Exercise 24 Page 656

Let's begin with drawing a 45^(∘)-45^(∘)-90^(∘) triangle. If we call the legs of this right isosceles triangle x, then the hypotenuse is xsqrt(2).

In our exercise we are asked to evaluate the sine of 45^(∘). Recall that the sine of ∠ A is the ratio of the length of the leg opposite ∠ A to the length of the hypotenuse. Using this definition, we can write the appropriate ratio for 45^(∘).

sin 45^(∘)=x/xsqrt(2)

ReduceFrac

a/b=.a /x./.b /x.

sin 45^(∘)=1/sqrt(2)

▼

Simplify right-hand side

ExpandFrac

a/b=a * sqrt(2)/b * sqrt(2)

sin 45^(∘)=1*sqrt(2)/sqrt(2)*sqrt(2)

ProdToPowTwoFac

a* a=a^2

sin 45^(∘)=1*sqrt(2)/(sqrt(2))^2

OneMult

1* a=a

sin 45^(∘)=sqrt(2)/(sqrt(2))^2

PowSqrt

( sqrt(a) )^2 = a

sin 45^(∘)=sqrt(2)/2

UseCalc

Use a calculator

sin 45^(∘)=0.7071...

RoundDec

Round to 2 decimal place(s)

sin 45^(∘)≈ 0.71

The sine of 45^(∘) is sqrt(2)2 or approximately 0.71.