Answer:
To prove that ( sin θ cos θ = cot θ ) is not a trigonometric identity.
Begin with the right hand side:
R.H.S = cot θ =
L.H.S = sin θ cos θ
so, sin θ cos θ ≠ 
So, the equation is not a trigonometric identity.
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<u>Anther solution:</u>
To prove that ( sin θ cos θ = cot θ ) is not a trigonometric identity.
Assume θ with a value and substitute with it.
Let θ = 45°
So, L.H.S = sin θ cos θ = sin 45° cos 45° = (1/√2) * (1/√2) = 1/2
R.H.S = cot θ = cot 45 = 1
So, L.H.S ≠ R.H.S
So, sin θ cos θ = cot θ is not a trigonometric identity.
9514 1404 393
Answer:
11.3 units
Step-by-step explanation:
The distance can be found using the distance formula:
d = √((x2 -x1)² +(y2 -y1)²)
d = √((6 -(-2))² +(-4-4)²) = √(8² +(-8)²) = √(64·2)
d = 8√2 ≈ 11.3
The distance is approximately 11.3 units.
You can say
x+y=5
x=5-y
You know the value of x, put it into the next equation.
3(5-y)-7y=19
15-3y-7y=19
15-19 = 10y
-4=10y
-4/10=y
-2/5=y
Put y value back into either equation to find value of x.
x+y=5
x-2/5=5
x=5+2/5
x=27/5
Therefore, x,y = (27/5, -2/5)
Hope I helped :)