Question

Find the value of sin5°+ sin?10° + sin15° + ... + sin 90°.​

Answer 1

Answer:

Firstly, question incorrect hai, square hogaya okk

Step-by-step explanation:

sin^2 5 + sin^2 10 + sin^2 15 + …… + sin^2 75 + sin ^2 80 +sin^2 85 + sin^2 90

(sin^2 5 + sin^2 85 )+( sin^2 10 + sin^2 80) +( sin^2 15 + sin^2 75) + ……..+ sin ^2 45 + sin^2 90

(sin^2 5 + cos^2 5 )+( sin^2 10 + cos^2 10) +( sin^2 15 + cos^2 15) + ……..+ sin ^2 45 + sin^2 90

(1 + 1 + 1 +1 + 1 + 1 +1 + 1 ) + 1/2 +1

19/2

Answer 2

$\sin^2 5+\sin^2 10+\sin^2 15+......+\sin^2 75+\sin ^2 80+\sin^2 85+\ sin^2 90=9.5$

Step-by-step explanation:

We have,

$\sin^2 5+\sin^2 10+\sin^2 15+......+\sin^2 75+\sin ^2 80+\sin^2 85+\ sin^2 90$

To find, the value of $\sin^2 5+\sin^2 10+\sin^2 15+......+\sin^2 75+\sin ^2 80+\sin^2 85+\ sin^2 90=?$

∴ $\sin^2 5+\sin^2 10+\sin^2 15+......+\sin^2 75+\sin ^2 80+\sin^2 85+\ sin^2 90$

$=\sin^2 5+\sin^2 10+\sin^2 15+......+\sin^2 (90-15)+\sin ^2 (90-10)+\sin^2 (90-5)+\ sin^2 90$

$=\sin^2 5+\sin^2 10+\sin^2 15+......+\cos^2 15+\cos^2 10+\cos^2 5+\ sin^2 90$

Using trigonometric identity,

$\sin (90-A)=\cos A$

$=(\sin^2 5+\cos^2 5)+(\sin^2 10+\cos^2 10)+(\sin^2 15+\cos^2 15)+.....+\sin^2 90$

$=1+1+1+.....+\sin^2 90$

Using trigonometric identity,

$\sin^2 A+\cos^2 A=1$

$[tex]=1+1+1+.....+1$

$=9\dfrac{1}{2}$

= 9.5

Hence, $\sin^2 5+\sin^2 10+\sin^2 15+......+\sin^2 75+\sin ^2 80+\sin^2 85+\ sin^2 90=9.5$