Algebra of the trigonometric functions.€¦ · Some trig identities. Let’s recall the things we...

Algebra of the trigonometric functions.

Algebra of the trigonometric functions. 1 / 7

Some trig identities.

Let’s recall the things we know about the trig functions

Theorem

(The most important identity)

cos(θ)2 + sin(θ)2 = 1

(The obvious identities)

tan(θ) = sin(θ)cos(θ) ,sec(θ) = 1

cos(θ) , csc(θ) = 1sin(θ) , cot(θ) = cos(θ)

sin(θ) ,

(Previously)

cos(−θ) = cos(θ), sin(−θ) = − sin(θ),cos(π − θ) = − cos(θ), sin(π − θ) = sin(θ),cos(π + θ) = − cos(θ), sin(π + θ) = − sin(θ),cos(π/2 − θ) = sin(θ), sin(π/2 − θ) = cos(θ).

Can you get a formula for cos(π/2 + θ) in terms of sin(θ) and cos(θ)?

Proposition

For every angle θ, cos(π/2 + θ) = − sin(θ), sin(π/2 + θ) = cos(θ)




Theorem

(The most important identity) cos(θ)2 + sin(θ)2 =

1




sin(θ) ,

(Previously)



Proposition





Theorem

(The most important identity) cos(θ)2 + sin(θ)2 = 1


tan(θ) =

sin(θ)cos(θ) ,sec(θ) = 1


sin(θ) ,

(Previously)



Proposition





Theorem



tan(θ) = sin(θ)cos(θ) ,

sec(θ) = 1cos(θ) , csc(θ) = 1

sin(θ) , cot(θ) = cos(θ)sin(θ) ,

(Previously)



Proposition





Theorem



tan(θ) = sin(θ)cos(θ) ,sec(θ) =

1cos(θ) , csc(θ) = 1

sin(θ) , cot(θ) = cos(θ)sin(θ) ,

(Previously)



Proposition





Theorem




cos(θ) ,

csc(θ) = 1sin(θ) , cot(θ) = cos(θ)

sin(θ) ,

(Previously)



Proposition





Theorem




cos(θ) , csc(θ) =

1sin(θ) , cot(θ) = cos(θ)

sin(θ) ,

(Previously)



Proposition





Theorem




cos(θ) , csc(θ) = 1sin(θ) ,

cot(θ) = cos(θ)sin(θ) ,

(Previously)



Proposition





Theorem




cos(θ) , csc(θ) = 1sin(θ) , cot(θ) =

cos(θ)sin(θ) ,

(Previously)



Proposition





Theorem





sin(θ) ,

(Previously)



Proposition





Theorem





sin(θ) ,

(Previously)

cos(−θ) =

cos(θ), sin(−θ) = − sin(θ),cos(π − θ) = − cos(θ), sin(π − θ) = sin(θ),cos(π + θ) = − cos(θ), sin(π + θ) = − sin(θ),cos(π/2 − θ) = sin(θ), sin(π/2 − θ) = cos(θ).


Proposition





Theorem





sin(θ) ,

(Previously)

cos(−θ) = cos(θ),

sin(−θ) = − sin(θ),cos(π − θ) = − cos(θ), sin(π − θ) = sin(θ),cos(π + θ) = − cos(θ), sin(π + θ) = − sin(θ),cos(π/2 − θ) = sin(θ), sin(π/2 − θ) = cos(θ).


Proposition





Theorem





sin(θ) ,

(Previously)

cos(−θ) = cos(θ), sin(−θ) =

− sin(θ),cos(π − θ) = − cos(θ), sin(π − θ) = sin(θ),cos(π + θ) = − cos(θ), sin(π + θ) = − sin(θ),cos(π/2 − θ) = sin(θ), sin(π/2 − θ) = cos(θ).


