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3 years ago

4 Tan A/1-Tan^4=Tan2A + Sin2A

Mathematics

1 answer:

Eva8 [605]3 years ago

7 0

tan(2A) + sin(2A) = sin(2A)/cos(2A) + sin(2A)

• rewrite tan = sin/cos

… = 1/cos(2A) (sin(2A) + sin(2A) cos(2A))

• expand the functions of 2A using the double angle identities

… = 2/(2 cos²(A) - 1) (sin(A) cos(A) + sin(A) cos(A) (cos²(A) - sin²(A)))

• factor out sin(A) cos(A)

… = 2 sin(A) cos(A)/(2 cos²(A) - 1) (1 + cos²(A) - sin²(A))

• simplify the last factor using the Pythagorean identity, 1 - sin²(A) = cos²(A)

… = 2 sin(A) cos(A)/(2 cos²(A) - 1) (2 cos²(A))

• rearrange terms in the product

… = 2 sin(A) cos(A) (2 cos²(A))/(2 cos²(A) - 1)

• combine the factors of 2 in the numerator to get 4, and divide through the rightmost product by cos²(A)

… = 4 sin(A) cos(A) / (2 - 1/cos²(A))

• rewrite cos = 1/sec, i.e. sec = 1/cos

… = 4 sin(A) cos(A) / (2 - sec²(A))

• divide through again by cos²(A)

… = (4 sin(A)/cos(A)) / (2/cos²(A) - sec²(A)/cos²(A))

• rewrite sin/cos = tan and 1/cos = sec

… = 4 tan(A) / (2 sec²(A) - sec⁴(A))

• factor out sec²(A) in the denominator

… = 4 tan(A) / (sec²(A) (2 - sec²(A)))

• rewrite using the Pythagorean identity, sec²(A) = 1 + tan²(A)

… = 4 tan(A) / ((1 + tan²(A)) (2 - (1 + tan²(A))))

• simplify

… = 4 tan(A) / ((1 + tan²(A)) (1 - tan²(A)))

• condense the denominator as the difference of squares

… = 4 tan(A) / (1 - tan⁴(A))

(Note that some of these steps are optional or can be done simultaneously)

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3 years ago

4 Tan A/1-Tan^4=Tan2A + Sin2A​

4 Tan A/1-Tan^4=Tan2A + Sin2A