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(1+cosπ/8)( 1+cos3π/8)( 1+cod 5π/8)(1+cos7π/8) prove that it is1/2

(1+cosπ/8)( 1+cos3π/8)(
1+cod 5π/8)(1+cos7π/8) prove that it is1/2

Grade:12

5 Answers

Arun
25750 Points
5 years ago

cos 5pi/8=cos(pi/2+pi/8)=-sin pi/8

cos 3pi/8=cos(pi/2-pi/8)=sin pi/8

cos 7pi/8=cos(pi-pi/8)=-cos pi/8. Now the expression (1+cos pi/8)(1+cos 3pi/8)(1+cos 5pi/8)(1+cos 7pi/8) can be written as,

(1+cos pi/8)(1+sin pi/8)(1-sin pi/8)(1-cos pi/8)

=[1-(cos pi/8)^2][1–(sin pi/8)^2]

=(sin pi/8)^2×(cos pi/8)^2

=1/4×(2 sin pi/8.cos pi/8)^2

=1/4(sin 2x pi/8)^2

=1/4(sin pi/4)^2

=1/4×1/[(2)^1/2]^2

=1/4×1/2

=1/8(proven)

 

Please check as answer is 1/8 . But you have given it 1/2

Narendra Beniwal
34 Points
5 years ago
cos 5pi/8=cos(pi/2+pi/8)=-sin pi/8
 
cos 3pi/8=cos(pi/2-pi/8)=sin pi/8
 
cos 7pi/8=cos(pi-pi/8)=-cos pi/8. Now the expression (1+cos pi/8)(1+cos 3pi/8)(1+cos 5pi/8)(1+cos 7pi/8) can be written as,
 
(1+cos pi/8)(1+sin pi/8)(1-sin pi/8)(1-cos pi/8)
 
=[1-(cos pi/8)^2][1–(sin pi/8)^2]
 
=(sin pi/8)^2×(cos pi/8)^2
 
=1/4×(2 sin pi/8.cos pi/8)^2
 
=1/4(sin 2x pi/8)^2
 
=1/4(sin pi/4)^2
 
=1/4×1/[(2)^1/2]^2
 
=1/4×1/2
 
=1/8
Narendra Beniwal
34 Points
5 years ago
cos 5pi/8=cos(pi/2+pi/8)=-sin pi/8
 
cos 3pi/8=cos(pi/2-pi/8)=sin pi/8
 
cos 7pi/8=cos(pi-pi/8)=-cos pi/8. Now the expression (1+cos pi/8)(1+cos 3pi/8)(1+cos 5pi/8)(1+cos 7pi/8) can be written as,
 
(1+cos pi/8)(1+sin pi/8)(1-sin pi/8)(1-cos pi/8)
 
=[1-(cos pi/8)^2][1–(sin pi/8)^2]
 
=(sin pi/8)^2×(cos pi/8)^2
 
=1/4×(2 sin pi/8.cos pi/8)^2
 
=1/4(sin 2x pi/8)^2
 
=1/4(sin pi/4)^2
= 1/4x(21/22
= 1/2
Arun
25750 Points
5 years ago
Dear Narendra
 
sin pi/4 = 1/ sqrt(2)
 
you have commited a mistake in the second last line of above answer. Answer should be 1/8
Vishakha
15 Points
3 years ago
Its correct this may be 1/7but it could hwhwhajwjjajqw  qqkkwnwjwnqjam q wow qjqjqmam awbhqnq q q qnkq qjqjw jsnsjsisjmsmsmsksmksms  snsksjnsjs snake so ssksmsjksks s  s mskwmskskkkwkqnsnsjsk  ajakmwmsmsmammss  snskksmsmsmms

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