Buckle up, 'cause we can't draw diagrams here and we have to explain everything. Let's jump. Draw a graph with an x-axis and a y-axis like usual. Don't use graph paper or a ruler unless you have to. Just eyeball the thing. We're going to draw a right triangle on the graph and here's how we'll do it. Start at the origin, (0, 0) and draw a line along the x-axis about "6 or 7 units" long. Now draw a line from the end of the first one straight up (at a right angle to the x-axis) and make it about "2 units" long. Lastly, draw the "slanted" line from the origin up to where the vertical line ended. That last line was the hypotenuse of your nice right triangle. Got a good picture? Super. Let's jump to some review. The trigonometry (trig) function called the "sine" (sin) is the relationship in any right triangle between the length of the opposite side (to an given angle in the triangle) and the length of the hypotenuse of that triangle. It's actually the length of the opposite side of the triangle divided by the length of the hypotenuse of the triangle. This number is a "pure" number without units because the units (inches, feet, miles - whatever) cancel out when the division is made. Now that we've reviewed the sine function, let's take it to our triangle. Look at the angle made by the first line you drew and the last one you drew (which was the hypotenuse). It's the angle with the origin of the graph (0, 0) as the vertex. It's gonna be 25 to 35 degrees or so, ballpark. We don't need to be exact. Now, the sine of that angle is the length of the opposite side divided by the length of the hypotenuse. It's some number between 0 and 1. The hypotenuse is obviously larger, and we'll end up with a fraction or, if you prefer, a decimal number. We don't need to know what it is because we are going to be looking at a "trend" or "shift" as we change our graph. We have some number as the sine, and we're good. Now let's modify our graph and draw a new triangle. Follow closely when we jump. We are going to "keep" the hypotenuse we drew. But we are going to "rotate it up" to make a new triangle. Note that we won't change its length. We're going to "open up" the angle between the x-axis and the hypotenuse. Let's do that by detatching the hypotenuse from the short vertical to the x-axis (which is that little second line we drew). Swing the hypotenuse up (that's counterclockwise from its first position) and put it about "half way" between where it was and where the y-axis is. Got it located? Now "drop a perpendicular" from the end of the hypotenuse to the x-axis, and make the line perpendicular to the x-axis. This forms a new right triangle. And this new triangle has a longer "second side" that is vertical to the x-axis. Let's look at our new triangle. The "new" angle formed by the x-axis and the new location of the hypotenuse is larger than it was. And the sine for that angle has changed. The sine is (again) the length of the opposite side over the length of the hypotenuse, and notice that the "new" opposite side is longer than the old one. (We can call that side, the one perpenducular to the x-axis, the "second side" here.) That means the "new" sine will be a larger fraction or a larger decimal (if you work it that way) than before. We don't know the exact number, but we only need to look at it in comparison to what it was. And it's bigger. So let's rotate the hypotenuse more. Start moving it in a slow but continuous motion in the counterclockwise direction. It's heading for the y-axis as you rotate it. Now focus. The new triangle formed as we rotate the hypotenuse (again, without changing its length) will have a longer and longer "perpendicular" to the x-axis as we move the hypotenuse. Pretend that the second side, the one we keep making longer as we rotate the hypotenuse up, is a rubber band stretching longer and longer as we rotate the hypotenuse. It still has to make a right angle where it is attached to the x-axis, so it must "slide along" that axis toward the origin to keep the angle at 90 degrees. Make sense? The triangle is "getting taller" as we rotate the hypotenuse. And the base is getting shorter and shorter. Through all this, the sine of the angle we are looking at is getting bigger and bigger. See how it works? One more jump. As the hypotenuse is rotated counter clockwise and approaches the y-axis, the length of that "second side" will continue to grow and will actually approach the length of the hypotenuse itself. (The triangle's base gets tinier and tinier through all this.) Our angle is getting bigger and bigger, too, and it is approaching 90 degrees. As the length of the second side approaches the length of the hypotenuse, the sine of the angle, that is, the length of the second side divided by the length of the hypotenuse, actually approaches one. That's because the second side is getting almost as long as the hypotenuse. Closer and closer to vertical we move that hypotenuse. At vertical, that is, when the hypotenuse is rotated to vertical, the triangle "disappears" from view, but imagine what is happeing as we approach this "vanishing point" where the triangle ceases to exist. At 90 degrees, the second side is the exact same length as the hypotenuse. That means the angle formed at the vertex becomes 90 degrees. And the base will be so short as to disappear as well. At the 90 degree point where the hypotenuse has been rotated up to lie along the y-axis, the length of the opposite will equal to the length of the hypotenuse. And the sine of the angle (which is 90 degrees) will be the length of the second side exactly 1 at this point. The sine of an angle varies as the measure of the angle, and as the angle increases in measure from 0 to 90 degrees, the sine of the angle varies from 0 to 1 as we discovered.

🙏

🤨

😮

😂

Q: How is sin 90 equal to 1?

Write your answer...

sin 90 is 1

It is: sin(90) = 1

sin(90) = 1

On the unit circle sin(90) degrees is at Y = 1 and as that is on the Y axis X will equal = 0. Ask yourself. Where would 90 degrees be on a 360 degree circle? Straight up.

No, it does not.

