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alekssr [168]
3 years ago
10

(xsinA-ucosA)^2+(xcosA+ysinA)^2=x^2+y^2​

Mathematics
2 answers:
Artist 52 [7]3 years ago
7 0

Answer:

u = x tan(A) - sec(A) sqrt(x^2 + y^2 - (x cos(A) + y sin(A))^2) or u = sec(A) sqrt(x^2 + y^2 - (x cos(A) + y sin(A))^2) + x tan(A)

Step-by-step explanation:

Solve for u:

(x sin(A) - u cos(A))^2 + (x cos(A) + y sin(A))^2 = x^2 + y^2

Subtract (x cos(A) + y sin(A))^2 from both sides:

(x sin(A) - u cos(A))^2 = x^2 + y^2 - (x cos(A) + y sin(A))^2

Take the square root of both sides:

x sin(A) - u cos(A) = sqrt(x^2 + y^2 - (x cos(A) + y sin(A))^2) or x sin(A) - u cos(A) = -sqrt(x^2 + y^2 - (x cos(A) + y sin(A))^2)

Subtract x sin(A) from both sides:

-u cos(A) = sqrt(x^2 + y^2 - (x cos(A) + y sin(A))^2) - x sin(A) or x sin(A) - u cos(A) = -sqrt(x^2 + y^2 - (x cos(A) + y sin(A))^2)

Divide both sides by -cos(A):

u = x tan(A) - sec(A) sqrt(x^2 + y^2 - (x cos(A) + y sin(A))^2) or x sin(A) - u cos(A) = -sqrt(x^2 + y^2 - (x cos(A) + y sin(A))^2)

Subtract x sin(A) from both sides:

u = x tan(A) - sec(A) sqrt(x^2 + y^2 - (x cos(A) + y sin(A))^2) or -u cos(A) = -x sin(A) - sqrt(x^2 + y^2 - (x cos(A) + y sin(A))^2)

Divide both sides by -cos(A):

Answer:  u = x tan(A) - sec(A) sqrt(x^2 + y^2 - (x cos(A) + y sin(A))^2) or u = sec(A) sqrt(x^2 + y^2 - (x cos(A) + y sin(A))^2) + x tan(A)

wolverine [178]3 years ago
6 0

Answer:

Step-by-step explanation:

(x sinA - y cosA)²+(x cosA + y sinA)²=

=x²sin²A+y²cos²A-2xysinA cosA +

+ x²cos²A+y²sin²A+2xy sin A cos A

2xy sin A cos A and -2xy sin A cos A gives 0 and when you sum other summands you get:

x²sin²A + y²cos²A + x²cos²A + y² sin²A =

= x²(sin²A+cos²A) + y²(sin²A+cos²A)

= x²*1+y²*1=x²+y²

So this equation is an identity. Is true. Because sin²x+cos²x=1.

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