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That is not correct: they are always LESS-THAN-OR-EQUAL to 1. They are also always greater-or-equal to minus 1.
The fact that this is so follows directly from the definition. Two different definitions are commonly used:
1) The sine is the y-coordinate of a unit circle (a circle of radius 1, with center at coordinates (0, 0)). The cosine is the x-coordinate of the same circle. The highest point on the circle has the y-coordinate 1 (at 90 degrees).
2) In a right triangle, the sine is the side opposite to the angle considered, divided by the hypothenuse. The hypothenuse is of course longer than the other sides (except in the extreme case of 0 or 90 degree angles - this would not be properly called a triangle).
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Which trigonometric functions always have values less than 1?
The sine and the cosine are always less than one.
What is cosine squared theta equal to'?
Cosine squared theta = 1 + Sine squared theta
Why is the sine and cosine of a number no less than 1 and no greater than 1?
lol! it can be less than 1 too, upto -1! it cannot be greater than 1 because hypotenuse is always longer than the adjacent and opposite side... (from pythagoras theorem)
Why sinθ value is less then and equal to one?
If you look at the definition of the sine function in a triangle, you'll discover that the maximum possible value of the sine function is ' 1 ' and the minimum possible value is ' -1 '. There's no angle that can have a sine greater than ' 1 ' or less than ' -1 '. So the absolute value of the sine of anything is always ' 1 ' or less.
How can you solve 1 plus cosx equals sinx?
1 + cos(x) = sin(x)==> You need to find an angle whose sine is 1 greater than its cosine.The numerical values of both the sine and cosine functions range from -1 to +1.No angle has a sin or cosine less than -1 or greater than +1. That'll help us putsome constraints on the equation, and see what may be going on.The equation also says: sin(x) - cos(x) = 1This would be a great place to flash a sketch of the graphs of the sin(x) and cos(x)functions up on the screen, and see where they differ by roughly 1, with the sinebeing the greater one. It's too bad that we can't do that. The best we can do is todraw them on our scratch pad here, look at them, and tell you what we see:-- The sine is greater than the cosine only between 45° and 225°,so any solutions must be in that range of angles.-- At 90°, the sine is 1 and the cosine is zero, so we have [ 1 + 0 = 1 ], and 90° definitely works.-- At 180°, the sine is zero and the cosine is -1, we have [ 1 + -1 = 0 ], and 180° works.-- If there were any range between 45° and 225° where the graphs of the sineand cosine functions were parallel curves, then any angle in that range mightalso be a solution.But there isn't any such place. 90° and 180° are the only points where the valuesare different by 1 and the sine is greater, so those are the only principle solutions(answers between zero and 360°.)
