Explore the properties of inverse trigonometric functions such as sine, cosine, and tangent, along with their proofs. Discover relationships between these functions and learn how to derive various trigonometric identities. Dive into the world of trigonometry and expand your knowledge with detailed explanations and examples.
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Presentation Transcript
Properties of Inverse Properties of Inverse Trigonometric Functions Trigonometric Functions 1. (I) sin-1(1/x) = cosec-1x. (II) cos-1(1/x) = sec-1x (III) tan-1(1/x) = cot-1x Proof: Let cosec-1x = y => x = cosec y => x = 1/ sin y => 1/x = sin y => sin-1(1/x) = y => sin-1(1/x) = cosec-1x In the same way we can proof other properties
2. (I) sin-1x + cos-1x = /2 , x[-1,1] (II) tan-1x + cot-1x = /2 , x R (III) cosec-1x + sec-1x = /2, |x|>= 1 Proof : Let sin-1x = y Then x = sin y => x = cos(( /2)- y) => cos-1x = ( /2)- y => cos-1x + y = /2 => cos-1x + sin-1x = /2 Hence sin-1x + cos-1x = /2 In the same way we can proof other properties
3. (I) tan-1x + tan-1y = tan-1[(x + y)/(1 - x. y) ] (II) tan-1x tan-1 y = tan-1[(x y) /(1 + x. y) ] Proof: Let tan-1x = and tan-1y = Then we have x = tan and y = tan Now tan( + ) = (tan + tan )/(1 tan . tan ) => tan( + ) = (x + y) /(1 x. y) => + = tan-1[(x + y) /(1 x. y) ] hence tan-1x + tan-1y = tan-1[(x + y) /(1 x. y) ] In the same way we can proof second property
4. (I) 2tan-1x = sin-1[2x/(1 + x2)] (II) 2tan-1x = cos-1[(1 x2)/(1 + x2)] (III) 2tan-1x = tan-1[2x/(1 x2)] Proof : Let tan-1x = y Now sin-1[2x/(1 + x2)] = sin-1[2tan y/(1 + tan2y)] (putting value of y) = sin-1(sin 2y) = 2y. = 2tan-1x Hence 2tan-1x = sin-1[2x/(1 + x2)] (using identity of sin 2y) ( as sin-1(sin x) = x)
5. (I) sin-1x = tan-1[x/(1 x2)] (II) cos-1x = tan-1[( 1 x2)/x] Proof : Let x = sin y => y = sin-1x Therefor 1 x2= 1 sin2y = cos2y = cos y => tan-1[x/ 1 x2] = tan-1(sin y/ cos y) = tan-1(tan y) => tan-1[x/ 1 x2] = y => tan-1[x/ 1 x2] = sin-1x Hence sin-1x = tan-1[x/ 1 x2] Similarly we can proof second property. (as tan-1(tan x) = x)
6. (I) sin-1x = cos-11 x2 (II) cos-1x = sin-1 1 x2 (III) cos ( sin-1x) = sin ( cos-1x) = 1 x2 Proof : Do yourself Use trigonometric identity of sin x and cos x 7. (I) sin-1x + sin-1y = sin-1(x. 1 y2+ y. 1 x2) (II) sin-1x - sin-1y = sin-1(x. 1 y2- y. 1 x2) (III) cos-1x + cos-1y = cos-1(x.y - 1 x2. 1 y2) (iv) cos-1x - cos-1y = cos-1(x.y + 1 x2. 1 y2)
Proof : (I) sin-1x + sin-1y = sin-1(x 1 y2+ y 1 x2) Let sin-1x = and sin-1y = => sin-1(x 1 y2+ y 1 x2 = sin and y = sin => cos = 1 sin2 = 1 x2 And cos = 1 sin2 = 1 y2 Now sin ( + ) = sin . cos + cos . sin sin( + ) = x 1 y2+ y 1 x2 ( + ) = sin-1(x 1 y2+ y 1 x2 sin-1x+ sin-1y = sin-1(x 1 y2+ y 1 x2)
Proof : (III) cos-1x + cos-1y = cos-1(x.y - 1 x2. 1 y2) Let cos-1x = and cos-1y = Then x = cos and y = cos => sin = 1 cos2 and sin = 1 cos2 => sin = 1 x2and sin = 1 y2 Now cos( + ) = cos .cos sin . sin - cos .cos cos( + ) = x. y - 1 x2. 1 y2 = > + = cos-1(x.y - 1 x2. 1 y2) cos-1x + cos-1y = cos-1(x.y - 1 x2. 1 y2) Similarly we can proof other properties
8. (I) 3 sin-1x = sin-1(3x 4x3) (II) 3 cos-1x = cos-1(4x3 3x) (III) 3 tan-1x = tan-1[(3x x3)/(1 3x2)] Proof : Let sin-1x = y => x = sin y sin 3y = 3 sin y 4 sin3y = 3x 4x3 => 3y = sin-1(3x 4x3) => 3sin-1x = sin-1(3x 4x3) Similarly we can proof other properties.
1. Prove that sin-1(3/5) - sin-1(8/17) = cos-1(84/85) Sol. We have to prove that sin-1(3/5) + sin-1(8/17) = cos-1(8/85) L. H. S. sin-1(3/5) + sin-1(8/17) Let sin-1 (3/5) = x => sin x= 3/5 . => cos x = 1 sin2x => cos x = 1 (3/5)2 => cos x = 1 (9/25) => cos x = (25 9)/ 25 = 16/25= 4/5 Again Let sin-1(8/17) = y => sin y = 8/17
=> cos y = 1 sin2y= 1 (8/17)2 Cos y = 1 ( 64/289) = (289 64)/289 => cos y = 225/289= 15/17 Now cos ( x - y) = cos x. cosy + sin x. siny = 4/5. 15/17 + 3/5. 8/17 . = 60/85 + 24/85 = (60+24)/85 = 84/85 . => x y = cos-1(84/85) => sin-1(3/5) sin-1(8/17) = cos-1(84/85) 2. Prove that tan-1(1/5) + tan-1(1/7) + tan-1(1/3) + tan-1(1/8) = /4 ( hint : use the property of tan-1x + tan-1y = tan-1[ ( x + y)/ (1 x. y) ] 3. cos-1(12/13) + sin-1(3/5) = sin-1(56/64) ( solve as ex. 1 )
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