Unit 2

Trigonometry


Hold on a minute! There wasn’t much direction to the proof in Example 1. How do you know when to use an identity? And, which one should you use? Why didn’t you start with the right side instead of the left? This makes no sense!

Proofs are different from a lot of other types of math you have learned because they don’t have a set pattern. Here are some suggestions you may find helpful when proving trigonometric identities.

  • Begin with the more complicated side. It is usually easier to simplify an expression than it is to expand one.
  • Make substitutions using identities. Keep your formula sheet handy!
  • Try changing all trigonometric ratios to sines and cosines.
  • If a second degree trigonometric ratio is used («math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mrow»«msup»«mi»sin«/mi»«mn»2«/mn»«/msup»«mi»x«/mi»«/mrow»«/mstyle»«/math», «math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mrow»«msup»«mi»sec«/mi»«mn»2«/mn»«/msup»«mi»x«/mi»«/mrow»«/mstyle»«/math», etc), try one of the Pythagorean identities.
  • If a double angle is used («math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mrow»«mi»cos«/mi»«mfenced»«mrow»«mn»2«/mn»«mi»x«/mi»«/mrow»«/mfenced»«/mrow»«/mstyle»«/math», «math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mrow»«mi»tan«/mi»«mfenced»«mrow»«mn»2«/mn»«mi»x«/mi»«/mrow»«/mfenced»«/mrow»«/mstyle»«/math», etc), try one of the double angle identities.
  • Try factoring an expression.
  • Try expanding an expression.
  • Try multiplying the numerator and denominator of an expression by its conjugate. (The conjugate of «math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mrow»«mi»x«/mi»«mo»+«/mo»«mi»y«/mi»«/mrow»«/mstyle»«/math» is «math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mrow»«mi»x«/mi»«mo»§#8722;«/mo»«mi»y«/mi»«/mrow»«/mstyle»«/math».) This is helpful when «math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mrow»«mn»1«/mn»«mo»§#177;«/mo»«mi»sin«/mi»«mi»x«/mi»«/mrow»«/mstyle»«/math» or «math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mrow»«mn»1«/mn»«mo»§#177;«/mo»«mi»cos«/mi»«mi»x«/mi»«/mrow»«/mstyle»«/math» is in the numerator or denominator.

But, the most important suggestion of all is to KEEP TRYING. Proofs often require several attempts, a lot of scrap paper, and a bit of luck—none of which are shown in the following examples! Do not become discouraged.

Prove the equation «math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mrow»«mfrac»«mn»1«/mn»«mrow»«mn»1«/mn»«mo»+«/mo»«mi»sin«/mi»«mi»x«/mi»«/mrow»«/mfrac»«mo»=«/mo»«mfrac»«mrow»«mi»sec«/mi»«mi»x«/mi»«mo»§#8722;«/mo»«mi»sin«/mi»«mi»x«/mi»«mi»sec«/mi»«mi»x«/mi»«/mrow»«mrow»«mi»cos«/mi»«mi»x«/mi»«/mrow»«/mfrac»«/mrow»«/mstyle»«/math» is an identity for all permissible values of «math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mi»x«/mi»«/mstyle»«/math».
This identity looks pretty complicated, so how should you start? Is multiplying the left side by the conjugate «math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mfrac»«mrow»«mn»1«/mn»«mo»§#8722;«/mo»«mi»sin«/mi»«mi»x«/mi»«/mrow»«mrow»«mn»1«/mn»«mo»§#8722;«/mo»«mi»sin«/mi»«mi»x«/mi»«/mrow»«/mfrac»«/mstyle»«/math» a good idea? Should you use the identity «math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mrow»«mi»sec«/mi»«mi»x«/mi»«mo»=«/mo»«mfrac»«mn»1«/mn»«mrow»«mi»cos«/mi»«mi»x«/mi»«/mrow»«/mfrac»«/mrow»«/mstyle»«/math» on the right side to reduce the number of different ratios used? Maybe factor out «math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mrow»«mi»sec«/mi»«mi»x«/mi»«/mrow»«/mstyle»«/math» on the right side to simplify the expression?

These are all good suggestions, and any of them can lead to a successful proof. Pick the one that seems the most reasonable for YOU, and get started. If something doesn’t work, try again, using a different approach.

Here is one way of proving the identity.