Proposition





Theorem





sin(θ) ,

(Previously)

cos(−θ) = cos(θ), sin(−θ) = − sin(θ),

cos(π − θ) = − cos(θ), sin(π − θ) = sin(θ),cos(π + θ) = − cos(θ), sin(π + θ) = − sin(θ),cos(π/2 − θ) = sin(θ), sin(π/2 − θ) = cos(θ).


Proposition





Theorem





sin(θ) ,

(Previously)

cos(−θ) = cos(θ), sin(−θ) = − sin(θ),cos(π − θ) =

− cos(θ), sin(π − θ) = sin(θ),cos(π + θ) = − cos(θ), sin(π + θ) = − sin(θ),cos(π/2 − θ) = sin(θ), sin(π/2 − θ) = cos(θ).


Proposition





Theorem





sin(θ) ,

(Previously)

cos(−θ) = cos(θ), sin(−θ) = − sin(θ),cos(π − θ) = − cos(θ),

sin(π − θ) = sin(θ),cos(π + θ) = − cos(θ), sin(π + θ) = − sin(θ),cos(π/2 − θ) = sin(θ), sin(π/2 − θ) = cos(θ).


Proposition





Theorem





sin(θ) ,

(Previously)

cos(−θ) = cos(θ), sin(−θ) = − sin(θ),cos(π − θ) = − cos(θ), sin(π − θ) =

sin(θ),cos(π + θ) = − cos(θ), sin(π + θ) = − sin(θ),cos(π/2 − θ) = sin(θ), sin(π/2 − θ) = cos(θ).


Proposition





Theorem





sin(θ) ,

(Previously)

cos(−θ) = cos(θ), sin(−θ) = − sin(θ),cos(π − θ) = − cos(θ), sin(π − θ) = sin(θ),

cos(π + θ) = − cos(θ), sin(π + θ) = − sin(θ),cos(π/2 − θ) = sin(θ), sin(π/2 − θ) = cos(θ).


Proposition





Theorem





sin(θ) ,

(Previously)

cos(−θ) = cos(θ), sin(−θ) = − sin(θ),cos(π − θ) = − cos(θ), sin(π − θ) = sin(θ),cos(π + θ) =

− cos(θ), sin(π + θ) = − sin(θ),cos(π/2 − θ) = sin(θ), sin(π/2 − θ) = cos(θ).


Proposition





Theorem





sin(θ) ,

(Previously)

cos(−θ) = cos(θ), sin(−θ) = − sin(θ),cos(π − θ) = − cos(θ), sin(π − θ) = sin(θ),cos(π + θ) = − cos(θ),

sin(π + θ) = − sin(θ),cos(π/2 − θ) = sin(θ), sin(π/2 − θ) = cos(θ).


Proposition





Theorem





sin(θ) ,

(Previously)

cos(−θ) = cos(θ), sin(−θ) = − sin(θ),cos(π − θ) = − cos(θ), sin(π − θ) = sin(θ),cos(π + θ) = − cos(θ), sin(π + θ) =

− sin(θ),cos(π/2 − θ) = sin(θ), sin(π/2 − θ) = cos(θ).


Proposition





Theorem





sin(θ) ,

(Previously)

cos(−θ) = cos(θ), sin(−θ) = − sin(θ),cos(π − θ) = − cos(θ), sin(π − θ) = sin(θ),cos(π + θ) = − cos(θ), sin(π + θ) = − sin(θ),

cos(π/2 − θ) = sin(θ), sin(π/2 − θ) = cos(θ).


Proposition





Theorem





sin(θ) ,

(Previously)

cos(−θ) = cos(θ), sin(−θ) = − sin(θ),cos(π − θ) = − cos(θ), sin(π − θ) = sin(θ),cos(π + θ) = − cos(θ), sin(π + θ) = − sin(θ),cos(π/2 − θ) =

sin(θ), sin(π/2 − θ) = cos(θ).


