Question

Question!

sin²5 + sin²10 + sin²15 +.... + sin²85° + sin²(90°) = ?

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Answer 1

$\large\underline{\sf{Solution-}}$

Consider

$\rm \: {sin}^{2}5 \degree + {sin}^{2}10 \degree + {sin}^{2}15 \degree + - - - + {sin}^{2}85 \degree + {sin}^{2}90$

We have to first pair the angles whom sum is 90°.

$\rm \: ({sin}^{2}5 \degree + {sin}^{2}85 \degree ) + ({sin}^{2}10 \degree + {sin}^{2}80 \degree ) + ({sin}^{2}15 \degree + {sin}^{2}75 \degree ) + \\ \rm \: ({sin}^{2}20 \degree + {sin}^{2}70 \degree ) + ({sin}^{2}25 \degree + {sin}^{2}65 \degree ) + ({sin}^{2}35 \degree + {sin}^{2}55 \degree ) + \\ \rm \: ({sin}^{2}40 \degree + {sin}^{2}50 \degree ) + ({sin}^{2}30\degree + {sin}^{2}60 \degree ) + {sin}^{2}45 \degree + {sin}^{2}90 \degree$

can be further rewritten as

$\rm \: ({sin}^{2}5 \degree + {sin}^{2}(90\degree - 5 \degree ) )+ ({sin}^{2}10 \degree + {sin}^{2}(90 \degree - 10 \degree )) + ({sin}^{2}15 \degree + {sin}^{2}(90 \degree - 15 \degree )) + \\ \rm \: ({sin}^{2}20 \degree + {sin}^{2}(90 \degree - 20 \degree )) + ({sin}^{2}25 \degree + {sin}^{2}(90\degree - 25 \degree )) + ({sin}^{2}35 \degree + {sin}^{2}(90 \degree - 35) ) + \\ \rm \: ({sin}^{2}40 \degree + {sin}^{2}(90\degree - 40 \degree )) + ({sin}^{2}30\degree + {sin}^{2}(90 \degree - 30 \degree )) + \dfrac{1}{2} + 1$

We know,

$\boxed{\tt{ \: \: sin(90 \degree - x) = cosx \: }}$

So, using this identity, we get

$\rm \: ({sin}^{2}5 \degree + {cos}^{2}5 \degree ) + ({sin}^{2}10 \degree + {cos}^{2}10 \degree ) + ({sin}^{2}15 \degree + {cos}^{2}15 \degree ) + \\ \rm \: ({sin}^{2}20 \degree + {cos}^{2}20 \degree ) + ({sin}^{2}25 \degree + {cos}^{2}25 \degree ) + ({sin}^{2}35 \degree + {cos}^{2}35 \degree ) + \\ \rm \: ({sin}^{2}40 \degree + {cos}^{2}40 \degree ) + ({sin}^{2}30\degree + {cos}^{2}30 \degree ) + \dfrac{3}{2}$

We know that,

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

So, using this identity, we get

$\rm \: = 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + \dfrac{3}{2}$

$\rm \: = \: 8 + \dfrac{3}{2}$

$\rm \: = \: \dfrac{16 + 3}{2}$

$\rm \: = \: \dfrac{19}{2}$

Hence,

$\boxed{\tt{ \rm \: {sin}^{2}5 \degree + {sin}^{2}10 \degree + - - - + {sin}^{2}85 \degree + {sin}^{2}90 = \dfrac{19}{2} }}$

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Additional Information:-

Relationship between sides and T ratios

sin θ = Opposite Side/Hypotenuse

cos θ = Adjacent Side/Hypotenuse

tan θ = Opposite Side/Adjacent Side

sec θ = Hypotenuse/Adjacent Side

cosec θ = Hypotenuse/Opposite Side

cot θ = Adjacent Side/Opposite Side

Reciprocal Identities

cosec θ = 1/sin θ

sec θ = 1/cos θ

cot θ = 1/tan θ

sin θ = 1/cosec θ

cos θ = 1/sec θ

tan θ = 1/cot θ

Co-function Identities

sin (90°−x) = cos x

cos (90°−x) = sin x

tan (90°−x) = cot x

cot (90°−x) = tan x

sec (90°−x) = cosec x

cosec (90°−x) = sec x

Fundamental Trigonometric Identities

sin²θ + cos²θ = 1

sec²θ - tan²θ = 1

cosec²θ - cot²θ = 1

Answer 2

GIVEN:

• sin²5° + sin²10° + sin²15° + sin²20° + sin²25° + sin²30° + sin²35° + sin²40° + sin²45° + sin²50° + sin²55° + sin²60° + sin²65° + sin²70° + sin²75° + sin²80° + sin²85 + cos²45°

Formula used:

$\\ { \underline{ \boxed{ \bf{ \pink{sin^{2} (90 - \theta) = cos ^{2} \theta}}}}}$

${ \underline{ \boxed{ \bf{ \red{ 1 \: sin^{2} \theta + cos ^{2} \theta}}}}}$

Solution:

$:\implies$sin2 5° + sin2 10° + sin2 15° + sin2 20° + sin2 25° + sin2 30° + sin2 35°+ sin2 40° + sin 2 45° + sin2 50° + sin2 55° + sin2 60° + sin2 65° + sin2 70° + sin2 75° + sin2 80° + sin2 85° + sin2 90°.

$:\implies$ (sin2 5° + sin2 85°) + (sin2 10° + sin2 80°) + (sin2 15° + sin2 75° ) + (sin2 20°+ sin2 70°) + (sin2 25° + sin2 65°) + (sin2 30° + sin2 60°) + (sin2 35° + sin2 55°) + (sin2 40° + sin2 50°) + sin 2 45° + sin2 90°.

$:\implies$ 1+1+1+1+1+1+1+1+1

$:\implies$ 9