How do I prove Tan x + cot x = 2 cosec(2x)?
My math teacher always told me to tackle the most complicated side of a trig identity first. (More complicated, by the way, usually means sums).Thus, let’s look at [math]\tan{x}+\cot{x}[/math].There’s not much relating them, so you can always use another trick: break everything down into sines and cosines.[math]\tan{x}+\cot{x}=\frac{\sin{x}}{\cos{x}}+\frac{\cos{x}}{\sin{x}}[/math].Still not much clever stuff we can do, so let’s do the dumb thing instead: add the fractions.[math]\frac{\sin{x}}{\cos{x}}+\frac{\cos{x}}{\sin{x}}=\frac{\sin^2{x}+\cos^2{x}}{\sin{x}\cos{x}}[/math]Aha! We can simplify this, because [math]\sin^2{x}+\cos^2{x}=1[/math]! (Interpret that exclamation point as a factorial if you want, it’s still true.)Thus, we’ve simplified the left side down to [math]\frac{1}{\sin{x}\cos{x}}[/math].In case you’re not seeing where to go from here, let’s look at what we’re trying to prove now.We already got to the point of showing that [math]\tan{x}+\cot{x}=\frac{1}{\sin{x}\cos{x}}[/math], so it suffices to show that [math]\frac{1}{\sin{x}\cos{x}}=2\csc{2x}[/math].Well, to show that expression is true, we can first prove [math]\sin{x}\cos{x}=\frac{1}{2}\sin{2x}[/math], and then take the reciprocal of both sides. This approach is motivated because both sides of the original equation are equal to the reciprocal of something nicer.But is it true? Yes!Since [math]\sin{2x}=2\sin{x}\cos{x}[/math] (by the angle addition or double sine formula), we know that [math]\sin{x}\cos{x}=\frac{1}{2}\sin{2x}[/math] is true, and we are done.All that is left is to connect our two trains of thought into a single coherent proof:[math]\tan{x}+\cot{x}=\frac{\sin{x}}{\cos{x}}+\frac{\cos{x}}{\sin{x}}[/math][math]=\frac{\sin^2{x}+\cos^2{x}}{\sin{x}\cos{x}}[/math][math]=\frac{1}{\sin{x}\cos{x}}[/math][math]=\frac{1}{\frac{1}{2}\sin{2x}}[/math][math]=2\frac{1}{\sin{2x}}[/math][math]=2\csc{2x}[/math].QED.
Simplifying basic trigonometric expressions?
1) 3(tan^2(θ) - sec^2(θ)) Using the Pythagorean identity, sin^2(θ) + cos^2(θ) = 1 and dividing each term by cos^2(θ), we get tan^2(θ) + 1 = sec^2(θ). So tan^2(θ) - sec^2(θ) = -1 3(-1) -3 2) [cos(θ)csc(θ)] / tan(θ) cos(θ)csc(θ)cot(θ) cos(θ) * (1/sin(θ)) * (cos(θ)/sin(θ)) cos^2(θ) / sin^2(θ) cot^2(θ) 3) [sinθ)csc(θ)] / cot(θ) sin(θ)csc(θ)tan(θ) sin(θ) * (1/sin(θ)) * tan(θ) tan(θ)
Help with using basic Trig identities?
Usually, the easiest way to simplify is to convert everything to sin and cos: tan(x) = sin(x)/cos(x) cot(x) = cos(x)/sin(x) sec(x) = 1/cos(x) csc(x) = 1/sin(x) Other identities to remember: sin²x + cos²x = 1 tan²x + 1 = sec²x cot²x + 1 = csc²x 1. cot(u) sin(u) = cos(u)/sin(u) * sin(u) = cos(u) 2. tan(π/2 − θ) = 1/tan(θ) = −5.32 cot(θ) = 1/tan(θ) = −5.32 3. (1+tanx) / (1+cotx) = (1 + sinx/cosx) / (1 + cosx/sinx) = sinx * cosx (1 + sinx/cosx) / (cosx * sinx (1 + cosx/sinx) = sinx (cosx + sinx) / (cosx (sinx + cosx)) = sinx / cosx = tanx 4. 1/(secx − 1) − 1/(secx + 1) = ((secx + 1) − (secx − 1)) / ((secx − 1)(secx + 1)) = 2 / (sec²x − 1) = 2 / tan²x = 2 cot²x 5. tan²x / (secx + 1) = (secx − 1) tan²x / ((secx − 1)(secx + 1)) = (secx − 1) tan²x / (sec²x − 1) = (secx − 1) tan²x / tan²x = secx − 1 Mαthmφm
What is the value of sin (n pi/2)?
Sin(nπ/2)=0. For n= 2,4,6,8,10,12 i.e evenfor odd no. it varies as for 2n+1 (n is a even no.) its one 1if 2n-1 (n is a even no.) its negative one -1Sin(nπ/2)=1. For n= 1,5,9,13,17Sin(nπ/2)= -1. For n= 3,7,11,15,19