Question

If x = r sinAcosC, y=rsinAsinC, z=rcosA, Prove that r²=x²+y²+z².

Answer 1

x = rsinAcosC
y= rsinAsinC
z= rcosA
squaring and adding all,

x² +y²+z² =( rsinAcosC)²+(rsinAsinC)+(rcosA)²
               =r²[ sin²Acos²C + sin²Asin²C + cos²A]
               =r² [sin²A(cos²C + sin²C) + cos²A]    ∵cos²α+sin²α =1
               =r²[sin²A+cos²A]    
               =r²

proved

Answer 2

Step-by-step explanation:

x= rsinAcosC ----1

y= rsinAsinC -----2

z= rcosA-----------3

squaring eqⁿ 1, 2 and 3, we get

x²= r²sin²Acos²C ----4

y²= r²sin²Asin²C -----5

z²= r²cos²A ------------6

Adding 4, 5 and 6, we get

x²+ y²+ z²= r²sin²Acos²C+ r²

sin²Asin²C+ r²cos²A

x²+ y²+ z²= r²sin²A(cos²C +sin²C)+

r²cos²A

x²+ y²+ z²= r²sin²A+ r²cos²A

-(sin²∅+cos²∅=1)

x²+ y²+ z²= r²(sin²A+ cos²A)

x²+ y²+ z²= r²

-(sin²∅+cos²∅=1)

Hence, proved. Hope this helps you.