# when to use sine vs cosine vs tangent

1. I'm a little confused about how you choose to use either sine or cosine or tangent over the others. Are they interchangeable given the same information you have about a right triangle? What are the circumstances that should dictate the use of one over the other? Or is it preference?

2. If you know the sides you can work out the angles. If you know the angle you can work out the length of the sides. Is this a correct assumption? So, is it correct so assume that if you know one of the angles besides the 90 degree angle and 1 length of one side you can determine the sine, cosine and tangent of that triangle? And if you know two sides you can always determine the angle? Is that a correct assumption?

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If you know only angles you can't tell anything about the sides. –  Shahab Feb 14 at 11:41
@Shahab Anything? Ratios, definitely. –  Sawarnik Feb 14 at 11:53
@Sawarnik: Yes I mean you cannot tell the length of the sides. –  Shahab Feb 14 at 11:54
@Shahab what if you have the angle and the length of 1 side? –  Jessica M. Feb 14 at 11:54
@JessicaM. Have you studied about things like law of cosines? Or have you just learned the definitions of trig functions in a right angle triangle. –  Sawarnik Feb 14 at 11:55

1 I'm a little confused about how you choose to use either sine or cosine or tangent over the others. Are they interchangeable given the same information you have about a right triangle? What are the circumstances that should dictate the use of one over the other? Or is it preference?

I assume that as in part 2 you have a right angled triangle. In this case you can work out the angle given two of the sides. Depending on which sides you have, you should choose sin, cos or tan, as shown in the diagram below. \begin{align} \sin d &= \frac{\text{opposite side}}{\text{hypoteneuse side}} \\ \cos d &= \frac{\text{adjacent side}}{\text{hypoteneuse side}} \\ \tan d &= \frac{\text{opposite side}}{\text{adjacent side}} \\ \end{align}

There are relationships between the different trig functions, e.g. $\sin d = \sqrt{1 - \cos^2 d}$. However it's a lot easier to find them directly given a standard right-angled triangle problem.

For reference, the identities that you can use are:

\begin{align} \sin^2 x+\cos^2 x&=1 \\ \tan^2 x+1&=\sec^2 x, \;\;\;\;\;\, \text{where } \sec x = 1/\cos x\\ 1+\cot^2 x&=\mathrm{cosec}^2 x, \;\;\; \text{where } \cot x = 1/\tan x, \mathrm{cosec}\,x=1/\sin x \end{align}

2 If you know the sides you can work out the angles.

Yes

If you know the angle you can work out the length of the sides. Is this a correct assumption?

No, see the right-hand side of the diagram. You can have similar triangles where the angles are all the same but the side lengths are scaled up or down by some factor, like Russian dolls.

If you think about it, this makes sense, because all the sin, cos, or tan gives you is the ratio between sides. E.g. for sin, how many times bigger is the opposite side than the hypoteneuse.

So, is it correct so assume that if you know one of the angles besides the 90 degree angle and 1 length of one side you can determine the sine, cosine and tangent of that triangle?

Strictly speaking, we talk about the sine, cosine and tangent of angles not triangles. A typical problem setting is that you are given one of the angles and one of the sides in a right-angled triangle. You can then work out the length of another side. E.g. if you have an angle $d$ and the opposite side you can rearrange the sine formula to find the hypoteneuse side.

\begin{align} \sin d &= \frac{\text{opposite side}}{\text{hypoteneuse side}} \\ \text{hypoteneuse side}\times\sin d &= \text{opposite side} \\ \text{hypoteneuse side} &= \frac{\text{opposite side}}{\sin d} \end{align}

Also note that if you have one angle and the 90 degree angle you can work out the third angle because the angles add to 180 degrees. And you can then work out all the side -- but you need at least one side to fix the scale factor as discussed above.

And if you know two sides you can always determine the angle? Is that a correct assumption?

Yes that's correct. See part 1.

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You beat me to the answer by few seconds :( –  Sawarnik Feb 14 at 12:21
You can add about these identites: $$\sin ^2x + \cos^2 x = 1$$ $$\tan^2x+1 = \sec^2x$$ $$1+\cot^2x=\csc^2x$$ –  Sawarnik Feb 14 at 12:22
@Sawarnik Sorry! And thankyou for the identities, I will add them. –  TooTone Feb 14 at 14:07
@TooTone Thank you very much for the great answer. It took me a little while to grasp the concepts. –  Jessica M. Feb 16 at 8:05
@JessicaM. thanks for the comment, happy to have helped. I personally didn't enjoy trigonometry at all at school, and I've only recently begun to appreciate it having studied maths at a more advanced level. It was nice to revisit what I learnt a long time ago given the view I have now. (If you keep on with maths you'll see sin and cos redefined, in, in my opinion, a more general, and elegant way, in terms of the y and x coordinates of a point as it moves around a circle.) –  TooTone Feb 17 at 0:08

Are they interchangeable, given the same information you have about a right triangle?

$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad$ Yes.

If $\sin x$ is known, then $\cos x=\sqrt{1-\sin^2x}$ and $\tan x=\dfrac{\sin x}{\cos x}=\dfrac{\sin x}{\sqrt{1-\sin^2x}}$ . Similarly for $\cos x$. If $\tan x$ is known, then $\cos x=\dfrac1{\sqrt{1+\tan^2x}}$ and $\sin x=\dfrac{\tan x}{\sqrt{1+\tan^2x}}$ . Basically, given

Pythagoras' theorem, if we know the ratio between any two sides, the other two ratios soon follow.

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thank you for your help. –  Jessica M. Feb 16 at 8:06