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sin(squared) 30 degrees + cos(squared)150 degrees + tan(squared)60 degrees= sec 300 degrees

cos 150 deg= 2(cos45 cos 330-sin 135 sin30)

cos 300= cos(squared)30-sin(squared)330

one fifth tan 150= tan 30

2007-08-31 02:02:26 · 2 answers · asked by khazzie_031239 1 in Education & Reference Homework Help

2 answers

sin² 30 + cos² 150 + tan² 60
= sin² 30 + cos²(180-30) + tan² 60
= sin² 30 + cos²30 + tan² 60
= 1 + tan² 60 = 1/cos²60 = sec²60
= sec²(360-60) = sec²300

2(cos45cos330 - sin135sin30)
= 2(cos45cos(360-30) - sin(180-45)sin30)
= 2(cos45cos30 - sin45sin30)
= 2cos(45+30) = 2 cos75

cos²30 - sin²330
= cos²30 - sin²(360-30)
= cos²30 - sin²30
= cos60 = cos(360-60)
=cos 300

(1/5)tan 150 ≠ tan 30

2007-08-31 02:32:33 · answer #1 · answered by Anonymous · 0 0

Sin^2 30 = 0.25, cos^2 150 = 0.75, tan^2 60 = 3 so .25 + .75 + 3 = 4 so if Sec 300 = 4 ::: FALSE

cos 150 = -0.866, 2(cos45 cos330 - sin135 sin330) = 0.517
False

cos 300 = 0.5, cos^2 30 = .75, sin^2 330 = .25
0.5 = 0.75 - 0.25 :: True

4th unsure

2007-08-31 02:23:16 · answer #2 · answered by merlin558nj 2 · 0 0

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