Sin 2 A - x and - r -14sin rA - ax 2- bx 3- cx 4- d x 5- then the value of - b - is

Given : nbsp;sin^2 A = x ∏_r = 1^4sin(rA) = sin A sin 2A sin 3A sin 4A Now, nbsp; sin 2 A = 2sin Acos A sin 3A = sin A(3 4sin^2 A) sin 4A = 2sin 2Acos 2A ⇒ ...

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Byju's Answer

Standard XII

Mathematics

Basic Trigonometric Identities

sin 2 A = x a...

Question

If ${sin}^{2} A = x$ and $4 \prod r = 1 sin (r A) = a x^{2} + b x^{3} + c x^{4} + d x^{5}$ , then the value of $| b |$ is

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Solution

Given : ${sin}^{2} A = x$
$4 \prod r = 1 sin (r A) = sin A sin 2 A sin 3 A sin 4 A$
Now,
$sin 2 A = 2 sin A cos A sin 3 A = sin A (3 - 4 {sin}^{2} A) sin 4 A = 2 sin 2 A cos 2 A \Rightarrow sin 4 A = 4 sin A cos A (1 - 2 {sin}^{2} A)$
So,
$4 \prod r = 1 sin (r A) = 8 {sin}^{4} A ({cos}^{2} A) (3 - 4 {sin}^{2} A) (1 - 2 {sin}^{2} A) = 8 x^{2} (1 - x) (3 - 4 x) (1 - 2 x) = 24 x^{2} - 104 x^{3} + 144 x^{4} - 64 x^{5}$

Hence, $| b | = 104$

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If sin2A=x and 4∏r=1sin(rA)=ax2+bx3+cx4+dx5, then the value of |b| is

Given : sin2A=x 4∏r=1sin(rA)=sinAsin2Asin3Asin4A Now, sin2A=2sinAcosAsin3A=sinA(3−4sin2A)sin4A=2sin2Acos2A⇒sin4A=4sinAcosA(1−2sin2A) So, 4∏r=1sin(rA)=8sin4A(cos2A)(3−4sin2A)(1−2sin2A)=8x2(1−x)(3−4x)(1−2x)=24x2−104x3+144x4−64x5 Hence, |b|=104

If ${sin}^{2} A = x$ and $4 \prod r = 1 sin (r A) = a x^{2} + b x^{3} + c x^{4} + d x^{5}$ , then the value of $| b |$ is