<span>sinx - cosx =sqrt(2)
Taking square on both sides:
</span>(sinx - cosx)^2 =sqrt(2)^2<span>
sin^2(x) -2cos(x)sin(x) + cos^2(x) = 2
Rearranging the equation:
sin^2(x)+cos^2(x) -2cos(x)sin(x)=2
As,
</span><span>sin^2(x)+cos^2(x) = 1
</span><span>So,
1-2sinxcosx=2
1-1-2sinxcosx=2-1
-</span><span>2sinxcosx = 1
</span><span>Using Trignometric identities:
-2(0.5(sin(x+x)+sin(x-x))=1
-sin2x+sin0=1
As,
sin 0 = 0
So,
sin2x+0 = -1
</span><span>sin2x = -1</span><span>
2x=-90 degrees + t360
Dividing by 2 on both sides:
x=-45 degrees + t180
or 2x=270 degrees +t360
x= 135 degrees + t180 where t is integer</span>
First what you need to do is find the HCF of both numbers.
After you've done that if it is 1 then they are what we call co-prime.
Now, if they are other than 1 they are not co-prime.
Answer: A. an Exponent
Step-by-step explanation:
Point Slope Form of a Line: y - y₁ = m(x - x₁)
Slope Formula:
y₂ - y₁ 4 - (-1) 5 The slope is 5/6.
------------- = -------------- = ------
x₂ - x₁ 8 - 2 6
(2, -1) & (8,4)
x₁ y₁ x₂ y₂
Using all of the information above...
y + 1 = 5/6(x - 2) is your equation in point slope form.
Steps for rewriting this equation in standard form:
Standard Form of a linear equation: ax + by = c
1). Put the equation from PS form into Slope Intercept Form, which is y=mx+b
y + 1 = 5/6(x - 2)
y + 1 = 5/6x - 5/3
- 1 - 1
------------------------
y = 5/6x - 8/3
2). Put the equation from Slope Intercept Form into Standard Form.
y = 5/6x - 8/3
- 5/6x - 5/6x
______________________
(-1) -5/6x + y(-1) = -8/3(-1)
5/6x - y = 8/3
The equation in standard form is 5/6x - y = 8/3.
I hope this helps you!
