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3 years ago

If cos θ = -0.6, and 180° < θ < 270°, find the exact value of sin 2θ.

Mathematics

1 answer:

sveta [45]3 years ago

7 0

Answer:

sin(2θ) = 0.96

Step-by-step explanation:

In the third quadrant, both sin(θ) and cos(θ) are negative. Then the double-angle trig identity tells us ...

sin(2θ) = 2·sin(θ)·cos(θ) = -2cos(θ)√(1 -cos(θ)²) . . . . using the negative root

Filling in the given value, we have

sin(2θ) = -2·(-0.6)(√(1-(-0.6)²) = 2·0.6·0.8 = 0.96

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r-ruslan [8.4K]

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ehidna [41]

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Explanation:

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Nataly [62]

<h3>Answers:</h3>

$f(g(x)) = \sqrt{x^2+5}+5\\\\g(f(x)) = x+30+10\sqrt{x-1}$

================================================

Work Shown:

Part 1

$f(x) = \sqrt{x-1}+5\\\\f(g(x)) = \sqrt{g(x)-1}+5\\\\f(g(x)) = \sqrt{x^2+6-1}+5\\\\f(g(x)) = \sqrt{x^2+5}+5\\\\$

Notice how I replaced every x with g(x) in step 2. Then I plugged in g(x) = x^2+6 and simplified.

------------------

Part 2

$g(x) = x^2+6\\\\g(f(x)) = \left(f(x)\right)^2+6\\\\g(f(x)) = \left(\sqrt{x-1}+5\right)^2+6\\\\g(f(x)) = \left(\sqrt{x-1}\right)^2+2*5*\sqrt{x-1}+\left(5\right)^2+6\\\\g(f(x)) = x-1+10\sqrt{x-1}+25+6\\\\g(f(x)) = x+30+10\sqrt{x-1}\\\\$

In step 4, I used the rule (a+b)^2 = a^2+2ab+b^2

In this case, a = sqrt(x-1) and b = 5.

You could also use the box method as a visual way to expand out $\left(\sqrt{x-1}+5\right)^2$

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