Sum and Difference Identities Using the sum and difference identities for sine, cosine, and tangent functions

Sum and Difference Identities for the Cosine Function If α and β represent the measures of two angles, then the following identities hold for all values of α and β. cos 𝛼+𝛽 =𝑐𝑜𝑠𝛼 𝑐𝑜𝑠𝛽−𝑠𝑖𝑛𝛼 𝑠𝑖𝑛𝛽 𝑐𝑜𝑠 𝛼−𝛽 =𝑐𝑜𝑠𝛼 𝑐𝑜𝑠𝛽+𝑠𝑖𝑛𝛼 𝑠𝑖𝑛𝛽

Find cos 15⁰ from values of functions of 30⁰ and 45⁰. =𝑐𝑜𝑠45° 𝑐𝑜𝑠30°+𝑠𝑖𝑛45° 𝑠𝑖𝑛30° = 2 2 ∗ 3 2 + 2 2 ∗ 1 2 = 6 + 2 4 𝑐𝑜𝑠15°≈0.9659

Find cos 75° from values of functions of 30° 𝑎𝑛𝑑 45°. 𝑐𝑜𝑠30 cos 45 −𝑠𝑖𝑛30 sin 45 3 2 ∗ 2 2 − 1 2 ∗ 2 2 6 4 − 2 4 1.0353 4 .2588

Terri Cox is an electrical engineer designing a three-phase AC-generator. Three-phase generators produce three currents fo electricity at one time. They can generate more power for the amount of materials used and lead to better transmission and use of power then single-phase generators can. The three phases of the generator Ms. Cox is making are expressed as 𝐼𝑐𝑜𝑠𝜃, 𝐼𝑐𝑜𝑠 𝜃+120° , 𝑎𝑛𝑑 𝐼𝑐𝑜𝑠(𝜃+240°). She must show that each phase is equal to the sum of the other two phases but opposite in sign. To do this, she will show that 𝐼𝑐𝑜𝑠𝜃+𝐼𝑐𝑜𝑠 𝜃+120° +𝐼𝑐𝑜𝑠 𝜃+240° =0. 𝐼𝑐𝑜𝑠𝜃+𝐼𝑐𝑜𝑠 𝜃+120 +𝐼𝑐𝑜𝑠(𝜃+240=0 𝐼𝑐𝑜𝑠𝜃+𝐼 𝑐𝑜𝑠𝜃 𝑐𝑜𝑠120−𝑠𝑖𝑛𝜃 𝑠𝑖𝑛120 +𝐼 𝑐𝑜𝑠𝜃 𝑐𝑜𝑠240−𝑠𝑖𝑛𝜃 𝑠𝑖𝑛240 =0 𝐼𝑐𝑜𝑠𝜃+𝐼 − 1 2 𝑐𝑜𝑠𝜃− 3 2 𝑠𝑖𝑛𝜃 + −1 2 𝑐𝑜𝑠𝜃− − 3 2 𝑠𝑖𝑛𝜃 =0 𝐼𝑐𝑜𝑠𝜃− 1 2 𝐼𝑐𝑜𝑠𝜃− 3 2 𝐼𝑠𝑖𝑛𝜃− 1 2 𝐼𝑐𝑜𝑠𝜃+ 3 2 𝐼𝑠𝑖𝑛𝜃=0 0=0

Sum and Difference Identities for Sine Function If α and β represent the measures of two angles, then the following identities hold for all values of α and β. sin 𝛼+𝛽 =𝑠𝑖𝑛𝛼 𝑐𝑜𝑠𝛽+𝑠𝑖𝑛𝛽 𝑐𝑜𝑠𝛼 sin(α−𝛽)=𝑠𝑖𝑛𝛼 𝑐𝑜𝑠𝛽−𝑠𝑖𝑛𝛽 𝑐𝑜𝑠𝛼

Find sin 75⁰ from values of functions of 30⁰ and 45⁰. = 1 2 ∗ 2 2 + 2 2 ∗ 3 2 = 2 4 + 6 4 𝑠𝑖𝑛75°≈.9659

Find sin 15⁰ from values of functions of 30⁰ and 45⁰. = 2 2 ∗ 3 2 − 2 2 ∗ 1 2 = 6 4 − 2 4 sin 15°≈0.2588

Sum and Difference Identities for the Tangent Function If α and β represent the measures of two angles, then the following identities hold for all values of α and β. tan 𝛼+𝛽 = 𝑡𝑎𝑛𝛼+𝑡𝑎𝑛𝛽 1−𝑡𝑎𝑛𝛼 𝑡𝑎𝑛𝛽 tan 𝛼−𝛽 = 𝑡𝑎𝑛𝛼−𝑡𝑎𝑛𝛽 1+𝑡𝑎𝑛𝛼 𝑡𝑎𝑛𝛽

Find tan 15⁰ from values of functions of 45⁰ and 30⁰ 𝑡𝑎𝑛15°= tan 45−30 = 𝑡𝑎𝑛45−𝑡𝑎𝑛30 1+ tan 45 tan 30 = 1− 3 3 1+ 3 3 = 3− 3 3 3+ 3 3 = 3− 3 3+ 3 3− 3 3− 3 = 9−6 3 +3 9−3 = 12−6 3 6 =2− 3 ≈0.2679

Find tan 105⁰ from values of functions of 45⁰ and 60⁰ tan 105= tan 45+ tan 60 1− tan 45 tan 60 = 1+ 3 1−(1)( 3 ) = 1+ 3 1− 3 1+ 3 1+ 3 = 1+2 3 +3 1−3 = 4+2 3 −2 =−2− 3 ≈−3.7321

Verify that cot 𝑥= tan 𝜋 2 −𝑥 cot 𝑥= sin 𝜋 2 −𝑥 cos 𝜋 2 −𝑥 cot 𝑥= sin 𝜋 2 𝑐𝑜𝑠𝑥 − cos 𝜋 2 sin 𝑥 cos 𝜋 2 cos 𝑥+ sin 𝜋 2 sin 𝑥 cot 𝑥= 1 cos 𝑥− 0 sin 𝑥 0 cos 𝑥+ 1 sin 𝑥 cot 𝑥= cos 𝑥 sin 𝑥 cot 𝑥= cot 𝑥