Finding Value of trignometric functions, given angle

Chapter 3 Class 11 Trigonometric Functions
Concept wise

## Value of cos 36°, sin 36° and sin 54°

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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