` cos 36^(@)` का मान क्या है ?

Options

A. ` (sqrt(5)-1)/4`
B. ` (sqrt(5)+1)/4`
C. `sqrt(10+2 sqrt(5))/4`
D. ` sqrt(10-2sqrt(5))/4`

Answer

B

Explanation

यहाँ हम लेते है , ` 5 theta = 180^(@) [ :' theta = 36^(@) ] `
` rArr 3 theta + 2 theta = 180^(@) 2 theta = 180^(@) - 3 theta`
दोनों पक्षों में cos लेने पर,
` cos 2 theta = cos ( 180^(@) - 3 theta)`
`rArr cos 2 theta = - cos 3theta`
` rArr 2 cos^(2) theta - 1 = - ( 4 cos^(3) theta - 3 cos theta)`
` rArr 2 cos^(2) theta -1 = - (4 cos^(3) theta - 3 cos theta ) `
` rArr 4 cos^(3) theta + 2 cos^(2) theta - 3 cos theta - 1 = 0`
`rArr 4 cos^(@) theta (cos theta+1) - 2 cos theta `
` rArr 4 cos^(2) theta ( cos theta +1) - 2 cos theta `
` rArr 4 cos^(@) theta ( cos thheta +1) -2 cos theta`
` ( cos theta +1) -1 (cos theta +1) =0`
` rArr ( cos theta +1) (4 cos^(@) theta -2 cos theta -1)=0`
` rArr ( cos theta +1) ( 4 cos^(2) theta - 2 cos theta-1)=0`
` rArr (cos theta +1) ( 4 cos^(2) theta -2cos theta -1)=0`
` rArr cos theta +1=0`
या ` 4 cos ^(@) theta - 2 cos theta -1 =0`
` rArr theta = -1`
` rArr cos theta =-1`
` rArr cos theta = (2pm sqrt(4-4xx(4)(-1)))/(2xx4)`
` rArr cos theta = (2 pmsqrt(4+16))/8 = (2 pm sqrt(20))/8`
` rArr cos theta = (2 pm 2sqrt(5))/8`
` rArr costheta = (1pm sqrt(5))/4 ( :' cos theta in [ -1,1])`
` theta = 36^(@) ` रखने पर
` cos 36^(@) = (sqrt(5)+1)/4 [ :' cos 36^(@) gt 0]`