Proposition





Theorem





sin(θ) ,

(Previously)

cos(−θ) = cos(θ), sin(−θ) = − sin(θ),cos(π − θ) = − cos(θ), sin(π − θ) = sin(θ),cos(π + θ) = − cos(θ), sin(π + θ) = − sin(θ),cos(π/2 − θ) = sin(θ),

sin(π/2 − θ) = cos(θ).


Proposition





Theorem





sin(θ) ,

(Previously)

cos(−θ) = cos(θ), sin(−θ) = − sin(θ),cos(π − θ) = − cos(θ), sin(π − θ) = sin(θ),cos(π + θ) = − cos(θ), sin(π + θ) = − sin(θ),cos(π/2 − θ) = sin(θ), sin(π/2 − θ) =

cos(θ).


Proposition





Theorem





sin(θ) ,

(Previously)



Proposition





Theorem





sin(θ) ,

(Previously)



Proposition

For every angle θ, cos(π/2 + θ) = − sin(θ), sin(π/2 + θ) =

cos(θ)




Theorem





sin(θ) ,

(Previously)



Proposition



Getting identities by putting everything in terms of sin andcos.

1 (I’ll do this one) Simplify tan(γ) + cot(γ).

tan(γ) + cot(γ) =

sin(γ)cos(γ) + cos(γ)

sin(γ)

=

sin(γ)2+cos(γ)2

cos(γ)·(sin(γ)

Use the most important identity

=

1cos(γ)·(sin(γ)

=

sec(γ) · csc(γ)

2 (group work) Simplify tan(γ) · cot(γ).




tan(γ) + cot(γ) = sin(γ)cos(γ) + cos(γ)

sin(γ)

=

sin(γ)2+cos(γ)2

cos(γ)·(sin(γ)


=

1cos(γ)·(sin(γ)

=

sec(γ) · csc(γ)






sin(γ)

= sin(γ)2+cos(γ)2

cos(γ)·(sin(γ)


=

1cos(γ)·(sin(γ)

=

sec(γ) · csc(γ)






sin(γ)

= sin(γ)2+cos(γ)2

cos(γ)·(sin(γ) Use the most important identity

=

1cos(γ)·(sin(γ)

=

sec(γ) · csc(γ)






sin(γ)

= sin(γ)2+cos(γ)2


= 1cos(γ)·(sin(γ)

=

sec(γ) · csc(γ)






sin(γ)

= sin(γ)2+cos(γ)2


= 1cos(γ)·(sin(γ)

= sec(γ) · csc(γ)



Proving something is an identity

1 (I’ll do this one) Prove that sec(α)2 + csc(α)2 = (sec(α) · csc(α))2

Expand sec(α)2 + csc(α)2

sec(α)2 + csc(α)2 =

1cos(α)2

+ 1sin(α)2

=

sin(α)2+cos(α)2

cos(α)2·sin(α)2

The important identity

=

1cos(α)2·sin(α)2

=

(sec(α) · csc(α))2

This is what we set out to prove!

2 (Group work) Prove that 1 + tan(θ)2 = sec(θ)2 is an identity bysimplifying one side of the equation.





sec(α)2 + csc(α)2 = 1cos(α)2

+ 1sin(α)2

=

sin(α)2+cos(α)2

cos(α)2·sin(α)2


=

1cos(α)2·sin(α)2

=









+ 1sin(α)2

= sin(α)2+cos(α)2

cos(α)2·sin(α)2


=

1cos(α)2·sin(α)2

=









+ 1sin(α)2

= sin(α)2+cos(α)2

cos(α)2·sin(α)2 The important identity

=

1cos(α)2·sin(α)2

=









+ 1sin(α)2

= sin(α)2+cos(α)2


= 1cos(α)2·sin(α)2

=









+ 1sin(α)2

= sin(α)2+cos(α)2


= 1cos(α)2·sin(α)2

= (sec(α) · csc(α))2




Proving something is NOT an identity

If you suspect that something might be an identity, but you are not sure,then try to check if for some easy to compute inputsExample: Is sin(α + β) = sin(α) + sin(β) an identity?

