if tan^2 α/2= 2tan^2 ∅/2 +1, then Prove that
cos α+ sin^2 ∅/2 =0
Answers
sin4(x) – cos4(x) = (sin2(x) + cos2(x))(sin2(x) – cos2(x))
The first factor, sin2(x) + cos2(x), is always equal to 1, so I can ignore it. This leaves me with:
sin2(x) – cos2(x)
Hmm... I'm not seeing much of anything here. But I do know, glancing back at the RHS of the identity, that I need more sines and fewer cosines. I think I'll try using the Pythagorean identity that simplified that first factor, but in a slightly different form. If sin2(x) + cos2(x) = 1, then cos2(x) = 1 – sin2(x), and:
sin2(x) – cos2(x) = sin2(x) – (1 – sin2(x)) = sin2(x) – 1 + sin2(x) = 2sin2(x) – 1
And that's what I needed. For my hand-in work, I'll put it all together:
sin4(x) – cos4(x) = (sin2(x) + cos2(x))(sin2(x) – cos2(x))
= 1(sin2(x) – cos2(x)) = sin2(x) – cos2(x) = sin2(x) – (1 – sin2(x))
= sin2(x) – 1 + sin2(x) = sin2(x) + sin2(x) – 1 = 2sin2(x) – 1
Answer:
Answer:
A) \frac{(856+167)^2+(856-167)^2}{856×856+167×167} A)
856×856+167×167
(856+167)
2
+(856−167)
2
{(a + b)}^{2} + {(a - b)}^{2} = 2( {a}^{2} + {b}^{2} )(a+b)
2
+(a−b)
2
=2(a
2
+b
2
)
so \: \frac{2( {856}^{2} + {167}^{2} ) }{( {856}^{2} + {167}^{2} )} = 2so
(856
2
+167
2
)
2(856
2
+167
2
)
=2
1st question answer is 2
2)\:a^2+b^2+c^2= > 792)a
2
+b
2
+c
2
=>79
and \: ab +bc+ca= > 45andab+bc+ca=>45
{(a + b + c)}^{2} = {a}^{2} + {b}^{2} + {c}^{2} + 2(ab + bc + ca)(a+b+c)
2
=a
2
+b
2
+c
2
+2(ab+bc+ca)
so \: {(a + b + c)}^{2} = 79 + 2(45) = 79 + 90 = 169so(a+b+c)
2
=79+2(45)=79+90=169
{(a + b + c)} = 13 \: or \: - 13(a+b+c)=13or−13
2nd question answer is 13 or -13
3)x + y + z = 143)x+y+z=14
{x}^{2} + {y}^{2} + {z}^{2} = 50x
2
+y
2
+z
2
=50
{(x + y + z)}^{2} = {x}^{2} + {y}^{2} + {z}^{2} + 2(xy + yz + xz)(x+y+z)
2
=x
2
+y
2
+z
2
+2(xy+yz+xz)
{14}^{2} = 50 + 2(xy + yz + xz)14
2
=50+2(xy+yz+xz)
196 - 50 = 2(xy + yz + zx)196−50=2(xy+yz+zx)
146 = 2(xy + yz + zx)146=2(xy+yz+zx)
(xy + yz + zx) = 73(xy+yz+zx)=73
3rd answer is 73.