# Calculate sin^2 1+sin^2 2 +...+ sin^2 90.

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### 6 Answers

We have to find the sum of the series (sin 1)^2 + (sin 2)^2 +...+ (sin 90)^2

(sin 1)^2 + (sin 2)^2 +...+ (sin 90)^2

=> (sin 1)^2 + (sin 2)^2 +...+ (sin 90)^2

=> (sin 1)^2 + (sin 2)^2 +... + (sin 44)^2 + (sin 45)^2 + (sin 46)^2 ...+ (sin 90)^2

=> (sin 1)^2 + (sin 2)^2 +... + (sin 44)^2 + (sin 45)^2 + (sin 46)^2 ...+ (sin 90)^2

use sin (90 - x) = cos x

=> (sin 1)^2 + (sin 2)^2 +... + (sin 44)^2 + (sin 45)^2 + (sin(90 - 44))^2 ...+ (sin (90 - 0))^2

=> (sin 1)^2 + (sin 2)^2 +... + (sin 44)^2 + (sin 45)^2 + (cos 44)^2 + (cos 43)^2 ...+ (cos 0)^2

=> (sin 1)^2 + (cos 1)^2 + (sin 2)^2 + (cos 2)^2 +...+ (sin 44)^2 + (cos 44)^2 + (sin 45)^2 + (cos 0)^0

=> 44*1 + (1/sqrt 2)^2 + 1

=> 44 + 1/2 + 1

=> 45.5

**The required sum is 45.5**

We'll calculate this sum based on the followings:

- Pythagorean identity: (sin x)^2 + (cos x)^2 = 1

- sin x = cos (90 - x)

According to these, we'll have:

S = (sin 1)^2 + (sin 2)^2 + ... + (sin 45)^2 + (sin 46)^2 + .. + (sin 89)^2 + 1

S = (sin 1)^2 + (sin 2)^2 + ... + (sin 45)^2 + [cos (90-46)]^2 + ...+ [cos (90-89)]^2 + 1

S = (sin 1)^2 + (sin 2)^2 + ... (sin 44)^2+ (sin 45)^2 + (cos 44)^2 + ... + (cos 1)^2 + 1

We'll group the terms, according to Pythagorean identity;

S = [(sin 1)^2 + (cos 1)^2] + ... + [(sin 44)^2 + (cos 44)^2] + (sin 45)^2 + 1

S = 1 + 1 + ... + 1 + (sqrt2/2)^2 + 1

S = 45 + 1/2

S = 91/2

**The required sum of the squares of sine functions is S = 91/2.**

We need 2 formulas :

sin^2 a + cos^2 =1

sin a = cos (90 - a)

S=(sin1)^2 + (sin2)^2 + ...+(sin45)^2+(sin46)^2+...+(sin89)^2+1

S=(sin1)^2 + (sin2)^2 + ...+(sin45)^2+[cos(90-46)]^2+...+[cos(sin90-89)]^2+1

then we group two by two just like in formula , only sin^2 45 is alone :

S=(sin1)^2+(cos1)^2+..+(sin44)^2+(cos44)^2+(sqrt 2/2)^2+1

S=1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1/2

S=45+1/2 => S=91/2

I wrote in this way, to understand better

Several properties:

sin^2(x) + cos^2(x) = 1

cos^2(x) = sin^2(90-x)

Replacing the second term of the first expression with the second expression we get:

sin^2(x) + sin^2(90-x) = 1

Now we replace the terms in the original expression:

[sin^2(1) + sin^2(89)]+[sin^2(2) + sin^2(88)]+...+[sin^2(44) + sin^2 (46)] + sin^2(45) + sin^2(90)

=>1+1+1+...+1+(1/2)+1

=>45 + 1/2 <=> 91/2

**The sum is equal to 91/2**

idk

Let

(1) S = sin^2(1) + sin^2(2)+ ...+sin^2(88) + sin^2(89)

we know sin(k) = cos(90-k) for k = 1, 2, ..., 89, so we have

S = cos^2(89) + cos^2(88) + ... + cos^2(2) + cos^1(1).

reverse the order of summation, we have

(2) S = cos^1(1) + cos^2(2) + ... + cos^2(88) + cos^2(89)

Add (1) and (2) together;

**2S** = (sin^2(1) + cos^2(1)) + (sin^2(2) + cos^2(2)) + ...+ (sin^2(89) + cos^2(89))

= 1 + 1+ ... + 1 (the total is 89)

= **89**

so devide both sides by 2, S = 89/2 = 45.5.