3 - sin4 3 -2- - - sin4 3 - - -2 - sin6 -2 - - -sin6 5 - - -

3 {sin^4 ( 3 π/2 α )+sin^4 (3 π+α) } 2 {sin^6 ( π/2+α )+sin^6(5 π α) } =3 {( cos α)^4+( sin α)^4 } 2{cos^6 α+sin^6 α} =3{ (cos^2 α+sin^2 α)^2 2 sin^2 α cos ...

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

Byju's Answer

Standard XIII

Mathematics

Basic Trigonometric Identities

3 [ sin4 3 π/...

Question

$3 [s i n^{4} (\frac{3 π}{2} - α) + s i n^{4} (3 π + α)] - 2 [s i n^{6} (\frac{π}{2} + α) + s i n^{6} (5 π - α)] =$

$0$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$1$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$3$

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

$s i n 4 α + s i n 6 α$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C
$3$

$3 {s i n^{4} (\frac{3 π}{2} - α) + s i n^{4} (3 π + α)} - 2 {s i n^{6} (\frac{π}{2} + α) + s i n^{6} (5 π - α)} = 3 {(- c o s α)^{4} + (- s i n α)^{4}} - 2 {c o s^{6} α + s i n^{6} α} = 3 {(c o s^{2} α + s i n^{2} α)^{2} - 2 s i n^{2} α c o s^{2} α} - 2 {(c o s^{2} α + s i n^{2} α)^{3} - 3 c o s^{2} α s i n^{2} α (c o s^{2} α + s i n^{2} α)} = 3 - 6 s i n^{2} α c o s^{2} α - 2 + 6 s i n^{2} α c o s^{2} α = 3 - 2 = 1$

Trick : Put $α = 0, \frac{π}{2}$ , the value of expression remains 1 i.e., it is independent of $α$ .

Suggest Corrections

Similar questions

Q. Evaluate :

3 [{sin}^{4} (\frac{3 π}{2} - α) + {sin}^{4} (3 π + α)] - 2 [{sin}^{6} (\frac{π}{2} + α) + {sin}^{6} (5 π - α)]

Q. The expression

3 [{sin}^{4} (\frac{3 π}{2} - α) + {sin}^{4} (3 π + α)] - 2 [{sin}^{6} (\frac{π}{2} + α) + {sin}^{6} (5 π - α)]

The expression $3 [{sin}^{4} (\frac{3 π}{2} - α) + {sin}^{4} (3 π + α)] - 2 [{sin}^{6} (\frac{π}{2} + α) + {sin}^{6} (5 π + α)]$ is equal to

Q. The expression,

3 [{sin}^{4} (\frac{3 π}{2} - α) + {sin}^{4} (3 π + α)] - 2 [{sin}^{6} (\frac{π}{2} + α) + {sin}^{6} (5 π - α)]

when simplified is equal to

Q. If

a = {sin}^{4} (\frac{3 π}{2} - α) + {sin}^{4} (3 π + α)

and

b = {sin}^{6} (\frac{π}{2} + α) + {sin}^{6} (5 π - α)

, then the value of

3 a - 2 b

Join BYJU'S Learning Program

3[sin4(3π2−α)+sin4(3π+α)]−2[sin6(π2+α)+sin6(5π−α)]=

$3 [s i n^{4} (\frac{3 π}{2} - α) + s i n^{4} (3 π + α)] - 2 [s i n^{6} (\frac{π}{2} + α) + s i n^{6} (5 π - α)] =$