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I was wondering if anyone here knows how to find the value of $\sin (90 - a)$ using a right-angled triangle. I can find the value using $\sin (a - b)$ but that's too lengthy and the value comes something like $\cos a$... the following might be helpful, but I'm not sure how.

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  • $\begingroup$ $sin(a-b)$ is the way to go, then use your knowledge of $\cos a$ and $\sin a$ when $a=90$. $\endgroup$ – Suzu Hirose Nov 13 '14 at 5:30
  • $\begingroup$ And notice that it's $90$ degrees. Angles are usually measured in radians in mathematics (and also on computers or handhelds, is case you are also doing numeric experiments). $\endgroup$ – Jean-Claude Arbaut Nov 13 '14 at 7:21
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There is a reason why we don't call the ratio $\dfrac{ADJACENT}{HYPOTENUSE}$ the HORIZONTALUS. It is because both of the non-right angles of a right triangle are related to each other, as they are called complimentary angles. Instead we call said ratio the $COSINE$, as an abbreviation of the COMPLIMENTS SINE. Suppose a right triangle has angles $\theta$, $\theta^\prime$ and $90^\circ$, then clearly $\theta+\theta^\prime$ must equal $90^\circ$ (because the sum of the angles of a triangle must be $180^\circ$). Both $\theta$ and $\theta^\prime$ share the same hypotenuse, but note that the side that is opposite $\theta$ is adjacent to $\theta^\prime$, and vice versa (see diagram).

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Therefore the following relationships must hold. $$\sin\theta^\prime =\dfrac{a}{c}=\cos\theta$$ $$\sec\theta^\prime =\dfrac{c}{b}= \csc\theta$$ $$\tan\theta^\prime =\dfrac{a}{b}= \cot\theta$$ But substituting $90^\circ-\theta$ for $\theta^\prime$ we have $$\sin (90^\circ-\theta) = \cos\theta$$ $$\sec (90^\circ-\theta) = \csc\theta$$ $$\tan (90^\circ-\theta) = \cot\theta$$

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  • $\begingroup$ Try Googling "triangular function"... then try Googling "circular function". Do you know why we always begin trigonometry with triangles and Satanic chants like "SOHCAHTOA"? I have no idea! $\endgroup$ – John Joy Nov 13 '14 at 15:32
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Hint: Think of a right angled triangle with angles 90, a and 90-a then use soh cah toa stuff and see what you get.

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  • $\begingroup$ how do i use that.. idk any of the sides of the triangle... $\endgroup$ – SRC SRC Nov 13 '14 at 5:47
  • $\begingroup$ Because if you let the hypotenuse be C the side adjacent to angle a be A and the side adjacent to angle 90-a be B you have sin(90-a)= A/C and cos(a)=A/C by trigonometric ratios. $\endgroup$ – John Marty Nov 13 '14 at 5:53
  • $\begingroup$ therefore sin(90-a)=cos(a)... i get it now.. thanks... is this the simplest way there is... or is there anything simpler than that..? and how about sin(180-a) or cos(180)? $\endgroup$ – SRC SRC Nov 13 '14 at 6:01
  • $\begingroup$ you can also do it by considering the unit circle. Sin(180-a) is not equal to cos(180). Perhaps you mean sin(180-x)=sin(x) This is true by the unit circle or you can use the fact that sin(a-b) =sin(a)cos(b)-cos(a) sin (b) so sin (180-x) = sin 180 cos(-x) - cos(180) sin (-x)= 0 - - -1 *sin (-x)= sin (x) $\endgroup$ – John Marty Nov 13 '14 at 6:08
  • $\begingroup$ you wouldn't mind if i ask you how to use the unit circle to solve that... thanks again... $\endgroup$ – SRC SRC Nov 13 '14 at 6:13
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Using the unit circle:

Draw a unit circle with a dot at $90^\circ$. Now see if the value of the sine varies as the dot moves $a$ clockwise ($90^\circ-a$) and $a$ counterclockwise $(90^\circ+a)$.

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