Exercise 1 Page 515

We want to find the values of the other five trigonometric functions of θ given that cos θ = 16 and 0 < θ < π2.

Finding sin θ

To do so, we will use one of the Pythagorean Identities. sin ^2 θ + cos^2 θ = 1We will substitute 16 for cos θ in the above equation and solve it for sin θ. Let's do it!

sin ^2 θ + cos^2 θ = 1

Substitute

cos θ= 1/6

sin ^2 θ + ( 1/6)^2 = 1

SubEqn

LHS- ( 1/6 )^2=RHS- ( 1/6 )^2

sin ^2 θ = 1 -( 1/6 )^2

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Simplify right-hand side

PowQuot

(a/b)^m=a^m/b^m

sin ^2 θ = 1 - 1^2/6^2

CalcPow

Calculate power

sin ^2 θ = 1- 1/36

OneToFrac

Rewrite 1 as 36/36

sin ^2 θ = 36/36 - 1/36

SubFrac

Subtract fractions

sin ^2 θ = 36-1/36

SubTerm

Subtract term

sin ^2 θ = 35/36

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Solve for sinθ

SqrtEqn

sqrt(LHS)=sqrt(RHS)

sin θ =± sqrt(35/36)

SqrtQuot

sqrt(a/b)=sqrt(a)/sqrt(b)

sin θ =± sqrt(35)/sqrt(36)

CalcRoot

Calculate root

sin θ =± sqrt(35)/6

Be aware that we are told that θ lies between 0 and π2. Therefore, θ is in Quadrant I.

In this quadrant, the sine of θ is positive. Therefore, we will only keep the positive solution. sin θ = sqrt(35)/6

Finding the Other Trigonometric Ratios

Having the two values of trigonometric functions allows us to find the four remaining trigonometric ratios. Remember to simplify fractions and rationalize denominators, if needed.

Function	Substitute	Simplify
tan θ=sin θ/cos θ	tan θ=sqrt(35)6/16	tan θ = sqrt(35)
cot θ=cos θ/sin θ	cot θ=16/sqrt(35)6	cot θ=sqrt(35)/35
csc θ=1/sin θ	csc θ=1/sqrt(35)6	csc θ=6sqrt(35)/35
sec θ=1/cos θ	sec θ=1/16	sec θ=6