Answer:
Please check the explanation.
Step-by-step explanation:
Given the vector
v = 6i + 2√3j
The Magnitude of a vector:




![\mathrm{Apply\:radical\:rule}:\quad \sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b}](https://tex.z-dn.net/?f=%5Cmathrm%7BApply%5C%3Aradical%5C%3Arule%7D%3A%5Cquad%20%5Csqrt%5Bn%5D%7Bab%7D%3D%5Csqrt%5Bn%5D%7Ba%7D%5Csqrt%5Bn%5D%7Bb%7D)


The Direction of a vector:
tan Ф = y/x
y=2√3
x = 6
tan Ф = y/x
= 2√3 / 6
= √3 / 3


Answer:
D = -7
Step-by-step explanation:
<u>EXPLANATION</u><u>:</u>
Given that
sin θ = 1/2
We know that
sin 3θ = 3 sin θ - 4 sin³ θ
⇛sin 3θ = 3(1/2)-4(1/2)³
⇛sin 3θ = (3/2)-4(1/8)
⇛sin 3θ = (3/2)-(4/8)
⇛sin 3θ = (3/2)-(1/2)
⇛sin 3θ = (3-1)/2
⇛sin 3θ = 2/2
⇛sin 3θ = 1
and
cos 2θ = cos² θ - sin² θ
⇛cos 2θ = 1 - sin² θ - sin² θ
⇛cos 2θ = 1 - 2 sin² θ
Now,
cos 2θ = 1-2(1/2)²
⇛cos 2θ = 1-2(1/4)
⇛cos 2θ = 1-(2/4)
⇛cos 2θ = 1-(1/2)
⇛cos 2θ = (2-1)/2
⇛cos 2θ = 1/2
Now,
The value of sin 3θ /(1+cos 2θ
⇛1/{1+(1/2)}
⇛1/{(2+1)/2}
⇛1/(3/2)
⇛1×(2/3)
⇛(1×2)/3
⇛2/3
<u>Answer</u> : Hence, the req value of sin 3θ /(1+cos 2θ) is 2/3.
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