Question

Prove that cos2a cos a/2 -cos3a cos9a/2=sin5a Sin 5a/2

Answer 1

Answer:

$cos2A\:cos\frac{A}{2}-cos3A\:cos\frac{9A}{2}=sin5A\:sin\frac{5A}{2}$

Step-by-step explanation:

Formula used:

$1.cos(A+B)+cos(A-B)=2\:cosA\:cosB$

$2.cosC-cosD=-2\:sin(\frac{C+D}{2})\:sin(\frac{C-D}{2})$

$3.sin(-A)=-sinA$

Now,

$cos2A\:cos\frac{A}{2}-cos3A\:cos\frac{9A}{2}$

Multiply and divide by 2, we get

$=\frac{1}{2}[2\:cos2A\:cos\frac{A}{2}-2\:cos3A\:cos\frac{9A}{2}]$

$=\frac{1}{2}[cos(2A+\frac{A}{2})+cos(2A-\frac{A}{2})-(cos(3A+\frac{9A}{2})+cos(3A-\frac{9A}{2}))]$

(by using formula (1))

$=\frac{1}{2}[cos\frac{5A}{2}+cos\frac{3A}{2}-cos\frac{15A}{2}-cos\frac{-3A}{2}]$

$=\frac{1}{2}[cos\frac{5A}{2}+cos\frac{3A}{2}-cos\frac{15A}{2}-cos\frac{3A}{2}]$

$=\frac{1}{2}[cos\frac{5A}{2}-cos\frac{15A}{2}]$

$=\frac{1}{2}[-2\:sin(\frac{\frac{5A}{2}+\frac{15A}{2}}{2})\:sin(\frac{\frac{5A}{2}-\frac{15A}{2}}{2})]$ (by using formula (2))

$=\frac{1}{2}[-2\:sin(\frac{\frac{20A}{2}}{2})\:sin(\frac{\frac{-10A}{2}}{2})]$

$=\frac{1}{2}[-2\:sin(\frac{10A}{2})\:sin(\frac{-5A}{2})]$ (by using formula (3))

$=\frac{1}{2}[2\:sin5A\:sin\frac{5A}{2}]$

$=sin5A\:sin\frac{5A}{2}$

Answer 2

Answer:

cos2Acos

2

A

−cos3Acos

2

9A

Multiply and divide by 2, we get

=\frac{1}{2}[2\:cos2A\:cos\frac{A}{2}-2\:cos3A\:cos\frac{9A}{2}]=

2

1

[2cos2Acos

2

A

−2cos3Acos

2

9A

]

=\frac{1}{2}[cos(2A+\frac{A}{2})+cos(2A-\frac{A}{2})-(cos(3A+\frac{9A}{2})+cos(3A-\frac{9A}{2}))]=

2

1

[cos(2A+

2

A

)+cos(2A−

2

A

)−(cos(3A+

2

9A

)+cos(3A−

2

9A

))]

(by using formula (1))

=\frac{1}{2}[cos\frac{5A}{2}+cos\frac{3A}{2}-cos\frac{15A}{2}-cos\frac{-3A}{2}]=

2

1

[cos

2

5A

+cos

2

3A

−cos

2

15A

−cos

2

−3A

]

=\frac{1}{2}[cos\frac{5A}{2}+cos\frac{3A}{2}-cos\frac{15A}{2}-cos\frac{3A}{2}]=

2

1

[cos

2

5A

+cos

2

3A

−cos

2

15A

−cos

2

3A

]

=\frac{1}{2}[cos\frac{5A}{2}-cos\frac{15A}{2}]=

2

1

[cos

2

5A

−cos

2

15A

]

=\frac{1}{2}[-2\:sin(\frac{\frac{5A}{2}+\frac{15A}{2}}{2})\:sin(\frac{\frac{5A}{2}-\frac{15A}{2}}{2})]=

2

1

[−2sin(

2

2

5A

+

2

15A

)sin(

2

2

5A

−

2

15A

)] (by using formula (2))

=\frac{1}{2}[-2\:sin(\frac{\frac{20A}{2}}{2})\:sin(\frac{\frac{-10A}{2}}{2})]=

2

1

[−2sin(

2

2

20A

)sin(

2

2

−10A

)]

=\frac{1}{2}[-2\:sin(\frac{10A}{2})\:sin(\frac{-5A}{2})]=

2

1

[−2sin(

2

10A

)sin(

2

−5A

)] (by using formula (3))

=\frac{1}{2}[2\:sin5A\:sin\frac{5A}{2}]=

2

1

[2sin5Asin

2

5A

]

=sin5A\:sin\frac{5A}{2}=sin5Asin

2

5A

Step-by-step explanation:

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