Difference between revisions of "Proofs of trig identities"
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+ | {{shortcut|[[Trig identity proof]]}} | ||
=Introduction= | =Introduction= | ||
<math>\sin</math> and <math>\cos</math> are easy to define. I prefer the unit circle definition as it makes these proofs easier to understand. | <math>\sin</math> and <math>\cos</math> are easy to define. I prefer the unit circle definition as it makes these proofs easier to understand. | ||
Line 10: | Line 11: | ||
<math>\csc = \frac{1}{\sin}</math> | <math>\csc = \frac{1}{\sin}</math> | ||
+ | |||
+ | Note: I've omitted <math>\theta</math> because it's unnecessary and might clog things up a little. | ||
With a bit of ingenuity, we can create the following diagram: | With a bit of ingenuity, we can create the following diagram: | ||
Line 30: | Line 33: | ||
label("B",B,dir(135+degrees(d))); | label("B",B,dir(135+degrees(d))); | ||
pair C = (1,0); | pair C = (1,0); | ||
− | label("C",C, | + | label("C",C,SE); |
pair D = (1,tan(d)); | pair D = (1,tan(d)); | ||
label("D",D,N); | label("D",D,N); | ||
Line 45: | Line 48: | ||
label("1",B--O); | label("1",B--O); | ||
draw(shift(dir(270)/24)*brace(C,O)); | draw(shift(dir(270)/24)*brace(C,O)); | ||
− | label("$\cot \theta$",shift(dir(270)/ | + | label("$\cot \theta$",shift(dir(270)/4)*brace(E,O),S); |
draw(shift(dir(d+90)/24)*brace(O,D)); | draw(shift(dir(d+90)/24)*brace(O,D)); | ||
− | label("$\ | + | label("$\sec \theta$",shift(dir(degrees(d)+90)/24)*brace(O,D),dir(degrees(d)+90)); |
draw(shift(dir(270)/4)*brace(E,O)); | draw(shift(dir(270)/4)*brace(E,O)); | ||
− | label("1",shift(dir(270)/ | + | label("1",shift(dir(270)/24)*brace(C,O),S); |
draw(shift(dir(270)/4)*O--shift(dir(270)/24)*O); | draw(shift(dir(270)/4)*O--shift(dir(270)/24)*O); | ||
draw(shift(dir(270)/4)*E--shift(dir(270)/24)*E); | draw(shift(dir(270)/4)*E--shift(dir(270)/24)*E); | ||
− | label("1",E--F, | + | label("1",E--F,SE); |
− | label("$\tan \theta$", | + | label("$\tan \theta$",C--D); |
draw(shift(dir(degrees(d)+90)/4)*brace(O,F)); | draw(shift(dir(degrees(d)+90)/4)*brace(O,F)); | ||
− | label("$\ | + | label("$\csc \theta$",shift(dir(degrees(d)+90)/4)*brace(O,F),dir(degrees(d)+90)); |
draw(shift(dir(degrees(d)+90)/4)*O--shift(dir(degrees(d)+90)/24)*O); | draw(shift(dir(degrees(d)+90)/4)*O--shift(dir(degrees(d)+90)/24)*O); | ||
draw(shift(dir(degrees(d)+90)/4)*F--shift(dir(degrees(d)+90)/24)*F); | draw(shift(dir(degrees(d)+90)/4)*F--shift(dir(degrees(d)+90)/24)*F); | ||
</asy> | </asy> | ||
+ | |||
+ | It has three main triangles: cos-sin-1, 1-tan-sec, and cot-1-csc. | ||
+ | |||
+ | We can note that the functions are correct by similar triangles. | ||
+ | <center>[[#toc|Back to Top]]</center> | ||
+ | |||
+ | =Symmetric identities= | ||
+ | |||
+ | If we draw a few copies of the triangle, we get: | ||
+ | |||
+ | <math>\sin(x)=\cos(90-x)=-\cos(90+x)=\sin(180-x)=-\sin(180+x)=\cos(270-x)=-\cos(270+x)=-\sin(-x)</math> | ||
+ | |||
+ | <math>\cos(x)=\sin(90-x)=\sin(90+x)=-\cos(180-x)=-\cos(180+x)=\sin(270-x)=-\sin(270+x)=\cos(-x)</math> | ||
+ | |||
+ | <math>\tan(x)=\cot(90-x)=-\cot(90+x)=-\tan(180-x)=\tan(180+x)=\cot(270-x)=-\cot(270+x)=-\tan(-x)</math> | ||
+ | |||
+ | The other three can be derived by taking the reciprocals of these three. | ||
+ | |||
+ | x is easier to type than theta | ||
+ | <center>[[#toc|Back to Top]]</center> | ||
+ | |||
+ | =Pythagorean identities= | ||
+ | |||