«math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mtable columnalign=¨right center left¨»«mtr»«mtd»«mi»right«/mi»«mo»§#160;«/mo»«mi»side«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mrow»«mi»sec«/mi»«mi»x«/mi»«mo»§#8722;«/mo»«mi»sin«/mi»«mi»x«/mi»«mi»sec«/mi»«mi»x«/mi»«/mrow»«mrow»«mi»cos«/mi»«mi»x«/mi»«/mrow»«/mfrac»«/mtd»«/mtr»«mtr»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mrow»«mi»sec«/mi»«mi»x«/mi»«mfenced»«mrow»«mn»1«/mn»«mo»§#8722;«/mo»«mi»sin«/mi»«mi»x«/mi»«/mrow»«/mfenced»«/mrow»«mrow»«mi»cos«/mi»«mi»x«/mi»«/mrow»«/mfrac»«/mtd»«/mtr»«mtr»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mrow»«mfenced»«mfrac»«mn»1«/mn»«mrow»«mi»cos«/mi»«mi»x«/mi»«/mrow»«/mfrac»«/mfenced»«mfenced»«mrow»«mn»1«/mn»«mo»§#8722;«/mo»«mi»sin«/mi»«mi»x«/mi»«/mrow»«/mfenced»«/mrow»«mrow»«mi»cos«/mi»«mi»x«/mi»«/mrow»«/mfrac»«/mtd»«/mtr»«mtr»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mrow»«mn»1«/mn»«mo»§#8722;«/mo»«mi»sin«/mi»«mi»x«/mi»«/mrow»«mrow»«msup»«mi»cos«/mi»«mn»2«/mn»«/msup»«mi»x«/mi»«/mrow»«/mfrac»«/mtd»«/mtr»«/mtable»«/mstyle»«/math»

«math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mtable columnalign=¨right center left¨»«mtr»«mtd»«mi»left«/mi»«mo»§#160;«/mo»«mi»side«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mn»1«/mn»«mrow»«mn»1«/mn»«mo»+«/mo»«mi»sin«/mi»«mi»x«/mi»«/mrow»«/mfrac»«/mtd»«/mtr»«mtr»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mn»1«/mn»«mrow»«mn»1«/mn»«mo»+«/mo»«mi»sin«/mi»«mi»x«/mi»«/mrow»«/mfrac»«mo»§#8226;«/mo»«mfrac»«mrow»«mn»1«/mn»«mo»§#8722;«/mo»«mi»sin«/mi»«mi»x«/mi»«/mrow»«mrow»«mn»1«/mn»«mo»§#8722;«/mo»«mi»sin«/mi»«mi»x«/mi»«/mrow»«/mfrac»«/mtd»«/mtr»«mtr»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mrow»«mn»1«/mn»«mo»§#8722;«/mo»«mi»sin«/mi»«mi»x«/mi»«/mrow»«mrow»«mn»1«/mn»«mo»§#8722;«/mo»«msup»«mi»sin«/mi»«mn»2«/mn»«/msup»«mi»x«/mi»«/mrow»«/mfrac»«/mtd»«/mtr»«mtr»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mrow»«mn»1«/mn»«mo»§#8722;«/mo»«mi»sin«/mi»«mi»x«/mi»«/mrow»«mrow»«msup»«mi»cos«/mi»«mn»2«/mn»«/msup»«mi»x«/mi»«/mrow»«/mfrac»«/mtd»«/mtr»«/mtable»«/mstyle»«/math»

The left side is equal to the right side, so the equation is an identity.

In this proof, make sure you understand

  • that factoring «math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mrow»«mi»sec«/mi»«mi»x«/mi»«mo»§#8722;«/mo»«mi»sin«/mi»«mi»x«/mi»«mi»sec«/mi»«mi»x«/mi»«/mrow»«/mstyle»«/math» is very similar to factoring «math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mrow»«mi»a«/mi»«mo»§#8722;«/mo»«mi»b«/mi»«mi»a«/mi»«/mrow»«/mstyle»«/math»,
  • how the identity «math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mrow»«mi»sec«/mi»«mi»x«/mi»«mo»=«/mo»«mfrac»«mn»1«/mn»«mrow»«mi»cos«/mi»«mi»x«/mi»«/mrow»«/mfrac»«/mrow»«/mstyle»«/math» was used,
  • how the identity «math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mrow»«msup»«mi»sin«/mi»«mn»2«/mn»«/msup»«mi»x«/mi»«mo»+«/mo»«msup»«mi»cos«/mi»«mn»2«/mn»«/msup»«mi»x«/mi»«mo»=«/mo»«mn»1«/mn»«/mrow»«/mstyle»«/math» was used, and
  • why multiplying by «math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mfrac»«mrow»«mn»1«/mn»«mo»§#8722;«/mo»«mi»sin«/mi»«mi»x«/mi»«/mrow»«mrow»«mn»1«/mn»«mo»§#8722;«/mo»«mi»sin«/mi»«mi»x«/mi»«/mrow»«/mfrac»«/mstyle»«/math» does not change the value of an expression.
Can you come up with a simpler proof than the one shown?

