Class 11th

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New answer posted

7 months ago

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P
Payal Gupta

Contributor-Level 10

38. L.H.S=cos4x+cos3x+cos2xsin4x+sin3x+sin2x.

=(cos4x+cos2x)+cos3x(sin4x+sin2x)+sin3x.

=2cos(4x+2x2)cos(4x2x2)+cos3x2sin(4x+2x2)cos(4x2x2)sin3x.

=2cos6x2cos2x2+cos3x2sin6x2cos2x2+sin3x

=2cos3xcosx+cos3x2sin3xcosx+3x

=cos3xsin3x*(2cosx+1)(2cosx+1)

=cot3x=R.H.S

New answer posted

7 months ago

0 Follower 2 Views

P
Payal Gupta

Contributor-Level 10

37. =sinxsin3xsin2xcos2x

=2cos (x+3x2)sinx3x2 (cos2xsin2x)

=2cos4x2sin (2x2)cos2x [? cos2x=cos2xsin2x]

=2cos2xsinxcos2x [? sin (x)=sinx]

= 2 sin x

= R.H.S.

New answer posted

7 months ago

0 Follower 2 Views

P
Payal Gupta

Contributor-Level 10

36. =sinx+sin3xcosx+cos3x

=2 (x+3x2)cos (x3x2)2cos (x+3x2)cos (x3x2)

=sin4x2cos4x2

=sin2xcos2x

=tan2x=R.H.S

New answer posted

7 months ago

0 Follower 2 Views

P
Payal Gupta

Contributor-Level 10

35. L.H.S: = L.H. S:=sinxsinycosx+cosy

= 2 c o s x + y 2 s i n x y 2 2 c o s x + y 2 c o s x y 2

= s i n ( x y 2 ) c o s ( x y 2 )

=tan (xy2)=R.H.S. = R.H.S.

New answer posted

7 months ago

0 Follower 2 Views

P
Payal Gupta

Contributor-Level 10

34.=sin5x+sin3xcos5x+cos3x.

=2sin (5x+3x2)cos (5x3x2)2cos (5x+3x2)cos (5x3x2)

=sin8x2cos2x2cos8x2cos2x2

=sin4xcos4x=tan4x=R.H.S

New answer posted

7 months ago

0 Follower 2 Views

P
Payal Gupta

Contributor-Level 10

33.=cos9xcos5xsin17xsin3x

=2sin (9x+5x2)sin (9x5x2)2cos (17x+3x2)sin (17x3x2)

=sin14x2sin4x2cos20x2sin14x2=sin7xsin2xcos10xsin7x=sin2xcos10x

= R.H.S

New answer posted

7 months ago

0 Follower 3 Views

P
Payal Gupta

Contributor-Level 10

32. L.H.S=cot4x (sin5x+sin3x)

=cot4x [2sin8x2cos2x2]

=cos4x4x*2sin4xcosx

=2·cos4x·cosx

R.H.S=cotx [sin5xsin3x]

=cotx [2cos8x2·sin2x2]

=cosxsinx·2cos4xsinx

=2·cos4x·cosx.

Hence, L.H.S. = R.H.S.

New answer posted

7 months ago

0 Follower 2 Views

P
Payal Gupta

Contributor-Level 10

31. L.H.S. = sin 2x + 2 sin 4x + sin 6x

Using sin A + sin B = 2 sin A + B/2 cos A - B/2  we have,

L.H.S. = (sin 2x + sin 6x) + 2 sin 4x

= 2 s i n ( 2 x + 6 x 2 ) c o s ( 2 x 6 x 2 ) + 2 s i n 4 x .

= 2 s i n 8 x 2 c o s ( 4 x 2 ) + 2 s i n 4 x .

= 2 s i n 4 x c o s 2 x + 2 s i n 4 x [ ? c o s ( x ) = x ]

= 2 s i n 4 x [ c o s 2 x + 1 ]

We know that,

cos2x=2cos2x1

cos2x+1=2cos2x

Hence,

L.H.S=2sin4x(2cos2x)

=4cos2xsin4x

= R.H.S

New answer posted

7 months ago

0 Follower 7 Views

P
Payal Gupta

Contributor-Level 10

30. L.H.S L.H.S=cos22xcos26x

= ( c o s 2 x + c o s 6 x ) ( c o s 2 x c o s 6 x ) [ ? a 2 b 2 = ( a b ) ( a + b ) ]

L.H.S = L.H.S=[2cos(2x+6x2)cos(2x6x2)][2sin2x+6x2sin(2x6x2)]
= [ 2 c o s 8 x 2 c o s ( 4 x 2 ) ] [ 2 s i n ( 8 x 2 ) s i n ( 4 x 2 ) ]
= 2 c o s 4 x c o s 2 x * 2 s i n 4 x · s i n 2 x [ ? c o s ( x ) = c o s x ; s i n ( x ) = s i n x ]
= 2 s i n 4 x c o s 4 x * 2 s i n 2 x c o s 2 x

As sin 2θ = 2 sinθ cosθ.

L.H.S. = sin (2*4x) sin(2*2x)

= sin 8x sin 4x

= R.H.S.

New answer posted

7 months ago

0 Follower 7 Views

P
Payal Gupta

Contributor-Level 10

29. L.H.S  L.H.S=sin26xsin24x.

= ( s i n 6 x + s i n 4 x ) ( s i n 6 x s i n 4 x ) . [ a 2 b 2 = ( a b ) ( a + b ) ]

So, L.H.S L.H.S.=(2sin(6x+4x2)cos(6x4x2))(2cos(6x+4x2)sin(6x4x2))

= ( 2 s i n 1 0 x 2 c o s 2 x 2 ) ( 2 c o s 1 0 x 2 s i n 2 x 2 )

= 2 s i n 5 x c o s x 2 c o s 5 x s i n x

= 2 s i n 5 x c o s 5 x 2 c s i n x c o s x .

As sin2θ=2sinθcosθ we can write.

L.H.S =sin(2*5x)=sin(2*x)

=sin10xsin2x= R.H.S. R.H.S

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