+ | Pythagorean identities are easy and there's no algebra involved. In fact, the name Pythagorean is a giveaway of what we should do! | ||
+ | |||
+ | ==<math>\cos^2+\sin^2=1</math>== | ||
+ | |||
+ | The proof here is very straightforward. We use the pythagorean theorem on <math>\triangle OAB</math> giving us <math>OA^2+AB^2=OB^2</math> or <math>\sin^2+\cos^2=1^2</math>. | ||
+ | |||
+ | ==<math>\tan^2+1=\sec^2</math>== | ||
+ | |||
+ | Same story here. Applying pythagorean to <math>\triangle OCD</math> gives us <math>OC^2+CD^2=OD^2</math> or <math>\tan^2+1^2=\sec^2</math>. | ||
+ | |||
+ | ==<math>1+\cot^2=\csc^2</math>== | ||
+ | |||
+ | Same. Pythagorean on <math>\triangle OEF</math> gives <math>OE^2+EF^2=OF^2</math> or <math>1^2+\cot^2=\csc^2</math>. | ||
+ | |||
+ | ==Conclusion== | ||
+ | |||
+ | Even though with the first one and the definitions, we can make the rest from algebra, having a geometric meaning is nice when we want to know what it actually means. | ||
+ | <center>[[#toc|Back to Top]]</center> | ||
+ | |||
+ | =Angle addition and subtraction= | ||
+ | |||
+ | ==<math>\sin(\alpha + \beta)</math>== | ||
+ | |||
+ | When does sin appear? When does sin appear? In the first triangle, of course. Let's make a diagram! | ||
+ | |||
+ | <asy> | ||
+ | unitsize(216); | ||
+ | real d = 1/cos(radians(35)); | ||
+ | real d1 = d * cos(radians(55)); | ||
+ | real d2 = d * sin(radians(55)); | ||
+ | pair O = (0,0); | ||
+ | pair A = (cos(radians(20)),0); | ||
+ | pair B = (cos(radians(20)),sin(radians(20))); | ||
+ | pair C = (cos(radians(20)),d2); | ||
+ | pair D = (d1,d2); | ||
+ | draw(O--A--B--O--D--B--O--D--C--B); | ||
+ | dot(O); | ||
+ | dot(B); | ||
+ | dot(A,red); | ||
+ | dot(C,green); | ||
+ | dot(D,blue); | ||
+ | label("O",O,SW); | ||
+ | label("$\alpha$",shift(dir(10)/5)*O); | ||
+ | label("$\beta$",shift(dir(37.5)/5)*O); | ||
+ | label("A",A,SE,red); | ||
+ | label("B",B,E); | ||
+ | label("C",C,NE,green); | ||
+ | label("D",D,dir(122.5),blue); | ||
+ | label("$\cos \alpha \cos \beta$",O--A,S); | ||
+ | label("$\sin \alpha \cos \beta$",A--B,E); | ||
+ | label("$\cos \beta$",O--B,dir(302.5)); | ||
+ | label("$\cos \alpha \sin \beta$",B--C,E); | ||
+ | label("$\sin \alpha \sin \beta$",C--D,N); | ||
+ | label("$\sin \beta$",B--D,dir(200)); | ||
+ | label("1",D--O,dir(325)); | ||
+ | </asy> | ||
+ | |||
+ | where <math>\triangle OAB \sim \triangle BCD</math> | ||
+ | |||
+ | The diagram illustrates the identities nicely. | ||
+ | |||
+ | The diagram shows the height of point <math>D</math> is <math>\sin(\alpha)+\frac{\cos \alpha \sin \beta}{\cos \beta}</math>. | ||
+ | However, the length of <math>OD</math> is <math>\frac{1}{\cos\beta}</math>. To compensate, we must divide by <math>\frac{1}{\cos\beta}</math> to make it the sine. After some *easy* algebra, we arrive at <math>\sin(\alpha+\beta)=\sin\alpha\cos\beta+\sin\beta\cos\alpha</math>. | ||
+ | |||
+ | ==<math>\cos(\alpha + \beta)</math>== | ||
+ | |||
+ | The diagram says that it is <math>\cos(\alpha)-\frac{\sin \alpha \sin \beta}{\cos \beta}</math>, but we need to divide by <math>\frac{1}{\cos\beta}</math> again. We arrive at <math>\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta</math>. | ||
+ | |||
+ | ==<math>\tan(\alpha + \beta)</math>== | ||
+ | |||
+ | This time, let's use the tan-1-sec triangle. | ||
+ | |||
+ | <asy> | ||
+ | unitsize(216); | ||
+ | real d = 1/cos(radians(35)); | ||
+ | real d1 = d * cos(radians(55)); | ||
+ | real d2 = d * sin(radians(55)); | ||
+ | pair O = (0,0); | ||
+ | pair A = (cos(radians(20)),0); | ||
