Trigonomic Identities

Trigonomic Identities

100

1 - cos^2(θ)

=sin^2(θ)

200

3 other ways to write tan(θ)

Way #1: sinθ/cosθ Way #2: 1/ (1/tanθ) =1/cotθ Way # 3: (1/cosθ)/(1/sinθ) =secθ/cscθ

300

tanθ/secθ

Step 1: (sinθ/cosθ) / (1/cosθ) Step 2: (sinθ/cosθ)(cosθ) Answer: =sinθ

400

Prove 1/cscθ + sin(-θ) = 0

Step 1: sinθ + sin(-θ) = 0 Step 2: sinθ - sinθ = 0 Step 3: 0=0 QED

500

Prove (cos^2(θ)+sin^2(θ))/(sinθ+cosθcotθ)=sinθ

Step 1: 1/(sinθ+cosθcotθ)=sinθ Step 2: 1/(sinθ+(cosθ)(cosθ/sinθ))=sinθ Step 3: 1/(sinθ(sinθ/sinθ)+(cos^2(θ)/sinθ))=sinθ Step 4: 1/((sin^2(θ)+cos^2(θ))/sinθ)=sinθ Step 5: 1/(1/sinθ)=sinθ Step 6: sinθ=sinθ QED

Sum and Difference Formulas

100

sin 105

Step 1) sin (45+60) Step 2) sin45cos60+sin60cos40 Step 3) (√2+√6)/4

200

Given sin x= -2/3 and cos y= 1/3, find tan (x+y)

Step 1) tanx+tany/1-tanxtany Step 2)((-2√5)/5+2√2)/(1-((-2√5)/5)(2√2)) Step 3) (5+4√10)/5

300

Prove tanx-tany= sin(x-y)/cosxcosy

Step 1) sin(x-y)/cosxcosy Step 2) (sinxcosy)-(sinycosx)/cosxcosy Step 3) (sinxcosy/cosxcosy)-(sinycosx/cosxcosy) Step 4) (sinx/cosx)-(siny/cosy) Step 5) tanx-tany=tanx-tany QED

400

cos 7π/18cos 2π/9+sin 7π/18sin 2π/9

Step 1) cos (7π/18-2π/9) Step 2) cos(7π/18-4π/18) Step 3) cos(3π/18) Step 4) cos(π/6) Step 5) √3/2

500

csc⁡(x-3π/2)=-secx

Step 1) csc(x-3π/2) Step 2) 1/ sinx-3π/2) Step 3) sinxcos 3π/2-sin3π/2cosx Step 4) sinx(0)-(-1)(cosx) Step 5) 1/cosx Step 6) -secx=-secx QED

Double-Angle Formulas

100

Another way to write tan(2θ)

(2tanθ)/(1-tan^2(θ))

200

(1-2sin^2(θ))/(1-tan^2(θ))=

Step 1: cos(2θ)/cos(2θ) Answer: =1

300

Prove csc(2θ)=(secθcscθ)/2

Step 1: 1/sin(2θ)=(secθcscθ)/2 Step 2: 1/(2cosθsinθ)= (secθcscθ)/2 Step 3: 1/(2(1/secθ)(1/cscθ))= (secθcscθ)/2 Step 4: (secθcscθ)/2= (secθcscθ)/2 QED

400

What is cos(2θ) if cosθ=-1/√6 and sinθ=3/√6?

Step 1: cos^2(θ)+sin^2(θ) Step 2: ((-1/√6)^2)-((3/√6)^2) Step 3: (1/6)-(9/6) Step 4: -8/6 Step 5: -4/3

500

Prove: sin(4θ)=(4sinθcosθ)(cos^2(θ)-sin^2(θ))

Step 1: 2sin(2θ)cos(2θ)=(4sinθcosθ)(cos^2(θ)-sin^2(θ)) Step 2: 2(2sinθcosθ)(cos^2(θ)-sin^2(θ))=(4sinθcosθ)(cos^2(θ)-sin^2(θ)) Step 3: (4sinθcosθ)(cos^2(θ)-sin^2(θ))=(4sinθcosθ)(cos^2(θ)-sin^2(θ)) QED

Half-Angle Formulas

100

sin(θ/2)

What is ±√((1-cos(θ))/2)

200

3 ways to express "tan((θ)/2)"

What are 1. ±√((1-cos(θ))/(1+cos(θ))) 2. (1-cos(θ))/(sin(θ)) 3. (sin(θ))/(1+cos(θ))

300

Prove that sec^2(θ/2) = 2/(1+cos(θ/2))

Step 1- sec^2(θ/2) = 2/(1+cos(θ/2)) Step 2- 2/(1+cos(θ/2)) = (1/(cos(θ/2)))^2 Step 3- 2/(1+cos(θ/2)) = (1/((1+cos (θ))/(2))) Step 4- 2/(1+cos(θ)) = 2/(1+cos(θ)) Q.E.D.