Prove the equation «math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mrow»«mfrac»«mrow»«mi»cos«/mi»«mn»2«/mn»«mi»x«/mi»«mo»+«/mo»«mi»sin«/mi»«mi»x«/mi»«/mrow»«mrow»«mi»sin«/mi»«mn»2«/mn»«mi»x«/mi»«mo»+«/mo»«mi»cos«/mi»«mi»x«/mi»«/mrow»«/mfrac»«mo»=«/mo»«mi»sec«/mi»«mi»x«/mi»«mo»§#8722;«/mo»«mi»tan«/mi»«mi»x«/mi»«/mrow»«/mstyle»«/math» is an identity for all permissible values of «math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mi»x«/mi»«/mstyle»«/math».
The left side looks more complicated and includes double angles, so using double angle identities to simplify the double angles is a good place to start. Here is one possible proof.

«math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mtable columnalign=¨right center left¨»«mtr»«mtd»«mi»left«/mi»«mo»§#160;«/mo»«mi mathvariant=¨normal¨» «/mi»«mi»side«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mrow»«mi»cos«/mi»«mn»2«/mn»«mi»x«/mi»«mo»+«/mo»«mi»sin«/mi»«mi»x«/mi»«/mrow»«mrow»«mi»sin«/mi»«mn»2«/mn»«mi»x«/mi»«mo»+«/mo»«mi»cos«/mi»«mi»x«/mi»«/mrow»«/mfrac»«/mtd»«/mtr»«mtr»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mrow»«mfenced»«mrow»«mn»1«/mn»«mo»§#8722;«/mo»«mn»2«/mn»«msup»«mi»sin«/mi»«mn»2«/mn»«/msup»«mi»x«/mi»«/mrow»«/mfenced»«mo»+«/mo»«mi»sin«/mi»«mi»x«/mi»«/mrow»«mrow»«mfenced»«mrow»«mn»2«/mn»«mi»sin«/mi»«mi»x«/mi»«mi»cos«/mi»«mi»x«/mi»«/mrow»«/mfenced»«mo»+«/mo»«mi»cos«/mi»«mi»x«/mi»«/mrow»«/mfrac»«/mtd»«/mtr»«mtr»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mrow»«mo»§#8722;«/mo»«mn»2«/mn»«msup»«mi»sin«/mi»«mn»2«/mn»«/msup»«mi»x«/mi»«mo»+«/mo»«mi»sin«/mi»«mi»x«/mi»«mo»+«/mo»«mn»1«/mn»«/mrow»«mrow»«mn»2«/mn»«mi»sin«/mi»«mi»x«/mi»«mi»cos«/mi»«mi»x«/mi»«mo»+«/mo»«mi»cos«/mi»«mi»x«/mi»«/mrow»«/mfrac»«/mtd»«/mtr»«mtr»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mrow»«mfenced»«mrow»«mn»1«/mn»«mo»§#8722;«/mo»«mi»sin«/mi»«mi»x«/mi»«/mrow»«/mfenced»«mfenced»«mrow»«mn»2«/mn»«mi»sin«/mi»«mi»x«/mi»«mo»+«/mo»«mn»1«/mn»«/mrow»«/mfenced»«/mrow»«mrow»«mi»cos«/mi»«mi»x«/mi»«mfenced»«mrow»«mn»2«/mn»«mi»sin«/mi»«mi»x«/mi»«mo»+«/mo»«mn»1«/mn»«/mrow»«/mfenced»«/mrow»«/mfrac»«/mtd»«/mtr»«mtr»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mrow»«mfenced»«mrow»«mn»1«/mn»«mo»§#8722;«/mo»«mi»sin«/mi»«mi»x«/mi»«/mrow»«/mfenced»«menclose notation=¨updiagonalstrike¨»«mfenced»«mrow»«mn»2«/mn»«mi»sin«/mi»«mi»x«/mi»«mo»+«/mo»«mn»1«/mn»«/mrow»«/mfenced»«/menclose»«/mrow»«mrow»«mi»cos«/mi»«mi»x«/mi»«menclose notation=¨updiagonalstrike¨»«mfenced»«mrow»«mn»2«/mn»«mi»sin«/mi»«mi»x«/mi»«mo»+«/mo»«mn»1«/mn»«/mrow»«/mfenced»«/menclose»«/mrow»«/mfrac»«/mtd»«/mtr»«mtr»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mrow»«mn»1«/mn»«mo»§#8722;«/mo»«mi»sin«/mi»«mi»x«/mi»«/mrow»«mrow»«mi»cos«/mi»«mi»x«/mi»«/mrow»«/mfrac»«/mtd»«/mtr»«/mtable»«/mstyle»«/math»

