McGraw Hill Glencoe Algebra 2, 2012
MH
McGraw Hill Glencoe Algebra 2, 2012 View details
2. Verifying Trigonometric Identities
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Exercise 50 Page 884

Practice makes perfect
a When an alternating current of frequency f and peak current I_0 passes through a resistance R, the power delivered to the resistance at time t is given by the following expression.
P=I_0 ^2 R sin^2 2Ď€ ft We want to rewrite this expression in terms of cos^2 2Ď€ ft. To do so, we will to transform the given expression using one of the Trigonometric Identities. Recall the Pythagorean Identity considering the sum of squares of sine and cosine.

cos^2θ+sin^2θ=1 Let's transform this identity a bit to obtain a formula for sin^2θ. cos^2θ+sin^2θ=1 ⇕ sin^2θ=1-cos^2θ In our case we have θ=2π ft. sin^2 2π ft=1-cos^2 2π ft Let's use this formula to transform the given expression. P=I_0 ^2 R sin^2 2π ft ⇕ P=I_0 ^2 R (1-cos^2 2π ft) We have obtained an expression in terms of cos^2 2π ft.

b Now we want to rewrite the given expression in terms of csc^2 2Ď€ ft. Recall one of the Reciprocal Identities considering the reciprocal of cosecant.
sinθ=1/cscθ, for cscθ ≠ 0In our case θ=2π ft. sin 2π ft=1/csc 2π ft, for csc 2π ft ≠ 0 ⇕ sin^2 2π ft=1/csc^2 2π ft, for csc 2π ft ≠ 0 Let's now use the above formula to transform the given expression.
P=I_0 ^2 R sin^2 2Ď€ ft
P=I_0 ^2 R * 1/csc^2 2Ď€ ft
P=I_0 ^2 R/csc^2 2Ď€ ft
We have obtained an expression in terms of csc^2 2Ď€ ft.