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If $\mathrm{cos}(t)=\frac{24}{25}$ and $t$ is in the fourth quadrant, find $\text{sin}(t).$
$\mathrm{sin}(t)=-\frac{7}{25}$
We have already learned some properties of the special angles, such as the conversion from radians to degrees. We can also calculate sines and cosines of the special angles using the Pythagorean Identity and our knowledge of triangles.
First, we will look at angles of $\text{\hspace{0.17em}}\mathrm{45\xb0}\text{\hspace{0.17em}}$ or $\text{\hspace{0.17em}}\frac{\pi}{4},$ as shown in [link] . A $\text{\hspace{0.17em}}\mathrm{45\xb0}\u2013\mathrm{45\xb0}\u2013\mathrm{90\xb0}\text{\hspace{0.17em}}$ triangle is an isosceles triangle, so the x- and y -coordinates of the corresponding point on the circle are the same. Because the x- and y -values are the same, the sine and cosine values will also be equal.
At $\text{\hspace{0.17em}}t=\frac{\pi}{4}$ , which is 45 degrees, the radius of the unit circle bisects the first quadrantal angle . This means the radius lies along the line $\text{\hspace{0.17em}}y=x.\text{\hspace{0.17em}}$ A unit circle has a radius equal to 1. So, the right triangle formed below the line $\text{\hspace{0.17em}}y=x\text{\hspace{0.17em}}$ has sides $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}y\text{}(y=x),$ and a radius = 1. See [link] .
From the Pythagorean Theorem we get
Substituting $\text{\hspace{0.17em}}y=x,$ we get
Combining like terms we get
And solving for $\text{\hspace{0.17em}}x,$ we get
In quadrant I, $\text{\hspace{0.17em}}x=\frac{1}{\sqrt{2}}.\text{\hspace{0.17em}}$
At $\text{\hspace{0.17em}}t=\frac{\pi}{4}\text{\hspace{0.17em}}$ or 45 degrees,
If we then rationalize the denominators, we get
Therefore, the $\text{\hspace{0.17em}}(x,y)\text{\hspace{0.17em}}$ coordinates of a point on a circle of radius $\text{\hspace{0.17em}}1\text{\hspace{0.17em}}$ at an angle of $\text{\hspace{0.17em}}\mathrm{45\xb0}\text{\hspace{0.17em}}$ are $\text{\hspace{0.17em}}\left(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right).\text{\hspace{0.17em}}$
Next, we will find the cosine and sine at an angle of $\text{\hspace{0.17em}}\mathrm{30\xb0},$ or $\text{\hspace{0.17em}}\frac{\pi}{6}\text{\hspace{0.17em}}$ . First, we will draw a triangle inside a circle with one side at an angle of $\text{\hspace{0.17em}}\mathrm{30\xb0},$ and another at an angle of $\text{\hspace{0.17em}}\mathrm{-30\xb0},$ as shown in [link] . If the resulting two right triangles are combined into one large triangle, notice that all three angles of this larger triangle will be $\text{\hspace{0.17em}}\mathrm{60\xb0},$ as shown in [link] .
Because all the angles are equal, the sides are also equal. The vertical line has length $\text{\hspace{0.17em}}2y,$ and since the sides are all equal, we can also conclude that $\text{\hspace{0.17em}}r=2y\text{\hspace{0.17em}}$ or $\text{\hspace{0.17em}}y=\frac{1}{2}r.\text{\hspace{0.17em}}$ Since $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}t=y\text{\hspace{0.17em}}$ ,
And since $\text{\hspace{0.17em}}r=1\text{\hspace{0.17em}}$ in our unit circle ,
Using the Pythagorean Identity, we can find the cosine value.
The $\text{\hspace{0.17em}}(x,y)\text{\hspace{0.17em}}$ coordinates for the point on a circle of radius $\text{\hspace{0.17em}}1\text{\hspace{0.17em}}$ at an angle of $\text{\hspace{0.17em}}\mathrm{30\xb0}\text{\hspace{0.17em}}$ are $\text{\hspace{0.17em}}\left(\frac{\sqrt{3}}{2},\frac{1}{2}\right).\text{\hspace{0.17em}}$ At $t=\frac{\pi}{3}$ (60°), the radius of the unit circle, 1, serves as the hypotenuse of a 30-60-90 degree right triangle, $\text{\hspace{0.17em}}BAD,$ as shown in [link] . Angle $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ has measure $\text{\hspace{0.17em}}\mathrm{60\xb0}.\text{\hspace{0.17em}}$ At point $\text{\hspace{0.17em}}B,$ we draw an angle $\text{\hspace{0.17em}}ABC\text{\hspace{0.17em}}$ with measure of $\text{\hspace{0.17em}}\mathrm{60\xb0}.\text{\hspace{0.17em}}$ We know the angles in a triangle sum to $\text{\hspace{0.17em}}\mathrm{180\xb0},$ so the measure of angle $\text{\hspace{0.17em}}C\text{\hspace{0.17em}}$ is also $\text{\hspace{0.17em}}\mathrm{60\xb0}.\text{\hspace{0.17em}}$ Now we have an equilateral triangle. Because each side of the equilateral triangle $\text{\hspace{0.17em}}ABC\text{\hspace{0.17em}}$ is the same length, and we know one side is the radius of the unit circle, all sides must be of length 1.
The measure of angle $\text{\hspace{0.17em}}ABD\text{\hspace{0.17em}}$ is 30°. So, if double, angle $\text{\hspace{0.17em}}ABC\text{\hspace{0.17em}}$ is 60°. $\text{\hspace{0.17em}}BD\text{\hspace{0.17em}}$ is the perpendicular bisector of $\text{\hspace{0.17em}}AC,$ so it cuts $\text{\hspace{0.17em}}AC\text{\hspace{0.17em}}$ in half. This means that $\text{\hspace{0.17em}}AD\text{\hspace{0.17em}}$ is $\text{\hspace{0.17em}}\frac{1}{2}\text{\hspace{0.17em}}$ the radius, or $\text{\hspace{0.17em}}\frac{1}{2}.\text{\hspace{0.17em}}$ Notice that $\text{\hspace{0.17em}}AD\text{\hspace{0.17em}}$ is the x -coordinate of point $\text{\hspace{0.17em}}B,$ which is at the intersection of the 60° angle and the unit circle. This gives us a triangle $\text{\hspace{0.17em}}BAD\text{\hspace{0.17em}}$ with hypotenuse of 1 and side $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ of length $\text{\hspace{0.17em}}\frac{1}{2}.\text{\hspace{0.17em}}$
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