Andrew s handout Trig identities. Fundamental identities These are the most fundamental identities, in the sense that ou should probabl memorize these and use them to derive the rest (or, if ou prefer, memorize all the identities, including these). There s not reall a good wa to derive these identities; all the derivations are (in m opinion) much harder to remember than the identities themselves are. Identit (The fundamental trig identit, a.k.a. the Pthagorean identit). Identit (Angle addition formulas). sin () + cos () = for an R sin( + ) = sin()cos() + cos()sin() cos( + ) = cos()cos() sin()sin(). Other identities coming from the Pthagorean identit These are reall simple, but I m repeating them for good measure. Take the Pthagorean identit, and divide both sides b cos (): sin () + cos () = sin () cos () + cos () cos () = cos () tan () + = sec () Or, solve for sin or cos (sometimes useful to solve a trig integral b u-substitution): sin() = ± cos () cos() = ± sin ().3 Consequences of the angle addition formulas First, the double angle formulas: sin() = sin( + ) = sin() cos() cos() = cos( + ) = cos () sin () The double angle formula for cos can be written in a few different was: Or, cos() = cos () sin () = cos () ( cos ()) = cos () cos() = cos () sin () = ( sin ()) sin () = sin ()
These two forms give us a nice wa of rewriting powers of sin or cos (just solve for cos () and sin () in the above identities):.4 Translations cos () = + cos() sin () = cos() The identities in this section can all be derived graphicall, b shifting the graphs of sin and cos, or b reflecting across the -ais, or b similar means. For eample, if ou graph sin and cos on the same graph, it should be clear that the curves are identical ecept that sin is shifted right b π units: π π π π sin() cos() What this means, is that an time we want to calculate sin(), we can instead calculate cos( π ): sin() cos() π π π π In other words, sin() = cos( π ). We can check this using the angle addition formula for cos(): cos( π ) = cos()cos( π ) sin()sin( π ) = cos() 0 sin() ( ) = sin()
Another important pair of identities captures the fact that π is half the wavelength of sin and cos, so sin( + π) = sin() and cos( + π) = cos(). Again, ou can check these identities with the angle addition formulas, but I find it s easiest to just remember these as shifting b half the wavelength negates the sine wave. π π π π sin() sin( + π) Finall, looking at the graphs of sin and cos, ou can see that sin() is an odd function and cos() is an even function. In particular, sin( ) = sin(): π π π π sin() sin( ) and cos( ) = cos(): cos() cos( ) π π π π 3
Simplifing epressions like sin(arctan()). Geometric approach This is the easiest wa to simplif these tpes of epressions for trig functions. Note though that this method doesn t work (or at least not easil) for hperbolic functions. We ll use the geometric definitions of sin and friends, as captured b the mnemonic SOH-CAH-TOA. We ll draw a right triangle so that one of its acute angles represents the inner inverse trig function, then solve the triangle to compute the value of the outer trig function. If that sounds confusing, don t worr, it s much easier to follow along with an eample. Eample. Simplif: sin(arctan()). Proof. For convenience, write θ = arctan(), so that tanθ = =. The reason for the sill notation is to connect to the definition of tan as the ratio of the lengths of the opposite and adjacent sides in a right triangle: θ Note that we could have assigned the side lengths to be anthing, as long as their ratio was equal to tan(θ) =. So, for eample, we could have drawn this triangle instead: θ Of course, there s no good reason for doing that, so we ll keep the original triangle. Now, we can easil solve for the length of the third side: + θ And now, using the ratio definition of sin, we can read off Since we defined θ = arctan(), we have shown that sin(θ) = sin(arctan()) = + +. Algebraic approach This approach also works with hperbolic functions. The idea is to use trig identities to replace the inner trig function b the outer (inverse) trig function. Eample. Simplif: sin(arctan()). 4
