If a - sin4 3-2- -sin43- and b -sin6 -2- -sin65- then the value of 3a-2b is

A = sin^4 ( 3π2 α )+sin^4(3π+α) ⇒ a=cos^4 α+sin^4α ⇒ a = 1 2sin^2αcos^2α b =sin^6 ( π2+α )+sin^6(5π α) ⇒ b = cos ^6α+sin^6α ⇒ b =1 3sin^2αcos^2α Now, nbsp; ...

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Byju's Answer

Standard XI

Mathematics

Sign of Trigonometric Ratios in Different Quadrants

If a = sin4 3...

Question

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 $3 a - 2 b$ is

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Solution

The correct option is B $1$
$a = {sin}^{4} (\frac{3 π}{2} - α) + {sin}^{4} (3 π + α) \Rightarrow a = {cos}^{4} α + {sin}^{4} α \Rightarrow a = 1 - 2 {sin}^{2} α {cos}^{2} α b = {sin}^{6} (\frac{π}{2} + α) + {sin}^{6} (5 π - α) \Rightarrow b = {cos}^{6} α + {sin}^{6} α \Rightarrow b = 1 - 3 {sin}^{2} α {cos}^{2} α$

Now,
$3 a - 2 b = 3 - 6 {sin}^{2} α {cos}^{2} α - 2 + 6 {sin}^{2} α {cos}^{2} α = 1$

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Similar questions

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

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{[1 - sin (π + α) + cos (π + α)]}^{2} + {[1 - sin (\frac{3 π}{2} + α) + cos (\frac{3 π}{2} - α)]}^{2} = a + b sin 2 α

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Sign of Trigonometric Ratios in Different Quadrants

Standard XI Mathematics

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If a=sin4(3π2−α)+sin4(3π+α) and b=sin6(π2+α)+sin6(5π−α), then the value of 3a−2b is

The correct option is B 1a=sin4(3π2−α)+sin4(3π+α)⇒a=cos4α+sin4α⇒a=1−2sin2αcos2αb=sin6(π2+α)+sin6(5π−α)⇒b=cos6α+sin6α⇒b=1−3sin2αcos2α Now, 3a−2b=3−6sin2αcos2α−2+6sin2αcos2α=1

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 $3 a - 2 b$ is