+ | pair B = (cos(radians(20)),sin(radians(20))); | ||
+ | pair C = (cos(radians(20)),d2); | ||
+ | pair D = (d1,d2); | ||
+ | draw(O--A--B--O--D--B--O--D--C--B); | ||
+ | dot(O); | ||
+ | dot(B); | ||
+ | dot(A,red); | ||
+ | dot(C,green); | ||
+ | dot(D,blue); | ||
+ | label("O",O,SW); | ||
+ | label("$\alpha$",shift(dir(10)/5)*O); | ||
+ | label("$\beta$",shift(dir(37.5)/5)*O); | ||
+ | label("A",A,SE,red); | ||
+ | label("B",B,E); | ||
+ | label("C",C,NE,green); | ||
+ | label("D",D,dir(122.5),blue); | ||
+ | label("1",O--A,S); | ||
+ | label("$\tan \alpha$",A--B,E); | ||
+ | label("$\sec \alpha$",O--B,dir(302.5)); | ||
+ | label("$\tan \beta$",B--C,E); | ||
+ | label(scale(0.75)*"$\tan \alpha \tan \beta$",C--D,N); | ||
+ | label(scale(0.75)*"$\sec \alpha \tan \beta$",B--D,dir(200)); | ||
+ | label(scale(0.75)*"$\sec \alpha \sec \beta$",D--O,dir(325)); | ||
+ | </asy> | ||
+ | |||
+ | Wait, is that just the same diagram? No! the labels have changed! | ||
+ | |||
+ | Note: I did some algebra when noting that sin * sec = tan and cos * sec = 1 | ||
+ | |||
+ | Looking at the diagram, the height of the new triangle is <math>\tan \alpha + \tan \beta</math>, but the width is only <math>1-\tan\alpha\tan\beta</math>, so we arrive at <math>\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}</math> | ||
+ | |||
+ | ==<math>\sec(\alpha + \beta)</math>== | ||
+ | |||
+ | I dunno why most people never use this, but it's right up there in the diagram. <math>\frac{\sec\alpha\sec\beta}{1-\tan\alpha\tan\beta}</math> | ||
+ | |||
+ | ==<math>\csc(\alpha + \beta)</math>== | ||
+ | |||
+ | Hey, if you don't need this, stop reading. | ||
+ | |||
+ | We need the third triangle here. I'm going to do something weird, you'll see why when I complete the diagram (I=[[User:Afly | afly]]. This page was all made by afly) | ||
+ | <asy> | ||
+ | unitsize(216); | ||
+ | real d = 1/cos(radians(35)); | ||
+ | real d1 = d * cos(radians(55)); | ||
+ | real d2 = d * sin(radians(55)); | ||
+ | pair O = (0,0); | ||
+ | pair A = (cos(radians(20)),0); | ||
+ | pair B = (cos(radians(20)),sin(radians(20))); | ||
+ | pair C = (cos(radians(20)),d2); | ||
+ | pair D = (d1,d2); | ||
+ | draw(O--A--B--O--D--B--O--D--C--B); | ||
+ | dot(O); | ||
+ | dot(B); | ||
+ | dot(A,red); | ||
+ | dot(C,green); | ||
+ | dot(D,blue); | ||
+ | label("O",O,SW); | ||
+ | label("$\alpha$",shift(dir(10)/5)*O); | ||
+ | label("$\beta$",shift(dir(37.5)/5)*O); | ||
+ | label("A",A,SE,red); | ||
+ | label("B",B,E); | ||
+ | label("C",C,NE,green); | ||
+ | label("D",D,dir(122.5),blue); | ||
+ | label(scale(0.75)*"$\cot \alpha \cot \beta$",O--A,S); | ||
+ | label("$\cot \beta$",A--B,E); | ||
+ | label(scale(0.75)*"$\csc \alpha \cot \beta$",O--B,dir(302.5)); | ||
+ | label("$\cot \alpha$",B--C,E); | ||
+ | label("1",C--D,N); | ||
+ | label(scale(0.75)*"$\csc \alpha$",B--D,dir(200)); | ||
+ | label(scale(0.75)*"$\csc \alpha \csc \beta$",D--O,dir(325)); | ||
+ | </asy> | ||
+ | |||
+ | Can you look at the diagram? what do you have? <math>\csc(\alpha+\beta)=\frac{\csc\alpha\csc\beta}{\cot\alpha\cot\beta-1}</math> | ||
+ | |||
+ | ==<math>\cot(\alpha + \beta)</math>== | ||
+ | |||
+ | Easy Peasy Lemon Squeezy. <math>\frac{\cot\alpha+\cot\beta}{\cot\alpha\cot\beta-1}</math> | ||
+ | |||
+ | =Double angle formulas= | ||
+ | This is a breeze. Just sub in for sum: | ||
+ | |||
+ | <math>\sin(2\theta)=2\sin\cos</math> | ||
+ | |||
+ | <math>\cos(2\theta)=\cos^2-\sin^2</math> | ||
+ | |||
+ | <math>\tan(2\theta)=\frac{2\tan}{1-\tan^2}</math> | ||