«math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mtable columnalign=¨right center left¨»«mtr»«mtd»«mi»right«/mi»«mo»§#160;«/mo»«mi»side«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mi»sec«/mi»«mi»x«/mi»«mo»§#8722;«/mo»«mi»tan«/mi»«mi»x«/mi»«/mtd»«/mtr»«mtr»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mn»1«/mn»«mrow»«mi»cos«/mi»«mi»x«/mi»«/mrow»«/mfrac»«mo»§#8722;«/mo»«mfrac»«mrow»«mi»sin«/mi»«mi»x«/mi»«/mrow»«mrow»«mi»cos«/mi»«mi»x«/mi»«/mrow»«/mfrac»«/mtd»«/mtr»«mtr»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mrow»«mn»1«/mn»«mo»§#8722;«/mo»«mi»sin«/mi»«mi»x«/mi»«/mrow»«mrow»«mi»cos«/mi»«mi»x«/mi»«/mrow»«/mfrac»«/mtd»«/mtr»«/mtable»«/mstyle»«/math»

The left side is equal to the right side, so the equation is an identity.

In this proof, make sure you understand

  • how the double angle identities were used to rewrite «math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mrow»«mi»cos«/mi»«mfenced»«mrow»«mn»2«/mn»«mi»x«/mi»«/mrow»«/mfenced»«/mrow»«/mstyle»«/math» and «math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mrow»«mi»sin«/mi»«mfenced»«mrow»«mn»2«/mn»«mi»x«/mi»«/mrow»«/mfenced»«/mrow»«/mstyle»«/math» in terms of «math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mrow»«mi»cos«/mi»«mi»x«/mi»«/mrow»«/mstyle»«/math» and «math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mrow»«mi»sin«/mi»«mi»x«/mi»«/mrow»«/mstyle»«/math»,
  • factoring «math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mrow»«mo»§#8722;«/mo»«mn»2«/mn»«msup»«mi»sin«/mi»«mn»2«/mn»«/msup»«mi»x«/mi»«mo»+«/mo»«mi»sin«/mi»«mi»x«/mi»«mo»+«/mo»«mn»1«/mn»«/mrow»«/mstyle»«/math» is similar to factoring «math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mrow»«mo»§#8722;«/mo»«msup»«mi»a«/mi»«mn»2«/mn»«/msup»«mo»+«/mo»«mi»a«/mi»«mo»+«/mo»«mn»1«/mn»«/mrow»«/mstyle»«/math»,
  • factoring «math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mrow»«mn»2«/mn»«mi»sin«/mi»«mi»x«/mi»«mi»cos«/mi»«mi»x«/mi»«mo»+«/mo»«mi»cos«/mi»«mi»x«/mi»«/mrow»«/mstyle»«/math» is similar to factoring «math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mrow»«mn»2«/mn»«mi»a«/mi»«mi»b«/mi»«mo»+«/mo»«mi»b«/mi»«/mrow»«/mstyle»«/math», and
  • «math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mrow»«mi»sec«/mi»«mi»x«/mi»«/mrow»«/mstyle»«/math» and «math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mrow»«mi»tan«/mi»«mi»x«/mi»«/mrow»«/mstyle»«/math» can both be written in terms of «math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mrow»«mi»cos«/mi»«mi»x«/mi»«/mrow»«/mstyle»«/math» and «math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mrow»«mi»sin«/mi»«mi»x«/mi»«/mrow»«/mstyle»«/math».
Again, can you come up with a simpler proof than the one shown?