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sveta [45]
3 years ago
7

Suppose theta= 11pi/12. How do you use the sum identity to find the exact value of sin theta?

Mathematics
1 answer:
SIZIF [17.4K]3 years ago
4 0
The better way is, first we have to find the equivalent in degrees

2\pi=360\º

\frac{11\pi}{12}=345\º

now we can change this value to -15\º

how do we get an angle like this?!

30\º-45\º=-15\º

then

sin(30\º-45\º)=sin(30\º)*cos(45\º)-sin(45\º)*cos(30\º)

\begin{Bmatrix}sin(30\º)&=&\frac{1}{2}\\\\sin(45\º)&=&cos(45\º)&=&\frac{\sqrt{2}}{2}}\end{matrix}\\\\cos(30\º)&=&\frac{\sqrt{3}}{2}\end{matrix}

now we replace this values

sin(-15\º)=\frac{1}{2}*\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}*\frac{\sqrt{3}}{2}

sin(-15\º)=\frac{\sqrt{2}}{4}-\frac{\sqrt{6}}{4}

\boxed{\boxed{sin(-15\º)=sin(345\º)=\frac{\sqrt{2}-\sqrt{6}}{4}}}
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3 years ago
The length of an edge of a cube is 5 inches. What is the volume of the cube? Enter your answer in the box.
adelina 88 [10]

Answer:

125 cubed inches

Step-by-step explanation:

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8 0
3 years ago
Which of the following is a polynomial with roots negative square root of 5, square root of 5, and 3?
Alex Ar [27]

Answer:

x^{3}-3x^{2}-5x+15

Step-by-step explanation:

The roots of the given polynomial are: -\sqrt{5}, \sqrt{5}, 3

Since, -\sqrt{5}, \sqrt{5}, 3 are the roots of the polynomial, according to the factor theorem, (x - (-\sqrt{5})), (x-\sqrt{5}), (x-3) would be the factors of the polynomial.

Since we have the factors of the polynomial, we can multiply them to get the desired polynomial.

Let the polynomial be represented by P(x), so

P(x) = (x - (-\sqrt{5}))(x-\sqrt{5})(x-3)\\\\ P(x)=(x +\sqrt{5})(x-\sqrt{5})(x-3)\\\\ P(x)=(x^{2}-(\sqrt{5} )^{2})(x-3)\\\\ P(x)=(x^{2}-5)(x-3)\\\\ P(x)=x^{3}-3x^{2}-5x+15

The polynomial represented by P(x) has the given roots.

5 0
3 years ago
Tim is placing bricks in a bucket. The
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Answer:

6

Step-by-step explanation:

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8 0
3 years ago
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