+ | |||
+ | ==Variations== | ||
+ | Since <math>\sin^2+\cos^2=1</math>, we can edit the double angle cosine formula a bit. Here are the three most helpful variants: | ||
+ | |||
+ | <math>\cos(2\theta)=2\cos^2-1</math> | ||
+ | |||
+ | <math>\cos(2\theta)=\cos^2-\sin^2</math> | ||
+ | |||
+ | <math>\cos(2\theta)=1-2\sin^2</math> | ||
+ | |||
+ | We can also solve for other expressions: | ||
+ | |||
+ | <math>\sin^2=\frac{1-\cos(2\theta)}{2}</math> | ||
+ | |||
+ | <math>\cos^2=\frac{\cos(2\theta)+1}{2}</math> | ||
+ | <center>[[#toc|Back to Top]]</center> | ||
+ | |||
+ | =Sum to Product to Sum= | ||
+ | These are the silliest identities I've ever seen. Do people really want to be surprised that much? | ||
+ | <asy> | ||
+ | unitsize(216); | ||
+ | real d = 1/cos(radians(35)); | ||
+ | real d1 = d * cos(radians(55)); | ||
+ | real d2 = d * sin(radians(55)); | ||
+ | pair O = (0,0); | ||
+ | pair A = (cos(radians(20)),0); | ||
+ | pair B = (cos(radians(20)),sin(radians(20))); | ||
+ | pair C = (cos(radians(20)),d2); | ||
+ | pair D = (d1,d2); | ||
+ | pair F = B+B-D; | ||
+ | draw(O--A--B--O--D--B--O--D--C--B--F--O); | ||
+ | dot(O); | ||
+ | dot(B); | ||
+ | dot(A,red); | ||
+ | dot(C,green); | ||
+ | dot(D,blue); | ||
+ | dot(F,blue); | ||
+ | label("O",O,SW); | ||
+ | label("$\alpha$",shift(dir(10)/5)*O); | ||
+ | label("$\beta$",shift(dir(37.5)/5)*O); | ||
+ | label("A",A,SE,red); | ||
+ | label("B",B,E); | ||
+ | label("C",C,NE,green); | ||
+ | label("D",D,dir(122.5),blue); | ||
+ | label("D'",F,dir(302.5),blue); | ||
+ | label("$\cos \alpha \cos \beta$",O--A,S); | ||
+ | label("$\sin \alpha \cos \beta$",A--B,E); | ||
+ | label("$\cos \beta$",O--B,dir(302.5)); | ||
+ | label("$\cos \alpha \sin \beta$",B--C,E); | ||
+ | label("$\sin \alpha \sin \beta$",C--D,N); | ||
+ | label("$\sin \beta$",B--D,dir(200)); | ||
+ | label("1",D--O,dir(325)); | ||
+ | </asy> | ||
+ | Note: D' is the reflection of D about line OB. | ||
+ | So, we have the angles <math>\alpha+\beta</math> and <math>\alpha-\beta</math> illustrated nicely in here. B is the midpoint of DD'. It is half the sum of D and D'. Calculate the coordinates of B two ways: One by the labels on triangle AOB, and one by finding the coordinates of D and D' by the sine and cosine of <math>\alpha+\beta</math> and <math>\alpha-\beta</math>, then averaging them. | ||
+ | |||
+ | Since half the difference of D to D' is the difference of B and one of them, it has the x coordinate equal to exactly the product of the sines, as illustrated above. | ||
+ | |||
+ | <math>\cos\alpha\cos\beta=\frac12(\cos(\alpha-\beta)+\cos(\alpha+\beta))</math> | ||
+ | <math>\sin\alpha\cos\beta=\frac12(\sin(\alpha-\beta)+\sin(\alpha+\beta))</math> | ||
+ | <math>\sin\alpha\sin\beta=\frac12(\cos(\alpha-\beta)-\cos(\alpha+\beta))</math> I almost thought I got this last one wrong, but no, it's right. | ||
+ | <center>[[#toc|Back to Top]]</center> | ||
+ | |||
+ | =Bonus: Product identity= | ||
+ | This is a special identity. I hope this helps you. | ||
+ | <math>\sin^2(\alpha+\beta)=(\sin\alpha\cos\beta+\cos\alpha\sin\beta)(\sin\alpha\cos\beta+\cos\alpha\sin\beta)=\sin^2\alpha\cos^2\beta+\cos^2\alpha\sin^2\beta+2\sin\alpha\sin\beta\cos\alpha\cos\beta</math> | ||
+ | |||
+ | and | ||
+ | |||
+ | <math>\cos^2(\alpha+\beta)=(\cos\alpha\cos\beta-\sin\alpha\sin\beta)(\cos\alpha\cos\beta-\sin\alpha\sin\beta)=\sin^2\alpha\cos^2\beta+\cos^2\alpha\sin^2\beta+(\sin\alpha\cos\alpha)^2+(\sin\beta\cos\beta)^2</math> | ||
+ | |||