400

tan(θ/2) = 67.5°

Step 1- let θ= 135° Step 2– tan (θ/2) = ±√((1-cos(135°))/(1+cos(135°))) Step 3- tan(θ/2) = (1+(√2)/2)/(1-(√2)/2) Step 4- tan(θ/2) = ((2+√2/2))/((2-√2)/2) Step 5- tan(θ/2) = (2+√2)/2 * 2/(2-√2) Step 6- tan(θ/2) = (2+√2)/(2-√2) Step 7- tan(θ/2) = ((2+√2) * (2 +√2))/((2-√2) * (2+√2)) Step 8- tan(θ/2) = (6+4√2)/2 Step 9- tan(θ/2) = 3+2√2

500

Prove tan(a/2) = csc(a) - cot(a)

Step 1- tan(a/2) = csc(a) - cot(a) Step 2- tan(a/2) = (1/(sin(a))) – ((cos(a))/(sin(a))) Step 3- tan(a/2) = (1-cos(a))/(sin(a)) Step 4- tan(a/2) = ±√(((1-cos(a))^2)/(sin^2(a))) Step 5- tan(a/2) = ±√(((1-cos(a))*(1-cos(a))/(1-cos^2(a))) Step 6- tan(a/2) = ±√((1-cos(a))*(1-cos(a)))/((1-cos(a))*(1+cos(a))) Step 7- tan(a/2) = ±√((1-cos(a))/(1-cos(a))) Step 8- tan(a/2) = tan(a/2) Q.E.D

Everything!

100

Prove that (sec(θ)-csc(θ))/(sec(θ)csc(θ)) = sin(θ)-cos(θ)

Step 1 - (sec(θ)-csc(θ))/(sec(θ)csc(θ)) = sin(θ)-cos(θ) --> Step 2 - ((1/(cos(θ))-(1/sin(θ)))/((1/(cos(θ))*(1/sin(θ))) = sin(θ)-cos(θ) --> Step 3 - ((1/cos(θ))-(1/sin(θ)))*(cos(θ)sin(θ)) = sin(θ)-cos(θ) --> Step 4 - ((sin(θ)/(cos(θ)sin(θ)))-(cos(θ)/(cos(θ)sin(θ))))*(cos(θ)sin(θ)) = sin(θ)-cos(θ) --> Step 5 - ((sin(θ)-cos(θ))/(cos(θ)sin(θ)))*(cos(θ)sin(θ)) = sin(θ)-cos(θ) --> Step 6 - sin(θ)-cos(θ) = sin(θ)-cos(θ) Q.E.D.

200

Prove that tan(θ)+(cos(θ)/(1+sin(θ))) = sec(θ)

Step 1 - tan(θ)+(cos(θ)/(1+sin(θ))) = sec(θ) Step 2 - (sin(θ))/(cos(θ))+(cos(θ))/(1+sin(θ)) = sec(θ) --> Step 3 - (sin(1+sin(θ)))/(cos(θ)*(1+sin(θ)))+(cos^2(θ))/(cos(θ)*(1+sin(θ))) = sec(θ) --> Step 4 - (sin(θ)+sin^2(θ)+cos^2(θ))/(cos(θ)*(1+sin(θ))) = sec(θ) --> Step 5 - (sin(θ)+1)/(cos(θ)*(1+sin(θ)) = sec(θ) --> Step 6 - 1/(cos(θ)) = sec(θ) --> Step 7 - sec(θ) = sec(θ) Q.E.D.

300

Prove that csc(2θ) = (sec(θ)csc(θ))/(2)

Step 1 - csc(2θ) = (sec(θ)csc(θ))/(2) --> Step 2 - csc(2θ) = (1/(cos(θ))*(1/(sin(θ)))/2 --> Step 3 - csc(2θ) = 1/(2(cos(θ)sin(θ))) --> Step 4 - csc(2θ) = 1/(sin(2θ)) --> Step 5 - csc(2θ) = csc(2θ) --> Q.E.D.

400

Prove that tan^2(θ)-sin^2(θ) = tan^2(θ)*sin^2(θ)

Step 1 - tan^2(θ)-sin^2(θ) = tan^2(θ)*sin^2(θ) Step 2 - tan^2(θ)-sin^2(θ) = tan^2(θ)*(1-cos^2(θ)) Step 3 - tan^2(θ)-cos^2(θ)*tan^2(θ) Step 4 - tan^2(θ)-((sin(θ))/(cos(θ)))^2*(cos^2(θ)) Step 5 - -sin^2(θ)+tan^2(θ) Step 6 - tan^2(θ)-sin^2(θ) = tan^2(θ)-sin^2(θ) Q.E.D.

500

Prove that (sin(3θ)+sin(7θ))/(cos(3θ)-cos(7θ)) = cot(2θ)

Step 1 - (sin(3θ)+sin(7θ))/(cos(3θ)-cos(7θ)) = cot(2θ) --> Step 2 - (sin(3θ)+sin(7θ))/(cos(3θ)-cos(7θ)) = (cos(2θ)/(sin(2θ))) --> Step 3 - (sin(3θ)+sin(7θ))/(cos(3θ)-cos(7θ)) = (cos(2θ)/(sin(2θ)))*(cos(3θ)-cos(7θ))/(cos(3θ)-cos(7θ)) Step 4 - (sin(3θ)+sin(7θ))/(cos(3θ)-cos(7θ)) = (cos(2θ))*((cos(3θ)-cos(7θ))/(cos(3θ)-cos(7θ)) Q.E.D.