Sin squared is equal to 1 - cos squared.

The answer is 1.

sinx(1-sinx)=0 sinx=0 or 1 x= 0, 90, 180, 270, 360...

sin(30) = sin(90 - 60) = sin(90)*cos(60) - cos(90)*sin(60) = 1*cos(60) - 0*sin(60) = cos(60).

No. Sin of any angle is always less than or equal to 1.

In radians; -0.8939966636 In degrees; -1, of course

cosecant = 1/sine csc 90 deg = 1/(sin 90 deg) = 1/1 = 1

cos(35)sin(55)+sin(35)cos(55) If we rewrite this switching the first and second terms we get: sin(35)cos(55)+cos(35)sin(55) which is a more common form of the sin sum and difference formulas. Thus this is equal to sin(90) and sin(90)=1

The solution is found by applying the definition of complementary trig functions: Cos (&Theta) = sin (90°-&Theta) cos (62°) = sin (90°-62°) Therefore the solution is sin 28°.

yes!

sin-30 = (-1) x 1/(square root of 2) -sin30 = -(1/square root of 2) They are equal

1/2 of sin(2x)

sin x + csc x = 2 sin x + 1/sin x = 2 (sin2 x + 1)/sin x = 2/1 (cross multiply) sin2 x + 1 = 2sin x (subtract 2sin x to both sides) sin2 x - 2sinx + 1 = 0 (sin x - 1)2 = 0 (take the square root of both sides) sin x - 1 = 0 (add 1 to both sides) sin x = 1 x = sin-1 1 = 90⁰ 40x = 40(90⁰) = 3,600⁰ sin 40x = sin (3,600⁰ ) = 0 50x = 50(90⁰) = 4500⁰ sin 50x = sin (4,500⁰) = 0, so that csc 50x is undefined, we cannot divide by 0. Then, sin40x + csc50x is undefined.

Cotangent = 1/Tangent : Cosecant = 1/Sine Then, cot + 1 = (1/tan) + 1 = (cos/sin) + (sin/sin) = (cos + sin)/ sin. Now, cos² + sin² = 1 so for the statement to be valid the final expression would have to be : (cos² + sin² ) / sin = 1/sin. As this is not the case then, cot + 1 ≠ cosec. In fact, the relationship link is cot² + 1 = cosec²

tan(9) + tan(81) = sin(9)/cos(9) + sin(81)/cos(81)= {sin(9)*cos(81) + sin(81)*cos(9)} / {cos(9)*cos(81)} = 1/2*{sin(-72) + sin(90)} + 1/2*{sin(72) + sin(90)} / 1/2*{cos(-72) + cos(90)} = 1/2*{sin(-72) + 1 + sin(72) + 1} / 1/2*{cos(-72) + 0} = 2/cos(72) since sin(-72) = -sin(72), and cos(-72) = cos(72) . . . . . (A) Also tan(27) + tan(63) = sin(27)/cos(27) + sin(63)/cos(63) = {sin(27)*cos(63) + sin(63)*cos(27)} / {cos(27)*cos(63)} = 1/2*{sin(-36) + sin(90)} + 1/2*{sin(72) + sin(36)} / 1/2*{cos(-36) + cos(90)} = 1/2*{sin(-36) + 1 + sin(36) + 1} / 1/2*{cos(-36) + 0} = 2/cos(36) since sin(-36) = -sin(36), and cos(-36) = cos(36) . . . . . (B) Therefore, by (A) and (B), tan(9) - tan(27) - tan(63) + tan(81) = tan(9) + tan(81) - tan(27) - tan(63) = 2/cos(72) â€“ 2/cos(36) = 2*{cos(36) â€“ cos(72)} / {cos(72)*cos(36)} = 2*2*sin(54)*sin(18)/{cos(72)*cos(36)} . . . . . . . (C) But cos(72) = sin(90-72) = sin(18) so that sin(18)/cos(72) = 1 and cos(36) = sin(90-36) = sin(54) so that sin(54)/cos(36) = 1 and therefore from C, tan(9) â€“ tan(27) â€“ tan(63) + tan(81) = 2*2*1*1 = 4

sin(0) = 0, sin(90) = 1, sin(180) = 0, sin (270) = -1 cos(0) = 1, cos(90) = 0, cos(180) = -1, cos (270) = 0 tan(0) = 0, tan (180) = 0. cosec(90) = 1, cosec(270) = -1 sec(0) = 1, sec(180) = -1 cot(90)= 0, cot(270) = 0 The rest of them: tan(90), tan (270) cosec(0), cosec(180) sec(90), sec(270) cot(0), cot(180) are not defined since they entail division by zero.

Note that an angle should always be specified - for example, 1 - cos square x. Due to the Pythagorean formula, this can be simplified as sin square x. Note that sin square x is a shortcut of (sin x) squared.

2 sin2(x) + sin(x) - 1 = 0(2 sin + 1) (sin - 1) = 0Either 2 sin(x) + 1 = 02sin(x) = -1sin(x) = -0.5x = 210Â°, 330Â°or sin(x) - 1 = 0sin(x) = 1x = 90Â°

Sin(x) cos(x) = 1/2 of sin(2x)

sin of angle a = opposite/hypotenuse = 1/3 sin-1(1/3) = 19.47122063 degrees