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8 Cards in this Set

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Simplify the expression (sec² x-1)(cot x) to a single trigonometric funtion.
We know that (sec² x-1)=tan²x
and cotx=1/tan x
(sec²x-1)(cot x)= (tan² x)(1/tan x) = tan (x)
Simplify sin² x + sin² x cot x
We know that cot x = cos x / sin x
sin² x + sin² x cot x=
sin² x + sin² x (cos² x/ sin² x) =
sin² x + cos² x= 1
If csc θ=7/6 and tan θ=6/3.6, find the values of the cos θ using a Pythagorean identity.
cscθ=7/6 therefore, sinθ=6/7 
a²+b²=c² 
a²+6²=7² 
a²+36=49 
a²=13 
a=√13 
a=3.6 
cosθ=3.6/7
cscθ=7/6 therefore, sinθ=6/7
a²+b²=c²
a²+6²=7²
a²+36=49
a²=13
a=√13
a=3.6
cosθ=3.6/7
Assume that the sum angle formula for sine is true. Namely, sin(θ+∅)=sinθcos∅+cosθsin∅.
To derive the difference angle formula for sine, write sin(θ-∅) as sin (θ+(-∅)) and apply the sum angle formula for sine to the angles θ and -∅.
Use the fact that sine is an odd function while cosine is an even funtion to simplify your answer. Conclude that sin(θ-∅)=sin(θ)cos(∅)-cos(θ)sin(∅).
sin(θ-∅)=
sin(θ+(-∅))=
sinθcos(-∅)+cosθsin(-∅)=
sinθcos∅-cosθsin∅
Assume that the sum angle formula for sine is true. Namely, sin(θ+∅)=sinθcos∅+cosθsin∅.
To derive the sum angle formula for cosine, use what you learned while deriving the difference angle formula to show that cos(θ+∅)=cosθcos∅-sinθsin∅.
(You may want to start with an exploration of sin(π/2 - (θ+∅))
To derive the sum angle formula for cosine, use the angle relations cos(A)=sin(π/2-A) and sin(A)= cos(π/2-A) for any angle A.
We begin by observing that we can write cos(θ+∅) as sin(π/2-(θ+∅)).
Moreover, we can write sin(π/2-(θ+∅))= sin((π/2-θ)-∅). Putting these equalities together, we have cos(θ+∅)= sin ((π/2-θ)-∅).
Using the difference angle formula for sine we have, cos(θ+∅)=
sin((π/2-θ)-∅)=
sin(π/2-θ)cos∅-cos(π/2-θ)sin∅=
cosθcos∅-sinθsin∅.
We conclude that cos(θ+∅)=cosθcos∅-sinθsin∅.
Assume that the sum angle formula for sine is true. Namely, sin(θ+∅)=sinθcos∅+cosθsin∅.
Derive the difference angle formula for cosine, cos(θ-∅)=cosθcos∅+sinθsin∅.
To derive the difference angle formula for cosine, we apply the strategy from previuosly and write
cos(θ-∅)= cos (θ+(-∅))= cosθcos(-∅)-sinθsin(-∅)
Then, using the fact that sine is odd and cosine is even, we reduce further to get
cos(θ-∅)=
cosθcos(-∅)-sinθsin(-∅)=
cosθcos∅+sinθsin∅
So we conclude that cos(θ-∅)=cosθcos∅+sinθsin∅
Assume that the sum angle formula for sine is true. Namely, sin(θ+∅)=sinθcos∅+cosθsin∅.
Derive the sum angle formula for tangent, tan(θ+∅)= tanθ+tan∅/ 1-tanθtan∅.
To derive the sum angle formula for tangent, we can write tan (θ+∅)= sin(θ+∅)/ cos(θ+∅).
Applying the sum angles for sine and cosine, we have tan(θ+∅)= sinθcos∅+cosθsin∅/ cosθcos∅-sinθsin∅.
In order to get tangent on the right hand side, we can substitute the equations sinθ=tanθcosθ and sin∅=tan∅cos∅. With this substitution, we have
sinθcos∅+cosθsin∅/ cosθcos∅-sinθsin∅ =
tanθcosθcos∅+cosθtan∅cos∅/ cosθcos∅-tanθcosθtan∅cos∅=
(cosθcos∅)(tanθ+tan∅)/ (cosθcos∅)(1-tanθtan∅)=
tanθ+tan∅/1-tanθtan∅
we can conclude that tan(θ+∅)= tanθ+tan∅/ 1-tanθtan∅.
Assume that the sum angle formula for sine is true. Namely, sin(θ+∅)=sinθcos∅+cosθsin∅.
Derive the difference angle formula for tangent,
tan(θ-∅)=tanθ-tan∅/ 1+tanθtan∅
To derive the difference angle formula for tangent, we apply the sum angle formula for tangent to θ+(-∅). Tan(θ-∅)=
tan(θ+(-∅))=
tanθ+tan(-∅)/ 1-tanθtan(-∅).
Since tangent is odd, we can simplify tan(-∅)= -tan∅. With this substitution, we have
tan(θ-∅)=
tanθ+tan(-∅)/ 1-tanθtan(-∅)=
tanθ-tan∅/ 1+tanθtan∅.
We can conclude that
tan(θ-∅)= tanθ-tan∅/ 1+tanθtan∅.