If tanA=2(x+1)/2x+1 find sinA and cosA in terms of x
Anish Mondal , 7 Years ago
Grade 9
Trigonometry> If tanA=2(x+1)/2x+1 find sinA and cosA in...
1 Answers
Arun
Last Activity: 7 Years ago
tanA = 2x(x+1) / 2x+1 ---- given
sec^2 A = 1+ tan^2 A --- identity.
sec^2 A = 1 + [4x^2(x+1)^2] / (2x+1)^2 = [(2x+1)^2 + 4x^2(x+1)^2] / (2x+1)^2
= [4x^4 + 8x^3 + 8x^2 + 4x +1 ] / (2x+1)^2
= (2x^2 + 2x + 1)^2 / (2x+1)^2 This gives --secA = (2x^2 + 2x + 1) / (2x + 1)
So cosA = 1/secA = (2x + 1) / (2x^2 + 2x + 1) --------- [1]
Now sin^2 A = (1 - cos^2 A ) = 1 - (2x+1)^2 / (2x^2 + 2x + 1)^2
= [(2x^2 + 2x + 1)^2 - (2x+1)^2] / (2x^2 + 2x + 1)^2
= (2x^2)*(2x^2 + 4x + 2 ) / (2x^2 + 2x + 1)^2 ---- using [a^2-b^2 =(a+b)*(a-b) ]
= (4x^2)*(x^2+2x+1) / (2x^2 + 2x + 1)^2 = [(2x )* (x+1)]^2 / (2x^2 + 2x + 1)^2
This gives sinA = (2x )* (x+1) / (2x^2 +2x +1) --------- [2]
sec^2 A = 1+ tan^2 A --- identity.
sec^2 A = 1 + [4x^2(x+1)^2] / (2x+1)^2 = [(2x+1)^2 + 4x^2(x+1)^2] / (2x+1)^2
= [4x^4 + 8x^3 + 8x^2 + 4x +1 ] / (2x+1)^2
= (2x^2 + 2x + 1)^2 / (2x+1)^2 This gives --secA = (2x^2 + 2x + 1) / (2x + 1)
So cosA = 1/secA = (2x + 1) / (2x^2 + 2x + 1) --------- [1]
Now sin^2 A = (1 - cos^2 A ) = 1 - (2x+1)^2 / (2x^2 + 2x + 1)^2
= [(2x^2 + 2x + 1)^2 - (2x+1)^2] / (2x^2 + 2x + 1)^2
= (2x^2)*(2x^2 + 4x + 2 ) / (2x^2 + 2x + 1)^2 ---- using [a^2-b^2 =(a+b)*(a-b) ]
= (4x^2)*(x^2+2x+1) / (2x^2 + 2x + 1)^2 = [(2x )* (x+1)]^2 / (2x^2 + 2x + 1)^2
This gives sinA = (2x )* (x+1) / (2x^2 +2x +1) --------- [2]
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