be written as

$sin^2 1^o + sin^2 3^o + sin^2 5^o +...+ cos ^2 5^o + cos ^2 3^o + cos ^2 1^o$

Rewriting the expression by grouping with the same magnitude of angles,

$(sin^2 1^o + cos ^2 1^o)+ (sin^2 3^o + cos ^2 3^o) +( sin^2 5^o + cos ^2 5^o) +...$

The expansion goes up to 45° and since the odd angles are given, the last angle will of 43°.

Thus we have the complete expansion as

$(sin^2 1^o + cos ^2 1^o)+ (sin^2 3^o + cos ^2 3^o) +( sin^2 5^o + cos ^2 5^o) +...( sin^2 43^o + cos ^2 43^o) + sin^2 45^o$

Using the trigonometric relation, $sin^{2} \theta+cos^{2} \theta=1$, we get

$1+ 1+1 +...1 + \bigg(\dfrac{1}{\sqrt{2} }\bigg)^2$

1 is repeated upto 22 terms, therefore

$1\times22 + \dfrac{1}{{2} }=22+\dfrac{1}{{2} }=\dfrac{45}{{2} }$

Therefore, the value of the given sin series expansion is 45/2.