+ | There's something we can cancel. | ||
+ | |||
+ | <math>\cos(2\alpha+2\beta)=\cos^2(\alpha+\beta)-\sin^2(\alpha+\beta)</math> | ||
+ | |||
+ | <math>=(\sin\alpha\cos\alpha)^2+(\sin\beta\cos\beta)^2-2\sin\alpha\cos\alpha\sin\beta\cos\beta</math> | ||
+ | |||
+ | If <math>f()=\sin\cos</math>, then it simplifies to | ||
+ | |||
+ | <math>(f(\alpha)-f(\beta))^2</math> | ||
+ | |||
+ | Notice <math>f\left(\frac{k\pi}{2}\right)=0</math>. If we let <math>\beta=0</math>: | ||
+ | |||
+ | <math>\cos(2\alpha+k\pi)=\sin^2\alpha\cos^2\alpha</math> | ||
+ | <center>[[#toc|Back to Top]]</center> | ||
+ | |||
+ | =Halved angles= | ||
+ | |||
+ | Starting with the identities from the double section: | ||
+ | |||
+ | <math>\sin^2=\frac{1-\cos(2)}{2}</math> | ||
+ | |||
+ | <math>\cos^2=\frac{1+\cos(2)}{2}</math> | ||
+ | |||
+ | We take the square root to obtain: | ||
+ | |||
+ | <math>\sin=\pm\sqrt{\frac{1-\cos(2)}{2}}</math> | ||
+ | |||
+ | <math>\cos=\pm\sqrt{\frac{1+\cos(2)}{2}}</math> | ||
+ | |||
+ | For tangent: | ||
+ | |||
+ | <math>\tan=\frac{\sin}{\cos}=\frac{\pm\sqrt{\frac{1-\cos(2)}{2}}}{\pm\sqrt{\frac{1+\cos(2)}{2}}}=\pm\sqrt{\frac{1-\cos(2)}{1+\cos(2)}}</math> | ||
+ | |||
+ | There are two nice variations to know. | ||
+ | |||
+ | <math>\pm\sqrt{\frac{1-\cos(2)}{1+\cos(2)}}\times\sqrt{\frac{1-\cos(2)}{1-\cos(2)}}=\pm\frac{1-\cos(2)}{\sin(2)}</math> | ||
+ | |||
+ | <math>\pm\sqrt{\frac{1-\cos(2)}{1+\cos(2)}}\times\sqrt{\frac{1+\cos(2)}{1+\cos(2)}}=\pm\frac{\sin(2)}{1+\cos(2)}</math> | ||
+ | <center>[[#toc|Back to Top]]</center> | ||
+ | |||
+ | =Triple angles and more= | ||
+ | ==Triple sums== | ||
+ | <math>\sin(\alpha+\beta+\gamma)=\sin(\alpha+(\beta+\gamma))</math> | ||
+ | <math>=\sin\alpha\cos(\beta+\gamma)+\cos\alpha\sin(\beta+\gamma)</math> | ||
+ | <math>=\sin\alpha(\cos\beta\cos\gamma-\sin\beta\sin\gamma)+\cos\alpha(\sin\beta\cos\gamma+\cos\beta\sin\gamma)</math> | ||
+ | <math>=\sin\alpha\cos\beta\cos\gamma+\cos\alpha\sin\beta\cos\gamma+\cos\alpha\cos\beta\sin\gamma-\sin\alpha\sin\beta\sin\gamma</math> | ||
+ | |||
+ | <math>\cos(\alpha+\beta+\gamma)=\cos(\alpha+(\beta+\gamma))</math> | ||
+ | <math>=\cos\alpha\cos(\beta+\gamma)-\sin\alpha\sin(\beta+\gamma)</math> | ||
+ | <math>=\cos\alpha(\cos\beta\cos\gamma-\sin\beta\sin\gamma)-\sin\alpha(\sin\beta\cos\gamma+\cos\beta\sin\gamma)</math> | ||
+ | <math>=\cos\alpha\cos\beta\cos\gamma-\cos\alpha\sin\beta\sin\gamma-\sin\alpha\cos\beta\sin\gamma-\sin\alpha\sin\beta\cos\gamma</math> | ||
+ | |||
+ | <math>\tan(\alpha+\beta+\gamma)=\tan(\alpha+(\beta+\gamma))</math> | ||
+ | <math>=\frac{\tan\alpha+\tan(\beta+\gamma)}{1-\tan\alpha\tan(\beta+\gamma)}</math> | ||
+ | <math>=\frac{\tan\alpha+\frac{\tan\beta+\tan\gamma}{1-\tan\beta\tan\gamma}}{1-\tan\alpha\frac{\tan\beta+\tan\gamma}{1-\tan\beta\tan\gamma}}</math> | ||
+ | <math>=\frac{\frac{\tan\alpha+\tan\beta+\tan\gamma-\tan\alpha\tan\beta\tan\gamma}{1-\tan\beta\tan\gamma}}{\frac{1-\tan\beta\tan\gamma-\tan\alpha\tan\beta-\tan\alpha\tan\gamma}{1-\tan\beta\tan\gamma}}</math> | ||
+ | <math>=\frac{\tan\alpha+\tan\beta+\tan\gamma-\tan\alpha\tan\beta\tan\gamma}{1-\tan\beta\tan\gamma-\tan\alpha\tan\gamma-\tan\alpha\tan\beta}</math> | ||
+ | |||
+ | ==Triple angles== | ||
+ | <math>\sin 3\theta=3\cos^2\theta\sin\theta-\sin^3\theta=3\sin\theta-3(1-\cos^2\theta)\sin\theta-\sin^3\theta=3\sin\theta-4\sin^3\theta</math> | ||
+ | |||
+ | <math>\cos 3\theta=\cos^3\theta-3\cos\theta\sin^2\theta=\cos^3\theta+3(1-\sin^2\theta)\cos\theta-3\cos\theta=4\cos^3\theta-3\cos\theta</math> | ||
