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can someone explain a simple trig concept for me ! (1 Viewer)

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sin(pi - x ) / cos[(pi/2) - x]

how does this equal to sinx/cosx??

like could someone explain this very simply because i have a real bad foundation with the complementary angles!
 

RivalryofTroll

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sin(pi - x ) / cos[(pi/2) - x]

how does this equal to sinx/cosx??

like could someone explain this very simply because i have a real bad foundation with the complementary angles!
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.
 

iSplicer

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sin(pi - x ) / cos[(pi/2) - x]

how does this equal to sinx/cosx??

like could someone explain this very simply because i have a real bad foundation with the complementary angles!
It's not = sinx/cosx, it's 1.

EDIT: See Rivalry's post =]
 
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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
 

Drongoski

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sin(pi - x) = sin x whatever the value of x (in radians) be it 0.3, -37.089 or 2000000. That's the beauty of the identity. For small x, we can interpret in terms of the 4 quadrants(ASTC) as pointed out above.
 

Sy123

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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.
 
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zeebobDD

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if the answer is tanx are you sure the bottom isnt sin(90-x)?
 

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