Does it check out for α = 0, β = 0?

sin(0 + 0) = sin(0) = 0 and sin(0) + sin(0) = 0 + 0 = 0sin(0) + sin(0) = 0 + 0 = 0Passes this test

Does it check out for α = π/2, β = π/2?

sin(π/2 + π/2) = sin(π) = 0sin(π/2) + sin(π/2) = 1 + 1 = 2This is not an identity!

In your groups, and in your notes:Prove that cos(α · β) = cos(α) · cos(β) is not an identity.




Does it check out for α = 0, β = 0?sin(0 + 0) = sin(0) = 0 and sin(0) + sin(0) = 0 + 0 = 0

sin(0) + sin(0) = 0 + 0 = 0Passes this test







Does it check out for α = 0, β = 0?sin(0 + 0) = sin(0) = 0 and sin(0) + sin(0) = 0 + 0 = 0sin(0) + sin(0) = 0 + 0 = 0

Passes this test







Does it check out for α = 0, β = 0?sin(0 + 0) = sin(0) = 0 and sin(0) + sin(0) = 0 + 0 = 0sin(0) + sin(0) = 0 + 0 = 0Passes this test








Does it check out for α = π/2, β = π/2?sin(π/2 + π/2) = sin(π) = 0

sin(π/2) + sin(π/2) = 1 + 1 = 2This is not an identity!






Does it check out for α = π/2, β = π/2?sin(π/2 + π/2) = sin(π) = 0sin(π/2) + sin(π/2) = 1 + 1 = 2

This is not an identity!






Does it check out for α = π/2, β = π/2?sin(π/2 + π/2) = sin(π) = 0sin(π/2) + sin(π/2) = 1 + 1 = 2This is not an identity!



Computing all the trig functions when you know only one.

1 (I’ll do this one) If tan(θ) = 3 and π < θ < 3π/3 then computecos(θ) and sin(θ)

2 (Groupwork) If sec(θ) = 2 and 0 < θ < π/2, compute cos(θ) andsin(θ).

Express tan(θ) = 3 in terms of sin and cos.

3 = sin(θ)cos(θ) so

3 · cos(θ) = sin(θ)

Now use the best identity:

1 = sin(θ)2 + cos(θ)2 =

(3 cos(θ)2) + cos(θ)2 so 10 cos(θ)2 = 1

cos(θ) = ±√1010

In which quadrant are we? Quadrant 3. cos(θ) and sin(θ) are negative.

cos(θ) = −√1010

sin(θ) = 3 cos(θ) =

−√1010






3 = sin(θ)cos(θ) so 3 · cos(θ) = sin(θ)


1 = sin(θ)2 + cos(θ)2 =

(3 cos(θ)2) + cos(θ)2 so 10 cos(θ)2 = 1

cos(θ) = ±√1010


cos(θ) = −√1010


−√1010








1 = sin(θ)2 + cos(θ)2 = (3 cos(θ)2) + cos(θ)2 so

10 cos(θ)2 = 1

cos(θ) = ±√1010


cos(θ) = −√1010


−√1010








1 = sin(θ)2 + cos(θ)2 = (3 cos(θ)2) + cos(θ)2 so 10 cos(θ)2 = 1

cos(θ) = ±√1010


cos(θ) = −√1010


−√1010









cos(θ) = ±√1010

In which quadrant are we?

Quadrant 3. cos(θ) and sin(θ) are negative.

cos(θ) = −√1010


−√1010









cos(θ) = ±√1010


cos(θ) = −√1010


−√1010









cos(θ) = ±√1010


cos(θ) = −√1010

sin(θ) = 3 cos(θ) = −√1010


The algebra tricks you know and love are still good.

1 (I’ll do this one) Prove that cos(θ)1+sin(θ) = 1−sin(θ)

cos(θ) .

2 (For you) Simplify each side to prove that1

1+cos(θ) + 11−cos(θ) = 2 + 2 · cot(θ)2.