+ | |||
+ | <math>\tan 3\theta=\frac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta}</math> | ||
+ | |||
+ | ==Third angles== | ||
+ | Let <math>\sin\theta = x</math> and <math>\sin 3\theta = y</math>. We get this depressed cubic: | ||
+ | |||
+ | <math>0=3x-4x^3-y</math> | ||
+ | |||
+ | First, divide both sides by -4 and rearrange: <math>x^3-\frac{3}{4}x+y=0</math>. The discriminant <math>\Delta = \frac{y^2}{4}-\frac{1}{64}=\frac{16y^2-1}{64}</math> | ||
+ | |||
+ | Then, <math>u=\frac{-4y\pm\sqrt{16y^2-1}}{8}</math> | ||
+ | |||
+ | The solutions are <math>\sqrt[3]{-4y+\sqrt{16y^2-1}}</math>, <math>\frac{\sqrt[3]{-4y+\sqrt{16y^2-1}}+\sqrt[3]{4y-\sqrt{16y^2-1}}}{2}</math>, and <math>\sqrt[3]{-4y-\sqrt{16y^2-1}}</math>. | ||
+ | |||
+ | A tiny adjustment gives us the cosine third-angle formulas: | ||
+ | |||
+ | <math>\sqrt[3]{4y+\sqrt{16y^2-1}}</math>, <math>\frac{\sqrt[3]{4y+\sqrt{16y^2-1}}+\sqrt[3]{4y-\sqrt{16y^2-1}}}{2}</math>, and <math>\sqrt[3]{4y-\sqrt{16y^2-1}}</math>. | ||
+ | |||
+ | For tangent: | ||
+ | |||
+ | <math>\frac{\sqrt[3]{-4y+\sqrt{16y^2-1}}}{\sqrt[3]{4y+\sqrt{16y^2-1}}}</math>, <math>\frac{\sqrt[3]{-4y+\sqrt{16y^2-1}}+\sqrt[3]{-4y-\sqrt{16y^2-1}}}{\sqrt[3]{4y+\sqrt{16y^2-1}}+\sqrt[3]{4y-\sqrt{16y^2-1}}}</math>, and <math>\frac{\sqrt[3]{-4y-\sqrt{16y^2-1}}}{\sqrt[3]{4y-\sqrt{16y^2-1}}}</math> | ||
+ | |||
+ | <center>[[#toc|Back to Top]]</center> | ||
+ | =All identities= | ||
+ | ==Definition== | ||
+ | <math>\tan = \frac{\sin}{\cos}</math> | ||
+ | |||
+ | <math>\cot = \frac{\cos}{\sin}</math> | ||
+ | |||
+ | <math>\sec = \frac{1}{\cos}</math> | ||
+ | |||
+ | <math>\csc = \frac{1}{\sin}</math> | ||
+ | ==Symmetric== | ||
+ | <math>\sin(x)=\cos(90-x)=-\cos(90+x)=\sin(180-x)=-\sin(180+x)=\cos(270-x)=-\cos(270+x)=-\sin(-x)</math> | ||
+ | |||
+ | <math>\cos(x)=\sin(90-x)=\sin(90+x)=-\cos(180-x)=-\cos(180+x)=\sin(270-x)=-\sin(270+x)=\cos(-x)</math> | ||
+ | |||
+ | <math>\tan(x)=\cot(90-x)=-\cot(90+x)=-\tan(180-x)=\tan(180+x)=\cot(270-x)=-\cot(270+x)=-\tan(-x)</math> | ||
+ | ==Pythagorean== | ||
+ | <math>\cos^2+\sin^2=1</math> | ||
+ | |||
+ | <math>\tan^2+1=\sec^2</math> | ||
+ | |||
+ | <math>1+\cot^2=\csc^2</math> | ||
+ | ==Sum== | ||
+ | ==Sum== | ||
+ | <math>\sin(\alpha+\beta)=\sin\alpha\cos\beta+\sin\beta\cos\alpha</math> | ||
+ | |||
+ | <math>\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta</math> | ||
+ | |||
+ | <math>\tan(\alpha + \beta)=\frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}</math> | ||
+ | |||
+ | <math>\sec(\alpha + \beta)=\frac{\sec \alpha \sec \beta}{1 - \tan \alpha \tan \beta}</math> | ||
+ | |||
+ | <math>\csc(\alpha + \beta)=\frac{\csc \alpha \csc \beta}{\cot \alpha + \cot \beta}</math> | ||
+ | |||
+ | <math>\cot(\alpha + \beta)=\frac{\cot \alpha \cot \beta - 1}{\cot \alpha + \cot \beta}</math> | ||
+ | |||
+ | Memory aids: sin is different, cos is same, tan is sum over one minus product, cot is product minus one over sum, sec is like tan but with product of sec on top, csc is like cot but with product of csc on top. | ||
+ | ==Double== | ||
+ | <math>\sin(2n)=2\sin n\cos n</math> | ||
+ | |||
+ | <math>\cos(2n)=\cos^2 n-\sin^2 n</math> | ||
+ | |||
+ | <math>\tan(2n)=\frac{2\tan n}{1-\tan^2 n}</math> | ||
+ | |||
+ | <math>\cos(2n)=2\cos^2 n-1</math> | ||
+ | |||
+ | <math>\cos(2n)=\cos^2 n-\sin^2 n</math> | ||
+ | |||
+ | <math>\cos(2n)=1-2\sin^2 n</math> | ||
+ | ==Sum <math>\iff</math> Product== | ||
+ | <math>\theta =\alpha +\beta</math> | ||
+ | |||
