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This is a list of equations in my book:

$$\sin(90-\theta)=\cos\theta$$ $$\cos(90-\theta)= \sin\theta$$ $$\tan(90-\theta)=\cot\theta$$

or this...

$$\sin(180-\theta)=\sin\theta$$ $$\cos(180-\theta)= -\cos\theta$$ $$\tan(180-\theta)=-\tan\theta$$

The list goes on and on $(90+\theta)$, $(180+\theta)$ etc

I can find these equations by drawing a unit circle and then figure the above but in our exams, we need to answer the questions related to these equations very quickly.

Example:

the value of $\cos(675)$

The answer is $\sin(45)$... I need to answer such a question very quickly. So, could someone please tell me how I could remember all these equations and apply them without using unit circles which take a long time to do.

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There is one, and only one, answer to this. Think of the functions graphically. Ask yourselves what they mean and how they relate to each other graphically? Always go back to the unit circle and relate the functions based on the geometry. I will now demonstrate.

First of all, for $90^\circ$ and $180^\circ$, they become very simple to remember. First, draw a unit circle:

Next, realize that if $\theta$ is the angle from the $+x$ axis, then $ \sin \theta $ gives you the $y$-coordinate, and $ \cos \theta $ gives you the $x$-coordinate. If you did not know this, think about why this is.

From here, it should be easy geometry. If you take an arbitrary acute angle $\theta$ (acute for simplicity), and you add $90^\circ$ onto it, where does it land? Draw this for yourself. Now, write the coordinates of the new endpoint in terms of the trigonometric functions and equate.

For $180^\circ$, it's even less geometry. It is simply a reflection across the origin. For example, I know that $ \sin \left( 180^\circ + \theta \right) = - \sin x $ because I know what the $180^\circ$ shift does to the vector/arrow I see in the unit circle.

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One thing that you could do is whenever you figure out a rule for one problem, write it down on a piece of scrap paper. That way, you only have to figure it out once.

Another trick is to realize that $90^\circ-\theta$ is $\theta$s complement. Thus $\sin(90^\circ-\theta)=\text{complement's sine }(\theta)$ or $\cos\theta$. Similarly, $\cot$ means the complement's tangent, and $csc$ mean the complement's secant.

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