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Alina [70]
3 years ago
15

On an interval of [0, 2π), can the sine and cosine values of a radian measure ever be equal? If so, enter the radian measure(s)

where the values are equal. If not, enter DNE. (Enter your answers as a comma-separated list.)
Mathematics
1 answer:
Lisa [10]3 years ago
6 0

Answer:

Yes, they are equal in the values (in radians):

π/4, 5π/4

If cos(x) and sin(x) are defined to you as nonnegative functions (in terms of lengths), then 3π/4 and 7π/4 are also included

Step-by-step explanation:

Remember that odd multiples of 45° are special angles, with the same sine and cosine values (you can prove this, for example, by considering a right triangle with an angle of 45° and hypotenuse with length 1, and finding the trigonometric ratios).

The radian measure of 45° corresponds to π/4, hence the odd multiples on the interval [0, 2π) are π/4, 3π/4, 5π/4, 7π/4.

If you define sin(x) and cos(x) using the cartesian coordinate system (via unit circle), then cos(3π/4)=-sin(3π/4) and cos(7π/4)=-sin(7π/4). In this case, only π/4 and 5π/4 are valid choices.  

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2. Draw the image of RST under the dilation with scale factor 2/3 and center of dilation (1,-1). Label the image RST .
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Answer:

From the graph: we have the coordinates of RST i.e,

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or

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or

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Now,  

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the image R' = (\frac{2}{3}(2)+\frac{1}{3} , \frac{2}{3}(1)-\frac{1}{3} ) or

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For S = (2, -2) ,

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(\frac{4}{3}+\frac{1}{3} , \frac{-4}{3}-\frac{1}{3} )

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and For T = (-1, -2)

The image T' =  (\frac{2}{3}(-1)+\frac{1}{3} , \frac{2}{3}(-2)-\frac{1}{3} ) or

(\frac{-2}{3}+\frac{1}{3} , \frac{-4}{3}-\frac{1}{3} )

⇒ T' = (\frac{-1}{3} , \frac{-5}{3})

Now, label the image of RST on the graph as shown below in the attachment:

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