+ | <math>\phi =\alpha -\beta</math> | ||
+ | |||
+ | <math>\alpha =\frac{\theta +\phi}{2}</math> | ||
+ | |||
+ | <math>\beta =\frac{\theta -\phi}{2}</math> | ||
+ | |||
+ | <math>2\sin\alpha\cos\beta\iff\sin\theta+\sin\phi</math> | ||
+ | |||
+ | <math>2\sin\alpha\sin\beta\iff\cos\phi-\cos\theta</math> | ||
+ | |||
+ | <math>2\cos\alpha\cos\beta\iff\cos\theta+\cos\phi</math> | ||
+ | ==Product== | ||
+ | <math>\cos(2\alpha+k\pi)=\sin^2\alpha\cos^2\alpha</math> | ||
+ | ==Halves== | ||
+ | <math>\sin\theta=\pm\sqrt{\frac{1-\cos(2\theta)}{2}}</math> | ||
+ | |||
+ | <math>\cos\theta=\pm\sqrt{\frac{1+\cos(2\theta)}{2}}</math> | ||
+ | |||
+ | <math>\tan\theta=\pm\sqrt{\frac{1-\cos(2\theta)}{1+\cos(2\theta)}}</math> | ||
+ | |||
+ | <math>\tan\theta=\pm\frac{1-\cos(2\theta)}{\sin 2\theta}</math> | ||
+ | |||
+ | <math>\tan\theta=\pm\frac{\sin 2\theta}{1+\cos(2\theta)}</math> | ||
+ | ==3 Sums== | ||
+ | <math>\sin(\alpha+\beta+\gamma)=\sin\alpha\cos\beta\cos\gamma+\cos\alpha\sin\beta\cos\gamma+\cos\alpha\cos\beta\sin\gamma-\sin\alpha\sin\beta\sin\gamma</math> | ||
+ | |||
+ | <math>\cos(\alpha+\beta+\gamma)=\cos\alpha\cos\beta\cos\gamma-\cos\alpha\sin\beta\sin\gamma-\sin\alpha\cos\beta\sin\gamma-\sin\alpha\sin\beta\cos\gamma</math> | ||
+ | |||
+ | <math>\tan(\alpha+\beta+\gamma)=\frac{\tan\alpha+\tan\beta+\tan\gamma-\tan\alpha\tan\beta\tan\gamma}{1-\tan\beta\tan\gamma-\tan\alpha\tan\gamma-\tan\alpha\tan\beta}</math> | ||
+ | ==Triple== | ||
+ | <math>\sin 3\theta=3\sin\theta-4\sin^3\theta</math> | ||
+ | |||
+ | <math>\cos 3\theta=4\cos^3\theta-3\cos\theta</math> | ||
+ | |||
+ | <math>\tan 3\theta=\frac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta}</math> | ||
+ | ==Thirds== | ||
+ | <math>\sin\theta=\sqrt[3]{-4\sin 3\theta+\sqrt{16\sin^2(3\theta)-1}},\frac{\sqrt[3]{-4\sin 3\theta+\sqrt{16\sin^2(3\theta)-1}}+\sqrt[3]{-4\sin 3\theta-\sqrt{16\sin^2(3\theta)-1}}}{2}, \ \text{or }\sqrt[3]{4\sin 3\theta-\sqrt{16\sin^2 3\theta-1}}</math> | ||
+ | |||
+ | <math>\cos\theta=\sqrt[3]{4\cos 3\theta+\sqrt{16\cos^2(3\theta)-1}},\frac{\sqrt[3]{4\cos 3\theta+\sqrt{16\cos^2(3\theta)-1}}+\sqrt[3]{4\cos 3\theta-\sqrt{16\cos^2(3\theta)-1}}}{2}, \ \text{or }\sqrt[3]{4\cos 3\theta-\sqrt{16\cos^2 3\theta-1}}</math> | ||
+ | |||
+ | <math>\tan\theta=\frac{\sqrt[3]{-4\sin 3\theta+\sqrt{16\sin^2(3\theta)-1}}}{\sqrt[3]{4\cos 3\theta+\sqrt{16\cos^2(3\theta)-1}}},\frac{\sqrt[3]{-4\sin 3\theta+\sqrt{16\sin^2(3\theta)-1}}+\sqrt[3]{-4\sin 3\theta-\sqrt{16\sin^2(3\theta)-1}}}{\sqrt[3]{4\cos 3\theta+\sqrt{16\cos^2(3\theta)-1}}+\sqrt[3]{4\cos 3\theta-\sqrt{16\cos^2(3\theta)-1}}}, \ \text{or }\frac{\sqrt[3]{-4\sin 3\theta-\sqrt{16\sin(3\theta)^2-1}}}{\sqrt[3]{4\cos 3\theta-\sqrt{16\cos^2(3\theta)-1}}}</math> | ||
+ | =See also= | ||
+ | [[Trigonometric identities]] | ||
+ | [[Category:Trigonometry]] | ||
+ | |||
+ | Created by [[User:Afly|Afly]] ([[User talk:Afly|talk]]) | ||
+ | <center>[[#toc|Back to Top]]</center> |
Latest revision as of 21:28, 29 August 2024
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Contents
[hide]Introduction
and are easy to define. I prefer the unit circle definition as it makes these proofs easier to understand. Next, we define some other functions:
Note: I've omitted because it's unnecessary and might clog things up a little.
With a bit of ingenuity, we can create the following diagram:
It has three main triangles: cos-sin-1, 1-tan-sec, and cot-1-csc.
We can note that the functions are correct by similar triangles.
Symmetric identities
If we draw a few copies of the triangle, we get:
