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Barry
Mathematics
19 August, 20:25
Write cos (3x) - cos x as a product
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Quinn
19 August, 20:40
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-2sin (x) * sin (2x)
or any equivalent form, such as 4cos (x) [cos² (x) - 1]
Step-by-step explanation:
simplify cos (3x) first:
cos (3x) = cos (2x+x) = cos (2x) cos (x) - sin (2x) sin (x)
using trig identities
= [2cos² (x) - 1]cos (x) - [2sin (x) cos (x) ]sin (x)
= 2cos³ (x) - cos (x) - 2sin² (x) cos (x)
substituting using trig identity sin² (x) + cos² (x) = 1
2cos³ (x) - cos (x) - 2[1-cos² (x) ]cos (x)
2cos³ (x) - cos (x) - 2cos (x) + 2cos³ (x)
4cos³ (x) - 3cos (x)
remember this cos (3x), we still have to subtract cos (x)
4cos³ (x) - 3cos (x) - cos (x) = 4cos³ (x) - 4cos (x)
we can factor 4cos (x) to write this as a product of:
4cos (x) [cos² (x) - 1]
further simplification if you want
trig identity sin² (x) + cos² (x) = 1
simplifying: sin² (x) = 1-cos² (x)
simplifying: - sin² (x) = cos² (x) - 1
4cos (x) [cos² (x) - 1]
4cos (x) [-sin² (x) ]
-4cos (x) sin² (x)
trig identity: sin (2a) = 2cos (a) sin (a)
-2sin (x) * 2cos (x) sin (x)
-2sin (x) * sin (2x)
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