Do have to MEMORIZE certain trigonometric relations?

Vikash

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My doubt is do we have to memorize trigonometric relations between sin(θ) and cos(θ) for example : cosθ =cos(-θ ), sin (90-θ )=cosθ , sin(-x) = -sinx and so on....or is there any other way we can UNDERSTAND to find the relations rather than memorizing. Help is really appreciated....
 
At the indicated thread, there is a pdf that answers your question.


As for understanding, the unit circle is a great tool
 
At the indicated thread, there is a pdf that answers your question.


As for understanding, the unit circle is a great tool
Kind of yes anyways thanks...
 
The hypotenuse is always positive. Now if an angle is in quadrant 1 (or 4), then the negative angle is in quadrant 4 (or 1). If an angle is in quadrant 2 (or 3) then the negative angle is in quadrant 3 (or 2). Since the sign of the adjacent side does not change when we consider an angle or its negative we have the cosine function being even. However the sine function will be odd since the opposite sides will have a sign change. I see this clearly so I do not have to remember this.

I remember that sin^2 x + cos^2 x = 1. I do not remember the other two phythagorean identities. I just divide both sides by cos^2 x or sin^2 x as needed.

If I remember the sin(x+y) then I do not have to remember the sin(2x) as I can just let y=x in the sin(x+y) formula. I also do not have to remember sin(x-y) as sin(x-y) = sin(x + (-y)) and I know the sin(-y) and cos(-y) using just angle y.

Same as above for cos(x+y).

You do not have to remember tan(x+/- y) as they follow from sin(x+/-y)/cos(x+/-y).

The list goes on!
 
The only one that is perhaps not obvious from the unit circle is (using radian measure)

[MATH]sin(\theta) - cos \left ( \dfrac{\pi}{2} - \theta \right ) = 0 = cos(\theta) - sin \left ( \dfrac{\pi}{2} - \theta \right ).[/MATH]
That one is obvious from using the triangular definition of the sine and cosine.
 
The only one that is perhaps not obvious from the unit circle is (using radian measure)

[MATH]sin(\theta) - cos \left ( \dfrac{\pi}{2} - \theta \right ) = 0 = cos(\theta) - sin \left ( \dfrac{\pi}{2} - \theta \right ).[/MATH]
That one is obvious from using the triangular definition of the sine and cosine.

I learned this identity long ago by the fact that cosine is the sine of an angle’s complement.
 
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