Question

${sin}^2{5}^{\circ }+{sin}^2{10}^{\circ }+.....+{sin}^2{180}^{\circ }=$

Answer 1

We have,

${\sin }^2{5}^o+{\sin }^2{10}^o+{\sin }^2{15}^o+......+{\sin }^2{180}^o$

Then we know that,

${\sin }^2\theta +{\cos }^2\theta =1$

Now,

${\sin }^2{5}^o+{\sin }^2{10}^o+{\sin }^2{15}^o+......+{\sin }^2{180}^o$

$={\sin }^2{5}^o+{\sin }^2{10}^o+{\sin }^2{15}^o+......+{\sin }^2{170}^o+{\sin }^2{175}^o+{\sin }^2{180}^o$

$={\sin }^2{5}^o+{\sin }^2{10}^o+{\sin }^2{15}^o+......+{\sin }^2{165}^o+{\sin }^2{170}^o+{\sin }^2{175}^o+{\sin }^2{180}^o$

$={\sin }^2{5}^o+{\sin }^2{10}^o+{\sin }^2{15}^o+......+{\sin }^2({180}^o-{15}^o)+{\sin }^2({180}^o-{10}^o)+{\sin }^2({180}^o-5^o)+{\sin }^{2}180$

$={\sin }^2{5}^o+{\sin }^2{10}^o+{\sin }^2{15}^o+......+co{s}^2{15}^o+co{s}^2{10}^o+co{s}^2{5}^o+{\sin }^{2}180$

$=({\sin }^2{5}^o+co{s}^2{5}^o)+({\sin }^2{10}^o+co{s}^2{10}^o)+({\sin }^2{15}^o+co{s}^2{15}^o)+.........+{\sin }^{2}180$

$=1+1+1+......89terms$

$=89$

Hence, this is the answer.