Question

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Answer 1

GIVEN :-

[ note : taking ∅ as a ]

• 2cos a - cos 3a - cos 5a - 16 cos³a sin² a

TO FIND :-

• value of :- 2cos a - cos 3a - cos 5a - 16 cos³a sin² a

SOLUTION :-

now , we have given that

$\implies \rm{2cos \: a - cos \: 3a - cos \: 5a - 16cos {}^{3}a \: sin {}^{2} a }$

now solving for cos 3a and cos 5a

$\implies \rm{2cos \: a -( cos \: 5a + cos \: 3a) - 16cos {}^{3}a \: sin {}^{2} a }$

now we know that

$\implies \boxed {\rm{cos \: x + cos \: y =2 \: cos(\dfrac{x + y}{2}) \: cos( \dfrac{x - y}{2}) }}$

hence,

$\implies \rm{\small{2cos \: a -(2 \: cos(\dfrac{5a + 3a}{2}) \: cos( \dfrac{5a- 3a}{2})) - 16cos {}^{3}a \: sin {}^{2} a }}\:$

[ here X = 5a and y = 3a ]

now , solving further :-

$\implies \rm{2cos \: a -(2 \: cos(\dfrac{8a}{2}) \: cos( \dfrac{2a}{2})) - 16cos {}^{3}a \: sin {}^{2} a }\:$

$\implies \rm{2cos \: a -(2 \: cos \: 4a \: . \: cos \: a) - 16cos {}^{3}a \: sin {}^{2} a }\:$

now taking common ,

4cos a from term 16 cos³a sin²a,

$\implies \rm{2cos \: a -(2 \: cos \: 4a \: . \: cos \: a) - 4cos \: a (\: 4cos^{2}a \: sin {}^{2} a )}\:$

now , we taking whole square from term

4cos²a sin²a =( 2cos a . sin a )²

$\implies \rm{2cos \: a -(2 \: cos \: 4a \: . \: cos \: a) - 4cos \: a (\: 2cos \: a \: sin \: a ) {}^{2} }\:$

now rearranging

2 × cos a × sin a = 2 × sin a × cos a

$\implies \rm{2cos \: a -(2 \: cos \: 4a \: . \: cos \: a) - 4cos \: a (\: 2sin \: a \: cos\: a ) {}^{2} }\:$

now we know that ,

$\implies \boxed {\rm{2 \: sin \: x \: . \: cos \: x =sin \: 2x }}$

hence ,

2sin a . cos a = sin 2a

$\implies \rm{2cos \: a -(2 \: cos \: 4a \: . \: cos \: a) - 4cos \: a \: (\:sin 2 \: a) {}^{2} }\:$

now rearranging terms ,

and removing brackets from

( sin 2a )² = sin² 2a

$\implies \rm{2cos \: a \:- 4cos \: a \: .\:sin {}^{2} \: 2 a -(2 \: cos \: 4a \: . \: cos \: a) }\:$

now taking 2cos a common,

$\implies \rm{2cos \: a (1- 2 \ \sin {}^{2} \: 2 a )-(2 \: cos \: 4a \: . \: cos \: a) }\:$

now we know that ,

$\implies \boxed {\rm{cos \: 2x \: = 1 - 2 \: sin {}^{2}x }}$

hence ,

1 - 2 sin² 2a = cos 4a

[ note that there is 2a instead of a hence it will be 4a instead of 2a ]

$\implies \rm{2cos \: a \: (cos \: 4a)-(2 \: cos \: 4a \: . \: cos \: a) }\:$

now if we resolve and rearrange that terms we see both are same hence both will cancel out and we get 0

$\implies \rm{\small{(2 \times cos \: a \: \times cos \: 4a)-(2 \times cos \: a \times cos \:4 a) = 0 }}\:$

hence ,

$\implies \boxed{ \boxed{\rm{ \small{2cos \: a - cos \: 3a - cos \: 5a - 16cos {}^{3}a \: sin {}^{2} a = 0}}}}$

so option c is correct

Answer 2

Answer:

a.2

b.1

c.0

d.4

Step-by-step explanation:

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