Use LEFT and RIGHT arrow keys to navigate between flashcards;
Use UP and DOWN arrow keys to flip the card;
H to show hint;
A reads text to speech;
8 Cards in this Set
- Front
- Back
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 |
|
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∅. |