can someone explain a simple trig concept for me ! (1 Viewer)
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RivalryofTroll
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sin(180-x) = sin x (according to ASTC)
cos(90-x) should be sinx.
therefore
sin(180-x)/cos(90-x) should be sinx/sinx which is 1.
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You could either remember it in terms of quadrants.
If you assume x is an acute angle, 180-x must be in the second quadrant and have a related angle (angle with the x axis) of "x" degrees. Also, sin is positive in the second quadrant.
So, sin(180-x) = +sin(x)
You can simplify draw up a right angle triangle, mark one angle as (180-x) and since angles in a triangle sum to 180 you know the other angle in the triangle must be "x" degrees. Then it is easy to show that cos(180-x)= sin(x).
Or, since this is in the 3 unit maths section, I will assume you know the sum and difference angle formulas.
If you forget the above in an exam, you can just go the long way and expand everything out.
= sin(180-x) / cos(90-x)
= [ sin(180)cos(x) - cos(180)sin(x) ] / [ cos(90)cos(x) +sin(90)sin(x) ]
= [ 0 - (-1)sin(x) ] / [ 0 + (1) sin(x) ]
= sin(x)/sin(x)
= 1
If you assume x is an acute angle, 180-x must be in the second quadrant and have a related angle (angle with the x axis) of "x" degrees. Also, sin is positive in the second quadrant.
So, sin(180-x) = +sin(x)
You can simplify draw up a right angle triangle, mark one angle as (180-x) and since angles in a triangle sum to 180 you know the other angle in the triangle must be "x" degrees. Then it is easy to show that cos(180-x)= sin(x).
Or, since this is in the 3 unit maths section, I will assume you know the sum and difference angle formulas.
If you forget the above in an exam, you can just go the long way and expand everything out.
= sin(180-x) / cos(90-x)
= [ sin(180)cos(x) - cos(180)sin(x) ] / [ cos(90)cos(x) +sin(90)sin(x) ]
= [ 0 - (-1)sin(x) ] / [ 0 + (1) sin(x) ]
= sin(x)/sin(x)
= 1
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The answer definitely should be 1. Either we are all wrong, and the textbook defies simple trig logic. Or the answer is wrong/misprint etc.
Test it, enter any radian value into the expression, and you should always get 1.
Another way of look at it is this.
If you graph sin(pi-x), the original sine graph will get translated to the left, however since it is only pi AND flipped downwards. But since Sine is symetrical and has a period of 2pi. The graph will look like the original sinx graph.
When you graph cos(pi/2-x), the original graph is shifted to the left pi/2 units. However if you enter it into a graphing program, then two graphs are the same thing. This is because cos is (you can imagine it this way) the same as sin except shifted to the left pi/2 units. (Same as sin(-x) when shifted to the right).
So since they are the same for all values of x, when you divide them, it will always equal to one.
However I just found this out. When dividing them, when x=0, the answer is 0/0, which is undefined. (same result for pi etc.)
But, since I accidentally learnt L Hopitals while looking for Ext1 vids from patrickJMT....
You dont need to know this, its just a proof. But its also proving that the expression is true for certain values of x (i.e. 0, pi etc.)
EDIT: Hang on, I dont know why I proved that, it still doesnt mean that for x=0, pi etc. there is a solution. So yeah there is no solution for those values.
Test it, enter any radian value into the expression, and you should always get 1.
Another way of look at it is this.
If you graph sin(pi-x), the original sine graph will get translated to the left, however since it is only pi AND flipped downwards. But since Sine is symetrical and has a period of 2pi. The graph will look like the original sinx graph.
When you graph cos(pi/2-x), the original graph is shifted to the left pi/2 units. However if you enter it into a graphing program, then two graphs are the same thing. This is because cos is (you can imagine it this way) the same as sin except shifted to the left pi/2 units. (Same as sin(-x) when shifted to the right).
So since they are the same for all values of x, when you divide them, it will always equal to one.
However I just found this out. When dividing them, when x=0, the answer is 0/0, which is undefined. (same result for pi etc.)
But, since I accidentally learnt L Hopitals while looking for Ext1 vids from patrickJMT....
You dont need to know this, its just a proof. But its also proving that the expression is true for certain values of x (i.e. 0, pi etc.)
EDIT: Hang on, I dont know why I proved that, it still doesnt mean that for x=0, pi etc. there is a solution. So yeah there is no solution for those values.
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