<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.
<u>also</u><u> read</u><u> similar</u><u> questions</u><u>:</u> If sin Θ = 2/3 and tan Θ < 0, what is the value of cos Θ?
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if cos θ -sin θ =1, find θ cos θ +sin θ?
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Answer: 7/17
So there are 7 blue marbles and there are 17 marbles in total.
That means the probability of picking one blue marble is 7/17
Answer:
(41/12)g+-23/12
Step-by-step explanation:
3(5g−1)
/4 - (2g+7)/6
(15g-3)/4 - (2g+7)/6---> distributive property
3(15g-3)/12 - 2(2g+7)/12--->make the denominators same
(45g-9)/12 - (4g+14)/12---distributive property
Answer:
Step-by-step explanation:
Problem A
t(1) = 2(1) + 5
t(2) = 2*2 + 5 = 9
t(3) = 2*3 + 5 = 11
t(4) = 2*4 + 5 = 13
So this is the explicit result. Now try it recursively.
t_3 = t_2 + 2
t_3 = 9 + 2
t_3 = 11 which is just what it should do.
t_n = t_(n - 1) + 2
Problem B
t(1) = 3 * 1/2
t(1) = 3/2
t(2) = 3*(1/2)^2
t(2) = 3 * 1/4
t(2) = 3/4
t(3) = 3*(1/2)^3
t(3) = 3 * 1/8
t(3) = 3/8
t(4) = 3 (1/2)^4
t(4) = 3 (1/16)
t(4) = 3/16
So in general
t_n = t_n-1 * 1/2
For example t(5)
t_5 = t_4 * 1/2
t_5 = 3 /16 * 1/2 = 3/32
Let the third side be x :
( 3√5 )^2 = 6^2 + x^2
45 = 36 + x^2
x^2 = 45 - 36
x^2 = 9
x = 3
