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If cos ( t ) = 24 25 and t is in the fourth quadrant, find sin ( t ) .

sin ( t ) = 7 25

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Finding sines and cosines of special angles

We have already learned some properties of the special angles, such as the conversion from radians to degrees, and we found their sines and cosines using right triangles. We can also calculate sines and cosines of the special angles using the Pythagorean Identity.

Finding sines and cosines of 45° Angles

First, we will look at angles of 45° or π 4 , as shown in [link] . A 45° 45° 90° 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.

Graph of 45 degree angle inscribed within a circle with radius of 1. Equivalence between point (x,y) and (x,x) shown.

At t = π 4 , which is 45 degrees, the radius of the unit circle bisects the first quadrantal angle. This means the radius lies along the line y = x . A unit circle has a radius equal to 1 so the right triangle formed below the line y = x has sides x and y   ( y = x ) , and radius = 1. See [link] .

Graph of circle with pi/4 angle inscribed and a radius of 1.

From the Pythagorean Theorem we get

x 2 + y 2 = 1

We can then substitute y = x .

x 2 + x 2 = 1

Next we combine like terms.

2 x 2 = 1

And solving for x , we get

x 2 = 1 2 x = ± 1 2

In quadrant I, x = 1 2 .

At t = π 4 or 45 degrees,

( x , y ) = ( x , x ) = ( 1 2 , 1 2 ) x = 1 2 , y = 1 2 cos  t = 1 2 , sin  t = 1 2

If we then rationalize the denominators, we get

cos  t = 1 2 2 2 = 2 2 sin  t = 1 2 2 2 = 2 2

Therefore, the ( x , y ) coordinates of a point on a circle of radius 1 at an angle of 45° are ( 2 2 , 2 2 ) .

Finding sines and cosines of 30° And 60° Angles

Next, we will find the cosine and sine at an angle of 30° , or π 6 . First, we will draw a triangle inside a circle with one side at an angle of 30° , and another at an angle of −30° , 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 60° , as shown in [link] .

Graph of a circle with 30-degree angle and negative 30-degree angle inscribed to form a triangle.
Image of two 30/60/90 triangles back to back. Label for hypotenuse r and side y.

Because all the angles are equal, the sides are also equal. The vertical line has length 2 y , and since the sides are all equal, we can also conclude that r = 2 y or y = 1 2 r . Since sin t = y ,

sin ( π 6 ) = 1 2 r

And since r = 1 in our unit circle,

sin ( π 6 ) = 1 2 ( 1 ) = 1 2

Using the Pythagorean Identity, we can find the cosine value.

cos 2 ( π 6 ) + sin 2 ( π 6 ) = 1 cos 2 ( π 6 ) + ( 1 2 ) 2 = 1 cos 2 ( π 6 ) = 3 4 Use the square root property . cos ( π 6 ) = ± 3 ± 4 = 3 2 Since  y  is positive, choose the positive root .

The ( x , y ) coordinates for the point on a circle of radius 1 at an angle of 30° are ( 3 2 , 1 2 ) . At t = π 3  (60° ), the radius of the unit circle, 1, serves as the hypotenuse of a 30-60-90 degree right triangle, B A D , as shown in [link] . Angle A has measure 60° . At point B , we draw an angle A B C with measure of 60° . We know the angles in a triangle sum to 180° , so the measure of angle C is also 60° . Now we have an equilateral triangle. Because each side of the equilateral triangle A B C is the same length, and we know one side is the radius of the unit circle, all sides must be of length 1.

Graph of circle with an isosceles triangle inscribed that has been divided in half.  The resulting triangle has a radius of 1 and a height of y.  The two bases for the triangles each have a length of x.

The measure of angle A B D is 30°. Angle A B C is double angle A B D , so its measure is 60°. B D is the perpendicular bisector of A C , so it cuts A C in half. This means that A D is 1 2 the radius, or 1 2 . Notice that A D is the x -coordinate of point B , which is at the intersection of the 60° angle and the unit circle. This gives us a triangle B A D with hypotenuse of 1 and side x of length 1 2 .

Questions & Answers

f(x)=x/x+2 given g(x)=1+2x/1-x show that gf(x)=1+2x/3
Ken Reply
proof
AUSTINE
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Manifoldee Reply
how to solve x²=2x+8 factorization?
Kristof Reply
x=2x+8
Manifoldee
×=2x-8 minus both sides by 2x
Manifoldee
so, x-2x=2x+8-2x
Manifoldee
then cancel out 2x and -2x, cuz 2x-2x is obviously zero
Manifoldee
so it would be like this: x-2x=8
Manifoldee
then we all know that beside the variable is a number (1): (1)x-2x=8
Manifoldee
so we will going to minus that 1-2=-1
Manifoldee
so it would be -x=8
Manifoldee
so next step is to cancel out negative number beside x so we get positive x
Manifoldee
so by doing it you need to divide both side by -1 so it would be like this: (-1x/-1)=(8/-1)
Manifoldee
so -1/-1=1
Manifoldee
so x=-8
Manifoldee
SO THE ANSWER IS X=-8
Manifoldee
so we should prove it
Manifoldee
x=2x+8 x-2x=8 -x=8 x=-8 by mantu from India
mantu
lol i just saw its x²
Manifoldee
x²=2x-8 x²-2x=8 -x²=8 x²=-8 square root(x²)=square root(-8) x=sq. root(-8)
Manifoldee
I mean x²=2x+8 by factorization method
Kristof
I think x=-2 or x=4
Kristof
x= 2x+8 ×=8-2x - 2x + x = 8 - x = 8 both sides divided - 1 -×/-1 = 8/-1 × = - 8 //// from somalia
Mohamed
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Find the possible value of 8.5 using moivre's theorem
Reuben Reply
which of these functions is not uniformly cintinuous on (0, 1)? sinx
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which of these functions is not uniformly continuous on 0,1
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Jamiz Reply
X=7
Muustapha
=7
mantu
x=7
mantu
3x-4x-7=0 -x=7 x=-7
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x=-7
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9x-16x-49=0 -7x=49 -x=7 x=7
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-x=7
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deals with circles, angles, and triangles. Usually in the form of Soh cah toa or sine, cosine, and tangent
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solve x
Rubben
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it will b -y+2=x
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If one side only of a triangle is given is it possible to solve for the unkown two sides?
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Practice Key Terms 3

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Source:  OpenStax, Algebra and trigonometry. OpenStax CNX. Nov 14, 2016 Download for free at https://legacy.cnx.org/content/col11758/1.6
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