Solution. We want to replace the outer sin b tan, to make use of the fact that tan(arctan()) =. We know the following: sin () + cos () = Since we want to get a relation between tan() and sin(), we divide b sin (): + tan () = sin () (Often, we write the above identit with cot and csc instead of involves sin and tan, it s better to use sin and tan eplicitl.) tan and sin, but since the epression we re simplifing We want to use this trig identit to replace sin(arctan()) b something involving tan(arctan()). From the identit, so and finall ( sin () = + ) tan = () sin (arctan()) = ( tan ) () + = tan () tan () tan () + tan (arctan()) tan (arctan()) + = + sin(arctan()) = +.3 Etension to hperbolic functions The advantage of the algebraic method from section. is that in etends easil to hperbolic functions (the geometric approach doesn t etend so well, since the geometr behind hperbolic functions is tied to hperbolas instead of more-familiar circles). Eample 3. Simplif: sinh(artanh()) (where artanh() denotes the inverse hperbolic tangent) Solution. The basic hperbolic relation is cosh () sinh () = Divide both sides b sinh (), b analog to the eample from section.: Thus, So, replacing b artanh(), we have: ( ) sinh () = tanh () = sinh (artanh()) = tanh () = sinh () ( tanh ) () = tanh () tanh () tanh (artanh()) tanh (artanh()) = tanh () Thus, sinh(artanh()) = 5
3 Trig and inverse trig derivatives and integrals 3. Trig Derivatives d I m going to assume ou know that d [sin()] = cos() and d d [cos()] = sin(). Then, it s eas to compute the rest of the trig derivatives if ou don t have them memorized. For eample, d d [tan()] = d [ ] sin() cos()(cos()) sin()( sin()) = d cos() cos = () cos () = sec () d d [csc()] = d [ ] = d sin() sin cos() = cot()csc() () 3. Inverse trig derivatives These are slightl harder than the trig derivatives, but onl slightl. Basicall, use the definition of the inverse trig function and implicit differentiation: Eample 4. Find d d [arccos()]. Solution. Let = arccos(), so that cos() =. We want to find d d. Let s use implicit differentiation: The onl thing left to do is to find the relation See section for details. d d [cos()] = d d [] sin() d d = d d = sin() d d = sin(arccos()) sin(arccos()) = = d d [arccos()] = 3.3 Trig integrals Two (well, reall four) trig integrals deserve special mention: tan()d = ln(cos())+c and sec()d = ln(tan()+ sec()) +C (and the similar integrals cot()d = ln(sin()) +C and csc()d = ln(cot() + csc()) +C). The tangent/cotangent integrals are eas: sin() tan()d = cos() d du = u = ln(u) +C = ln(cos()) +C u = cos(),du = sin()d = ln(sec()) +C (if ou prefer ln(sec()) to ln(cos())) 6
The other non-obvious trig integral is sec()d (or csc()d). You might just want to memorize sec()d = ln(tan() + sec()) +C On the other hand, in case ou re interested, here s a reasonabl-memorable calculation (courtes of Wikipedia, https://en.wikipedia.org/wiki/integral_of_the_secant_function): sec()d = = = = cos() cos () d cos() sin () d dt t t = sin(),dt = cos()d ( +t + ) dt B partial fractions; eas. t = ln( +t) ln( t) +C = ( ) +t ln +C t = ( ) + sin() ln +C Note: alread a reasonable answer. sin() = ( ( + sin() ln sin() + sin() ) +C Multipl b the conjugate often useful! + sin() ( + sin()) = ln sin () ( + sin()) = ln cos () ( ) + sin() = ln cos() = ln(sec() + tan()) Before ou give up and just memorize, sec()d = ln(sec() + tan()), ou might want to tr working through this calculation b hand (without just coping m notes or the Wikipedia page!) to see if that s easier for ou to remember. 7
Identit Mnemonic () sin () + cos () = Memorize () sin () = cos () Restatement of () (3) cos () = sin () Restatement of () (4) tan () + = sec () Divide both sides of () b cos () (5) sin( + ) = sin() cos() + cos() sin() Memorize (6) cos( + ) = cos() cos() sin() sin() Memorize (7) sin() = sin() cos() Derive (easil!) from (5) (8) cos() = cos () sin () Derive (easil!) from (6) (9) cos() = cos () Combine (8) and () (0) cos() = sin () Combine (8) and (3) () cos () = ( + cos()) Solve (9) for cos () () sin () = ( cos()) Solve (0) for sin () (3) sin( π ) = cos() Look at the graphs (see above) (4) sin( + π) = sin() Half-wavelength (5) cos( + π) = cos() Half-wavelength (6) cos( π ) = sin() Combine (3) and (4) (tr it!) (7) cos( + π ) = sin() Restatement of (3) (8) sin( + π ) = cos() Restatement of (6) (9) tan( ± π ) = tan() Combine (7)+(8) or (3)+(6) (0) tan( + π) = tan() Combine (4) and (5) Table : Trig identities and mnemonics Result Method () derivative of tan(), sec(), etc. write as a fraction in sin() and cos(); use quotient rule () derivative of arctan(), etc. write e.g. tan() = ; find d/d via implicit differentiation (3) tan()d = ln(sec()) +C write as a fraction sin()/cos(); sustitute u = cos() (4) sec()d = ln(sec() + tan()) +C Memorize, or use sec() = cos()/( sin ()), substitute u = sin(), and then use partial fractions Table : Calculus with trigonometric functions 8