Prove that-sin 2-8-sin 23 -8-sin 25 -8-sin 27 -8-2

LHS, = sin^2 π/8+sin^2 3 π/8 + sin^2 5 π/8 + sin^2 7 π/8 = sin^2 π/8 + sin^2 (π/2 π/8 )+sin^2 5 π/8 + sin^2 (π π/8 ) = sin^2 π/8 + cos^2 π/8 + sin^2 (π 3 π/ ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Sum of Trigonometric Ratios in Terms of Their Product

Prove that:si...

Question

Prove that:

$s i n^{2} \frac{π}{8} + s i n^{2} \frac{3 π}{8} + s i n^{2} \frac{5 π}{8} + s i n^{2} \frac{7 π}{8} = 2$

Open in App

Solution

LHS,

$= s i n^{2} \frac{π}{8} + s i n^{2} \frac{3 π}{8} + s i n^{2} \frac{5 π}{8} + s i n^{2} \frac{7 π}{8} = s i n^{2} \frac{π}{8} + s i n^{2} (\frac{π}{2} - \frac{π}{8}) + s i n^{2} \frac{5 π}{8} + s i n^{2} (π - \frac{π}{8}) = s i n^{2} \frac{π}{8} + c o s^{2} \frac{π}{8} + s i n^{2} (π - \frac{3 π}{8}) + s i n^{2} \frac{π}{8} = 1 + s i n^{2} \frac{3 π}{8} + s i n^{2} \frac{π}{8} = 1 + s i n^{2} (\frac{π}{2} - \frac{π}{8}) + s i n^{2} \frac{π}{8} = 1 + c o s^{2} \frac{π}{8} + s i n^{2} \frac{π}{8} = 1 + 1 = 2$

= RHS

Suggest Corrections

Similar questions

Q. Prove sin2 π/8 + sin2 3π/8 + sin2 5π/8 + sin2 7π/8 = 2

Q. Solve :

{sin}^{2} \frac{π}{8} + {sin}^{2} \frac{3 π}{8} + {sin}^{2} \frac{5 π}{8} + s i n^{2} \frac{7 π}{8} = 2

Q. The value of

{sin}^{2} \frac{π}{8} + {sin}^{2} \frac{3 π}{8} + {sin}^{2} \frac{5 π}{8} + {sin}^{2} \frac{7 π}{8}

Q. If

{sin}^{2} \frac{π}{8} + {sin}^{2} \frac{3 π}{8} + {sin}^{2} \frac{5 π}{8} + {sin}^{2} \frac{7 π}{8} = m

.Find

m

Prove that:

$s i n^{2} (\frac{π}{8} + \frac{A}{2}) - s i n^{2} (\frac{π}{8} - \frac{A}{2}) = \frac{1}{\sqrt{2}} s i n A$

Join BYJU'S Learning Program

Prove that: sin2 π8+sin23π8+sin25π8+sin27π8=2

LHS, =sin2 π8+sin23π8+sin25π8+sin27π8=sin2 π8+sin2(π2−π8)+sin25π8+sin2(π−π8)=sin2π8+cos2π8+sin2(π−3π8)+sin2π8=1+sin23π8+sin2π8=1+sin2(π2−π8)+sin2π8=1+cos2π8+sin2π8=1+1=2 = RHS

Prove that:

$s i n^{2} \frac{π}{8} + s i n^{2} \frac{3 π}{8} + s i n^{2} \frac{5 π}{8} + s i n^{2} \frac{7 π}{8} = 2$