Question

show that 4sin 5 theta /2 •cos 3 theta /2•cos 3 theta =sin theta -sin2theta +sin4 theta +sin 7theta
​

Answer 1

Answer:

ANSWER

⇒sin7θ+sinθ+sin4θ

⇒2sin4θ⋅cos3θ+sin4θ=0

⇒sin4θ[2cos3θ+1]=0

∴sin4θ=0

4θ=nπ,wheren∈I

∴θ=

4

nπ

cosθ=−

2

1

⇒cosθ=cos(

3

2π

)

⇒cos3θ=2kπ+

3

2π

,k∈I

θ=2k

3

π

±

9

2π

Answer 2

It is proved  that $4 sin\frac{5\theta}{2} .cos\frac{3\theta}{2} .cos3\theta= sin\theta-sin 2\theta + sin4\theta+sin7\theta$.

Step-by-step explanation:

Given:

$4 sin\frac{5\theta}{2} .cos\frac{3\theta}{2} .cos3\theta= sin\theta-sin 2\theta + sin4\theta+sin7\theta$

To Find:

It is to be proved $4 sin\frac{5\theta}{2} .cos\frac{3\theta}{2} .cos 3\theta= sin\theta-sin 2\theta + sin4\theta+sin7\theta$.

Identities Used:

$2 sinx .cosy= sin(x+y) + sin(x-y)$

$2cosx.siny= sin(x+y) -sin( x-y)$

Solution:

As given-$4 sin\frac{5\theta}{2} .cos\frac{3\theta}{2} .cos 3\theta= sin\theta-sin 2\theta + sin4\theta+sin7\theta$

LHS $=4 sin\frac{5\theta}{2} .cos\frac{3\theta}{2} .cos 3\theta$

       $=2( 2sin\frac{5\theta}{2} .cos\frac{3\theta}{2}) .cos 3\theta$

       $=2( sin( \frac{5\theta }{2} + \frac{3\theta}{2} )+sin(\frac{5\theta }{2} - \frac{3\theta}{2} )) .cos 3\theta$

       $=2( sin(\frac{5\theta+3\theta}{2}) +sin(\fr{\frac{5\theta-3\theta}{2} ) ).cos 3\theta$

       $=2( sin4 \theta +sin \theta ).cos 3\theta$

      $=2 sin4 \theta .cos3\theta + 2cos3 \theta. sin\theta$

     $= sin( 4\theta+ 3\theta)+sin( 4\theta- 3\theta) +sin( 3\theta+ \theta)-sin( 3\theta- \theta)$

     $= sin 7\theta+sin\theta +sin4\theta-sin 2\theta$

     $= sin\theta -sin 2\theta +sin4\theta+sin 7\theta$

     $= RHS$

LHS=RHS

Hence, it is proved that $4 sin\frac{5\theta}{2} .cos\frac{3\theta}{2} .cos3\theta= sin\theta-sin 2\theta + sin4\theta+sin7\theta$.