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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:
[tex]\cos^2x\,+\,\sin^2x\,=\,1[/tex]
(it comes from Pythagoras' theorem: [itex]\mathrm{adj}^2\,+\,\mathrm{opp}^2\,=\,\mathrm{hyp}^2[/itex])
Divide the equation by [itex]\cos^2x[/itex], and rearrange terms to get:
[tex]\sec^2x\,-\,\tan^2x\,=\,1[/tex]
Divide it instead by [itex]\sin^2x[/itex], and rearrange terms to get:
[tex]\mathrm{cosec}^2x\,-\,\cot^2x\,=\,1[/tex]
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:
[tex]\cos^2x\,+\,\sin^2x\,=\,1[/tex]
(it comes from Pythagoras' theorem: [itex]\mathrm{adj}^2\,+\,\mathrm{opp}^2\,=\,\mathrm{hyp}^2[/itex])
Divide the equation by [itex]\cos^2x[/itex], and rearrange terms to get:
[tex]\sec^2x\,-\,\tan^2x\,=\,1[/tex]
Divide it instead by [itex]\sin^2x[/itex], and rearrange terms to get:
[tex]\mathrm{cosec}^2x\,-\,\cot^2x\,=\,1[/tex]
Extended explanation
[tex]\cos2x\,=\,\cos^2x\,-\,\sin^2x[/tex]
[tex]1\,+\,\cos2x\,=\,2\,\cos^2x[/tex]
[tex]1\,-\,\cos2x\,=\,2\,\sin^2x[/tex]
[tex]\sin{2x}\,=\,2\,\sin{x}\,\cos{x}[/tex]
[tex]\sin(x\,+\,y)\,=\,\sin x\cos y\,+\,\cos x\sin y[/tex]
[tex]\sin(x\,-\,y)\,=\,\sin x\cos y\,-\,\cos x\sin y[/tex]
[tex]\cos(x\,+\,y)\,=\,\cos x\cos y\,-\,\sin x\sin y[/tex]
[tex]\cos(x\,-\,y)\,=\,\cos x\cos y\,+\,\sin x\sin y[/tex]
[tex]1\,+\,\cos2x\,=\,2\,\cos^2x[/tex]
[tex]1\,-\,\cos2x\,=\,2\,\sin^2x[/tex]
[tex]\sin{2x}\,=\,2\,\sin{x}\,\cos{x}[/tex]
[tex]\sin(x\,+\,y)\,=\,\sin x\cos y\,+\,\cos x\sin y[/tex]
[tex]\sin(x\,-\,y)\,=\,\sin x\cos y\,-\,\cos x\sin y[/tex]
[tex]\cos(x\,+\,y)\,=\,\cos x\cos y\,-\,\sin x\sin y[/tex]
[tex]\cos(x\,-\,y)\,=\,\cos x\cos y\,+\,\sin x\sin y[/tex]
You must learn all the equations above.
[tex]A\sin x\,+\,B\cos x\,=\,\sqrt{(A^2+B^2)}\sin (x\,+\,\tan^{-1}(B/A))[/tex]
. . . . . . . . . . . . . [tex]=\,\sqrt{(A^2+B^2)}\cos (x\,-\,\tan^{-1}(A/B))[/tex]
[tex]\sin x\,+\,\sin y\,=\,2 \sin \frac{x\,+\,y}{2} \cos \frac{x\,-\,y}{2}[/tex]
[tex]\sin x\,-\,\sin y\,=\,2 \sin \frac{x\,-\,y}{2} \cos \frac{x\,+\,y}{2}[/tex]
[tex]\cos x\,+\,\cos y\,=\,2 \cos \frac{x\,+\,y}{2} \cos \frac{x\,-\,y}{2}[/tex]
[tex]\cos x\,-\,\cos y\,=\,-2 \sin \frac{x\,+\,y}{2} \sin \frac{x\,-\,y}{2}[/tex]
. . . . . . . . . . . . . [tex]=\,\sqrt{(A^2+B^2)}\cos (x\,-\,\tan^{-1}(A/B))[/tex]
[tex]\sin x\,+\,\sin y\,=\,2 \sin \frac{x\,+\,y}{2} \cos \frac{x\,-\,y}{2}[/tex]
[tex]\sin x\,-\,\sin y\,=\,2 \sin \frac{x\,-\,y}{2} \cos \frac{x\,+\,y}{2}[/tex]
[tex]\cos x\,+\,\cos y\,=\,2 \cos \frac{x\,+\,y}{2} \cos \frac{x\,-\,y}{2}[/tex]
[tex]\cos x\,-\,\cos y\,=\,-2 \sin \frac{x\,+\,y}{2} \sin \frac{x\,-\,y}{2}[/tex]
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!