Math, asked by runingwaterfall, 2 months ago

If 11x=π,prove that cos x cos 2x cos 3x cos 4x cos 5x =1/32


Answers

Answered by mohammedaatif323
0

Answer:

iven:x=

15

π

\textbf{To prove:}To prove:

cosx\;cos2x\;cos3x\;cos4x\;cos5x\;cos6x\;cos7x=\displaystyle\frac{1}{2^7}cosxcos2xcos3xcos4xcos5xcos6xcos7x=

2

7

1

\textbf{That is, to prove}That is, to prove

cos12^{\circ}\;cos24^{\circ}\;cos36^{\circ}\;cos48^{\circ}\;cos60^{\circ}\;cos72^{\circ}\;cos84^{\circ}=\displaystyle\frac{1}{2^7}cos12

cos24

cos36

cos48

cos60

cos72

cos84

=

2

7

1

\text{Consider,}Consider,

cos12^{\circ}\;cos24^{\circ}\;cos36^{\circ}\;cos48^{\circ}\;cos60^{\circ}\;cos72^{\circ}\;cos84^{\circ}cos12

cos24

cos36

cos48

cos60

cos72

cos84

=cos12^{\circ}\;cos24^{\circ}\;cos36^{\circ}\;cos48^{\circ}\displaystyle(\frac{1}{2})cos72^{\circ}\;cos84^{\circ}=cos12

cos24

cos36

cos48

(

2

1

)cos72

cos84

=\displaystyle(\frac{1}{2})[cos48^{\circ}\;cos12^{\circ}\;cos72^{\circ}][\;cos36^{\circ}\;cos24^{\circ}\;cos84^{\circ}]=(

2

1

)[cos48

cos12

cos72

][cos36

cos24

cos84

]

\text{We know that,}We know that,

\boxed{\bf\,cos(60-A)\;cosA\;cos(60+A)=\frac{1}{4}cos\,3A}}

=\displaystyle(\frac{1}{2})[\frac{1}{4}\;cos3(12^{\circ})][\frac{1}{4}\;cos3(24^{\circ})]=(

2

1

)[

4

1

cos3(12

)][

4

1

cos3(24

)]

=\displaystyle(\frac{1}{2})[\frac{1}{4}\;cos36^{\circ}][\frac{1}{4}\;cos72^{\circ}]=(

2

1

)[

4

1

cos36

][

4

1

cos72

]

=\displaystyle(\frac{1}{32})[cos36^{\circ}][cos72^{\circ}]=(

32

1

)[cos36

][cos72

]

=\displaystyle(\frac{1}{32})[\frac{\sqrt{5}+1}{4}][\frac{\sqrt{5}-1}{4}]=(

32

1

)[

4

5

+1

][

4

5

−1

]

=\displaystyle(\frac{1}{32})[\frac{(\sqrt{5})^2-1^2}{16}]=(

32

1

)[

16

(

5

)

2

−1

2

]

=\displaystyle(\frac{1}{32})[\frac{5-1}{16}]=(

32

1

)[

16

5−1

]

=\displaystyle(\frac{1}{32})[\frac{4}{16}]=(

32

1

)[

16

4

]

=\displaystyle(\frac{1}{32})[\frac{1}{4}]=(

32

1

)[

4

1

]

=\displaystyle(\frac{1}{2^5})[\frac{1}{2^2}]=(

2

5

1

)[

2

2

1

]

=\bf\displaystyle\frac{1}{2^7}=

2

7

1

\therefore\bf\,cos12^{\circ}\;cos24^{\circ}\;cos36^{\circ}\;cos48^{\circ}\;cos60^{\circ}\;cos72^{\circ}\;cos84^{\circ}=\displaystyle\frac{1}{2^7}∴cos12

cos24

cos36

cos48

cos60

cos72

cos84

=

2

7

1

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