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Like if we have a right angle triangle and we want to find the sine, then we can simply just divide the perpendicular by hypotenuse. So why we need theta in sin theta= opp/hyp

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closed as unclear what you're asking by Henrik, M. Winter, JonMark Perry, Namaste, Simply Beautiful Art Sep 29 '17 at 19:41

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Here's a MathJax tutorial :) $\endgroup$ – Shaun Sep 29 '17 at 18:24
  • $\begingroup$ Because sometimes we do not know the length of these both sides. If you know just one you need another information to find the other one (and therewith the ratio), e.g. an angle. $\endgroup$ – M. Winter Sep 29 '17 at 18:25
  • $\begingroup$ $\sin\theta$ is the value of "opposite/hypothenuse" for a triangle with the angle $\theta$. Without $\theta$ there is no information what triangle we are talking about. Different triangles have different such ratios. The remarkable fact is that the ratio only depends on $\theta$ and we need no other information like e.g. the size if the triangle. So be lucky to only have a $\theta$ in your $\sin$ and nothing more ;) $\endgroup$ – M. Winter Sep 29 '17 at 18:32
  • $\begingroup$ "... and we want to find the sin" Halt! sin of what? $\endgroup$ – Namaste Sep 29 '17 at 18:36
  • $\begingroup$ I will point out as well that although triangles may have motivated the sine function and may have been the original use, the way sine is currently defined does not rely on relationships of triangles at all. This is part of why we can talk about $\sin(520^\circ)$ despite an angle of $520^\circ$ not making much sense in the context of what a triangle is. We define the sine of an "angle" in radians as $\sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\dots\pm\frac{x^{2n+1}}{(2n+1)!}\mp\dots$ $\endgroup$ – JMoravitz Sep 29 '17 at 18:45
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Consider any right triangle with angle $\theta$. Then the ratio $\dfrac{\text{opposite}}{\text{hypotenuse}}$, is same for all those triangles. This is a theorem which follows from similarity of triangles. What $\sin \theta$ does is that it gives a name to that ratio. That value is independent of the actual values of $\text{opposite}$ etc, which is a nice fact in itself.

This fact reduces the amount of data you need to carry to know that ratio. For general triangles, to know this ratio, you would need to know both opposite and hypotenuse (or some other things, but always you'd need atleast 2). But for right angled triangles, you need to know only one thing. Note that neither opposite nor hypotenuse are sufficient. For example $1-1-\sqrt{2}$, and $1-2-\sqrt{5}$ have the same value for $\text{opposite}$, i.e., $1$. but the ratio varies. But, it turns out that keeping that angle $\theta$ is sufficient, as stated in the first paragraph.

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Different triangles may have different sine ratios, so the angle $\theta$ helps you bookkeep this dependency.

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Your opposite stand for opposite side with respect to the angle. Writing only sin or cos it's the same of writing only $\sqrt{\phantom{h}}$. You need an argument for each function/operator standing for the quantity about you compute the function/operator.

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