The other three can be derived by taking the reciprocals of these three.
x is easier to type than theta
Pythagorean identities
Pythagorean identities are easy and there's no algebra involved. In fact, the name Pythagorean is a giveaway of what we should do!
The proof here is very straightforward. We use the pythagorean theorem on giving us or .
Same story here. Applying pythagorean to gives us or .
Same. Pythagorean on gives or .
Conclusion
Even though with the first one and the definitions, we can make the rest from algebra, having a geometric meaning is nice when we want to know what it actually means.
Angle addition and subtraction
When does sin appear? When does sin appear? In the first triangle, of course. Let's make a diagram!
where
The diagram illustrates the identities nicely.
The diagram shows the height of point is . However, the length of is . To compensate, we must divide by to make it the sine. After some *easy* algebra, we arrive at .
The diagram says that it is , but we need to divide by again. We arrive at .
This time, let's use the tan-1-sec triangle.
Wait, is that just the same diagram? No! the labels have changed!
Note: I did some algebra when noting that sin * sec = tan and cos * sec = 1
Looking at the diagram, the height of the new triangle is , but the width is only , so we arrive at
I dunno why most people never use this, but it's right up there in the diagram.
Hey, if you don't need this, stop reading.
We need the third triangle here. I'm going to do something weird, you'll see why when I complete the diagram (I= afly. This page was all made by afly)
Can you look at the diagram? what do you have?
Easy Peasy Lemon Squeezy.
Double angle formulas
This is a breeze. Just sub in for sum:
Variations
Since , we can edit the double angle cosine formula a bit. Here are the three most helpful variants:
We can also solve for other expressions:
Sum to Product to Sum
These are the silliest identities I've ever seen. Do people really want to be surprised that much? Note: D' is the reflection of D about line OB. So, we have the angles and illustrated nicely in here. B is the midpoint of DD'. It is half the sum of D and D'. Calculate the coordinates of B two ways: One by the labels on triangle AOB, and one by finding the coordinates of D and D' by the sine and cosine of and , then averaging them.
Since half the difference of D to D' is the difference of B and one of them, it has the x coordinate equal to exactly the product of the sines, as illustrated above.
I almost thought I got this last one wrong, but no, it's right.
Bonus: Product identity
This is a special identity. I hope this helps you.
and
There's something we can cancel.
If , then it simplifies to
Notice . If we let :
Halved angles
Starting with the identities from the double section:
We take the square root to obtain:
For tangent:
There are two nice variations to know.
Triple angles and more
Triple sums
Triple angles
Third angles
Let and . We get this depressed cubic:
First, divide both sides by -4 and rearrange: . The discriminant
Then,
The solutions are , , and .
A tiny adjustment gives us the cosine third-angle formulas:
, , and .
For tangent:
, , and
All identities
Definition
Symmetric
Pythagorean
Sum
Sum
Memory aids: sin is different, cos is same, tan is sum over one minus product, cot is product minus one over sum, sec is like tan but with product of sec on top, csc is like cot but with product of csc on top.
Double
Sum Product
Product
Halves
3 Sums
Triple
Thirds