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cos sin question (1 Viewer)

emilios

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You kind of just have to know that cos(90-theta)=sin(theta) and vice versa, BUT there is intuition behind this.

Draw up any right angled triangle and label one of the angles as theta. Since all the angles of a triangle add up to 180, the remaining angle must be 90-theta. Now take notice that if you take the sine of one angle [Opposite/Hypotenuse) , it will give you the exact same result as taking the cosine of the other angle (Adjacent/Hypotenuse). That's the basic idea of the result.

So your problem is reduced down to sin(theta)/cos(theta) = tan(theta)
 

scischool

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Aug 31, 2011
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On @emilios's point, the quickest way to solve this is to use the fact that cos(x) and sin(x) are related to each other by shifting cos(x) to the right by 90 deg, i.e. cos(90-theta) = cos(theta-90) = sin(theta). And for the denominator, sin(90-theta) = cos(theta).

But this is stuff that you covered before Year 12, so it could be a little hazy. You can use the compound-angle results: cos(A-B) = cos(A)cos(B)+sin(A)sin(B) & sin(A-B) = sin(A)cos(B)-sin(B)cos(A), to arrive simply the numerator as:

cos(90)cos(theta)+sin(90)sin(theta) = sin(theta), since cos(90)=0 & sin(90)=1.

In the denominator,

sin(90)cos(theta)-sin(theta)cos(90) = cos(theta).

Hence, they divide out to give tan(theta).

If you'd like more practice of harder HSC-style questions, check out the Sci School HSC programs in the first week of the holidays. Sample notes showing the style of the lecture slides are online. :read: :cool:
 

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