Trigonometric Equations In quadratic form, using identities or linear in sine and cosine.
Standard A5: Prove Trigonometric Identities and use them ... · These alternative forms are very...
Transcript of Standard A5: Prove Trigonometric Identities and use them ... · These alternative forms are very...
Standard A5: Prove Trigonometric Identities and use them to simplify Trigonometric
equations
y
x
sinyr
θ=
r
cosxr
θ= tanyx
θ=
θ
sin2θ + cos2θ = 1
2 2 2
2 2 2x y rr r r
+ =
This is the first of three Pythagorean Iden..es
2 2 2x y r+ =
2 2
1x yr r
⎛ ⎞ ⎛ ⎞+ =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠−1 1
−1
1
x
y
r
( ),x y
θ
Reciprocal Trig Functions
θθ
sin1csc =
sin yr
θ = 1 1sin y
rθ=⎛ ⎞⎜ ⎟⎝ ⎠
1sin
ryθ
=
Which according to page 21 is cscθ
How did we get this, you wonder?
θθ
sin1csc =
Reciprocal Trig Functions
θθ
sin1csc =
θθ
cos1sec =
θθ
tan1cot =
And don’t forget…
tanθ = sinθ
cosθ cotθ = cosθ
sinθ
sin2θ + cos2θ = 1
xxx
xx
22
2
2
2
sin1
sincos
sinsin =+
tan2θ +1= sec2θ
1+ cot2θ = csc2θ
These are the three Pythagorean Iden..es
sin2θ + cos2θ = 1
tan2θ +1= sec2θ
1+ cot2θ = csc2θ
These and the other identities on Pg 21 and 22…
…will have to be memorized
555
The Pythagorean identities have alternative versions as well:
sin2θ + cos2θ =11− cos2θ = sin2θ1− sin2θ = cos2θ
tan2θ +1= sec2θsec2θ −1= tan2θsec2θ − tan2θ =1
1+ cot2θ = csc2θcsc2θ −1= cot2θcsc2θ − cot2θ =1
These alternative forms are very useful because they are “difference of squares” binomials that can be factored. For example,
1− sin2θ = 1− sinθ( ) 1+ sinθ( )
LEARNING OUTCOME Prove basic trigonometric identities. The proofs of the identities are algebraic in nature, meaning that the use of multiplication, addition, and common denominators will cause one side of the equation to simplify to the other. Or, similar to proofs in Geometry, that both sides simplify to the same thing. EX 1 Prove csc x tan xcosx =1
csc x tan xcosx =1
sin xi
sin xcosx
icosx =
1
Notice that the answer is the process, not the final line; the final line was given.
555
The Pythagorean identities have alternative versions as well:
sin2θ + cos2θ =11− cos2θ = sin2θ1− sin2θ = cos2θ
tan2θ +1= sec2θsec2θ −1= tan2θsec2θ − tan2θ =1
1+ cot2θ = csc2θcsc2θ −1= cot2θcsc2θ − cot2θ =1
These alternative forms are very useful because they are “difference of squares” binomials that can be factored. For example,
1− sin2θ = 1− sinθ( ) 1+ sinθ( )
LEARNING OUTCOME Prove basic trigonometric identities. The proofs of the identities are algebraic in nature, meaning that the use of multiplication, addition, and common denominators will cause one side of the equation to simplify to the other. Or, similar to proofs in Geometry, that both sides simplify to the same thing. EX 1 Prove csc x tan xcosx =1
csc x tan xcosx =1
sin xi
sin xcosx
icosx =
1
Notice that the answer is the process, not the final line; the final line was given.
Pg 22 is important as well
Show that
θθθ coscotsin = Rewrite in terms of sine and cosine
θθθθ cos
sincossin =
θθ coscos =
These are proofs but not as rigorous. Here are some tips on how you can approach them.
• Write everything in terms of sine and cosine • Look for squares - Check for Pythagorean Identity substitutions (squared trig
functions). If a direct substitution is there, use it. • Parentheses - Distribute if parentheses get in the way. Factor if parentheses can be
helpful • Common Denominators - If you have fractions that need to be added or subtracted,
look for common denominators
cos2 x(1+ tan2 x) = cos2 x(sec2 x) = cos2 x
1cos2 x
= 1
cos x(sec x + tan x) = cos xsec x + cos x tan x = 1+ sin x
1cos2 x
+ 1sin2 x
= sin2 xsin2 xcos2 x
+ cos2 xsin2 xcos2 x
= sin2 x + cos2 xsin2 xcos2 x
= 1sin2 xcos2 x
This often works though not always. Still, it can be a good way to start as you saw in the first example.
Show that
θ
θ
θ cot
cos1sin1
=
cosθsinθ
= cotθ
θθθ cot
seccsc =
Show that
θθθ csc)cot1(sin 2 =+ Notice the identity first
θθθ csc)(cscsin 2 =
θθ
θ cscsin1sin 2 =
θθ
cscsin1 =