1.) Apply the following method you used in #2! You should notably understand this already! Goodness grief!
2.) let A = arctan x
let B = arccos x
then we are looking for
sin(A + B) .
now,
tan A = x ,
sin A = x/√(x^2+1)
cos A = 1/√(x^2+1) ... create a right triangle where the opposite side is x, and the adjacent side is 1 ... the hypotenuse is √(x^2+1)
cos B = x
sin B = √(1 - x^2)
meanwhile
sin(A + B)
= sin A cos B + cos A sin B
= x/√(x^2+1) * x + 1/√(x^2+1) * √(1 - x^2)
= [x^2 + √(1 - x^2)] / (x^2 + 1) .
Write the trigonometric expression as an algebraic expression.
1.) SIN(ARCSIN X+ARCCOS X)
ANSWER: 1
2.) SIN(ARCTAN 2X-ARCCOS X)
ANSWER: 2x²-SquareRoot of 1-x²/Square
Root of 4x²+1.
1 answer