- #1
- 19,865
- 10,850
Definition/Summary
In a right-angled triangle, with a hypotenuse ("hyp"), and with sides adjacent ("adj") and opposite ("opp") to the acute angle we are interested in, the six basic functions are defined as follows:
sin = opp/hyp, cos = adj/hyp, tan = opp/adj,
cosec = 1/sin, sec = 1/cos, cot = 1/tan.
Equations
Memorize this equation:
\cos^2x\,+\,\sin^2x\,=\,1
(it comes from Pythagoras' theorem: \mathrm{adj}^2\,+\,\mathrm{opp}^2\,=\,\mathrm{hyp}^2)
Divide the equation by \cos^2x, and rearrange terms to get:
\sec^2x\,-\,\tan^2x\,=\,1
Divide it instead by \sin^2x, and rearrange terms to get:
\mathrm{cosec}^2x\,-\,\cot^2x\,=\,1
Extended explanation
These last four equations are too difficult to remember
, but when needed you can work them out as follows 
…
* This entry is from our old Library feature. If you know who wrote it, please let us know so we can attribute a writer. Thanks!
In a right-angled triangle, with a hypotenuse ("hyp"), and with sides adjacent ("adj") and opposite ("opp") to the acute angle we are interested in, the six basic functions are defined as follows:
sin = opp/hyp, cos = adj/hyp, tan = opp/adj,
cosec = 1/sin, sec = 1/cos, cot = 1/tan.
Equations
Memorize this equation:
\cos^2x\,+\,\sin^2x\,=\,1
(it comes from Pythagoras' theorem: \mathrm{adj}^2\,+\,\mathrm{opp}^2\,=\,\mathrm{hyp}^2)
Divide the equation by \cos^2x, and rearrange terms to get:
\sec^2x\,-\,\tan^2x\,=\,1
Divide it instead by \sin^2x, and rearrange terms to get:
\mathrm{cosec}^2x\,-\,\cot^2x\,=\,1
Extended explanation
\cos2x\,=\,\cos^2x\,-\,\sin^2x
1\,+\,\cos2x\,=\,2\,\cos^2x
1\,-\,\cos2x\,=\,2\,\sin^2x
\sin{2x}\,=\,2\,\sin{x}\,\cos{x}
\sin(x\,+\,y)\,=\,\sin x\cos y\,+\,\cos x\sin y
\sin(x\,-\,y)\,=\,\sin x\cos y\,-\,\cos x\sin y
\cos(x\,+\,y)\,=\,\cos x\cos y\,-\,\sin x\sin y
\cos(x\,-\,y)\,=\,\cos x\cos y\,+\,\sin x\sin y
1\,+\,\cos2x\,=\,2\,\cos^2x
1\,-\,\cos2x\,=\,2\,\sin^2x
\sin{2x}\,=\,2\,\sin{x}\,\cos{x}
\sin(x\,+\,y)\,=\,\sin x\cos y\,+\,\cos x\sin y
\sin(x\,-\,y)\,=\,\sin x\cos y\,-\,\cos x\sin y
\cos(x\,+\,y)\,=\,\cos x\cos y\,-\,\sin x\sin y
\cos(x\,-\,y)\,=\,\cos x\cos y\,+\,\sin x\sin y
You must learn all the equations above. 
A\sin x\,+\,B\cos x\,=\,\sqrt{(A^2+B^2)}\sin (x\,+\,\tan^{-1}(B/A))
. . . . . . . . . . . . .[/color] =\,\sqrt{(A^2+B^2)}\cos (x\,-\,\tan^{-1}(A/B))
\sin x\,+\,\sin y\,=\,2 \sin \frac{x\,+\,y}{2} \cos \frac{x\,-\,y}{2}
\sin x\,-\,\sin y\,=\,2 \sin \frac{x\,-\,y}{2} \cos \frac{x\,+\,y}{2}
\cos x\,+\,\cos y\,=\,2 \cos \frac{x\,+\,y}{2} \cos \frac{x\,-\,y}{2}
\cos x\,-\,\cos y\,=\,-2 \sin \frac{x\,+\,y}{2} \sin \frac{x\,-\,y}{2}
. . . . . . . . . . . . .[/color] =\,\sqrt{(A^2+B^2)}\cos (x\,-\,\tan^{-1}(A/B))
\sin x\,+\,\sin y\,=\,2 \sin \frac{x\,+\,y}{2} \cos \frac{x\,-\,y}{2}
\sin x\,-\,\sin y\,=\,2 \sin \frac{x\,-\,y}{2} \cos \frac{x\,+\,y}{2}
\cos x\,+\,\cos y\,=\,2 \cos \frac{x\,+\,y}{2} \cos \frac{x\,-\,y}{2}
\cos x\,-\,\cos y\,=\,-2 \sin \frac{x\,+\,y}{2} \sin \frac{x\,-\,y}{2}
These last four equations are too difficult to remember
, but when needed you can work them out as follows …They all have a 2, an (x+y)/2, and an (x-y)/2, and …
Sum or difference of sin always has a cos and a sin, just as in sin(x±y).
Sum or difference of cos always has two coses or two sines, just as in cos(x±y).
And a sum doesn't depend on the order, so it has to have cos the difference, which also doesn't; while a difference does, so it has to have sin the difference, which also does.
Sum or difference of sin always has a cos and a sin, just as in sin(x±y).
Sum or difference of cos always has two coses or two sines, just as in cos(x±y).
And a sum doesn't depend on the order, so it has to have cos the difference, which also doesn't; while a difference does, so it has to have sin the difference, which also does.
* This entry is from our old Library feature. If you know who wrote it, please let us know so we can attribute a writer. Thanks!