Start with the right hand side and multiply and divide by the correct thingto make the denominator a difference of squares

cos(θ)

1 + sin(θ)=

cos(θ) · (1 − sin(θ))

(1 + sin(θ)) · (1 − sin(θ))=

cos(θ) · (1 − sin(θ))

1 − sin(θ)2

Use the best identity cos(θ)1+sin(θ) = cos(θ)·(1−sin(θ))

cos(θ)2

Cancellation: cos(θ)1+sin(θ) = 1−sin(θ)

cos(θ)But that’s what we wanted to prove!




cos(θ) .


1+cos(θ) + 11−cos(θ) = 2 + 2 · cot(θ)2.


cos(θ)

1 + sin(θ)=

cos(θ) · (1 − sin(θ))

(1 + sin(θ)) · (1 − sin(θ))=

cos(θ) · (1 − sin(θ))

1 − sin(θ)2

Use the best identity

cos(θ)1+sin(θ) = cos(θ)·(1−sin(θ))

cos(θ)2






cos(θ) .


1+cos(θ) + 11−cos(θ) = 2 + 2 · cot(θ)2.


cos(θ)

1 + sin(θ)=

cos(θ) · (1 − sin(θ))

(1 + sin(θ)) · (1 − sin(θ))=

cos(θ) · (1 − sin(θ))

1 − sin(θ)2

Use the best identity cos(θ)1+sin(θ) =

cos(θ)·(1−sin(θ))cos(θ)2






cos(θ) .


1+cos(θ) + 11−cos(θ) = 2 + 2 · cot(θ)2.


cos(θ)

1 + sin(θ)=

cos(θ) · (1 − sin(θ))

(1 + sin(θ)) · (1 − sin(θ))=

cos(θ) · (1 − sin(θ))

1 − sin(θ)2


cos(θ)2






cos(θ) .


1+cos(θ) + 11−cos(θ) = 2 + 2 · cot(θ)2.


cos(θ)

1 + sin(θ)=

cos(θ) · (1 − sin(θ))

(1 + sin(θ)) · (1 − sin(θ))=

cos(θ) · (1 − sin(θ))

1 − sin(θ)2


cos(θ)2

Cancellation:

cos(θ)1+sin(θ) = 1−sin(θ)





cos(θ) .


1+cos(θ) + 11−cos(θ) = 2 + 2 · cot(θ)2.


cos(θ)

1 + sin(θ)=

cos(θ) · (1 − sin(θ))

(1 + sin(θ)) · (1 − sin(θ))=

cos(θ) · (1 − sin(θ))

1 − sin(θ)2


cos(θ)2

Cancellation: cos(θ)1+sin(θ) =

1−sin(θ)cos(θ)

But that’s what we wanted to prove!




cos(θ) .


1+cos(θ) + 11−cos(θ) = 2 + 2 · cot(θ)2.


cos(θ)

1 + sin(θ)=

cos(θ) · (1 − sin(θ))

(1 + sin(θ)) · (1 − sin(θ))=

cos(θ) · (1 − sin(θ))

1 − sin(θ)2


cos(θ)2


cos(θ)

But that’s what we wanted to prove!




cos(θ) .


1+cos(θ) + 11−cos(θ) = 2 + 2 · cot(θ)2.


cos(θ)

1 + sin(θ)=

cos(θ) · (1 − sin(θ))

(1 + sin(θ)) · (1 − sin(θ))=

cos(θ) · (1 − sin(θ))

1 − sin(θ)2


cos(θ)2




One of the homework problems

Prove the identity:

tan(θ) + sec(θ) − 1

tan(θ) − sec(θ) + 1=

1 + sin(θ)

cos(θ)


Algebra of the trigonometric functions.€¦ · Some trig identities. Let’s recall the things we...

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Transcript of Algebra of the trigonometric functions.€¦ · Some trig identities. Let’s recall the things we...