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Value of sin 18 °

Find values of sin 18, cos 18, cos 36, sin 36, sin 54, cos 54
Find values of sin 18, cos 18, cos 36, sin 36, sin 54, cos 54 - Part 2

Value of cos 18° and sin 72°

Find values of sin 18, cos 18, cos 36, sin 36, sin 54, cos 54 - Part 3

Value of cos 36°, sin 36° and sin 54°

Find values of sin 18, cos 18, cos 36, sin 36, sin 54, cos 54 - Part 4 Find values of sin 18, cos 18, cos 36, sin 36, sin 54, cos 54 - Part 5

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Transcript

What is value of sin 18 Let θ = 18° 5θ = 5 × 18° = 90° 2θ + 3θ = 90° 2θ = 90° – 3θ sin 2θ = sin (90° – 3θ) sin 2θ = cos 3θ 2 sin θ cos θ = 4 cos3 θ – 3 cos θ 2 sin θ cos θ – 4 cos3 θ + 3 cos θ = 0 cos θ (2 sin θ – 4 cos2 θ + 3) = 0 2 sin θ – 4 cos2 θ + 3 = 0 2 sin θ – 4 (1 – sin2 θ) + 3 = 0 2 sin θ – 4 + 4sin2 θ + 3 = 0 4sin2 θ + 2 sin θ – 1 = 0 Let sin θ = x 4x2 + 2x – 1 = 0 x = (−𝑏±√(𝑏^2 − 4𝑎𝑐))/2𝑎 x = (−2 ± √(2^2 − 4(4)(−1) ))/(2(4)) = (−2 ± √(4 + 16 ))/8 = (−2 ± √20)/8 = (−2 ± √(4 × 5))/8 x = (−2 ±2√5)/8 = (−1 ± √5)/4 ∴ sin θ = sin 18° = (−𝟏 ± √𝟓)/𝟒 Since sin is positive sin 18° = (−𝟏 + √𝟓)/𝟒 cos 18 We know that , 〖𝑠𝑖𝑛〗^2 18°+cos^2⁡〖18°〗=1 ∴ cos^2⁡〖18°〗=1−〖𝑠𝑖𝑛〗^2 18° 𝑐𝑜𝑠⁡〖18°〗=√(1−〖𝑠𝑖𝑛〗^2 18°" " ) 𝑐𝑜𝑠⁡〖18°〗=√(1−((−𝟏 + √𝟓)/𝟒)^2 )=√(1−((𝟏 + 𝟓 − 𝟐√𝟓)/𝟏𝟔) )=√((16 − 6 + 2 √5)/16) 𝒄𝒐𝒔⁡〖𝟏𝟖°〗=√((𝟏𝟎 + 𝟐 √𝟓)/𝟒) 𝒔𝒊𝒏 𝟕𝟐°=sin⁡〖(90−18°)〗= 𝑐𝑜𝑠⁡〖18°〗=√((10 + 2 √5)/4) cos 36 and sin 36 𝑐𝑜𝑠⁡〖36°〗=1−2 sin^2⁡〖18°〗 ∴ 𝑐𝑜𝑠⁡〖36°〗=1−2((√5 − 1)/4)^2=1((5 + 1 2√6)/8)=1−((6−2√5)/8) =(8 − 6 + 2√5)/8=(2 + 2√5)/8=(√5 + 1)/4 〖𝑠𝑖𝑛^2〗⁡〖36°〗+cos^2⁡〖36°〗=1 〖𝑠𝑖𝑛^2〗⁡〖36°〗=1 −cos^2⁡〖36°〗 ∴ 〖𝑠𝑖𝑛^2〗⁡〖36°〗=1 −((√5 + 1)/4)^2=1−((5 + 1 + 2√5)/16)=(16 − 6 − 2√5)/16 〖𝑠𝑖𝑛^2〗⁡〖36°〗 = (10 − 2√5)/16 𝑠𝑖𝑛⁡〖36°〗=√(10 − 2√5) /4 sin 54° sin 54°=sin⁡(90°−36°)=cos⁡36°=(√5 + 1)/4

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.