Question

cos²73+cos²47-sin²43+sin²107is equal to

Answer 1

make sure you know the basic trig identities, the ones i used in this question include 
cos(a + b) = cosacosb - sinasinb 
cos(a - b) = cosacosb + sinasinb 
sin^2(a) = 1 - cos^2(a) (by rearranging sin^2(a) + cos^2(a) = 1) 
also know that cos(120) = -cos(180-120) = -cos(60) = -1/2 
i also used (a + b)(a - b) = a^2 - b^2 

now, let's show a few things 
cos(73 - 47)cos(73 + 47) 
= ((cos(73)cos(47) + sin(73)sin(47))(cos(73)cos(47) - sin(73)sin(47)) 
= cos^2(73)cos^2(47) - sin^2(73)sin^2(47) 
= cos^2(73)cos^2(47) - [(1 - cos^2(73))(1 - cos^2(47))] 
= cos^2(73)cos^2(47) - [1 - cos^2(47) - cos^2(73) + cos^2(73)cos^2(47)] 
= cos^2(73)cos^2(47) - 1 + cos^2(47) + cos^2(73) - cos^2(73)cos^2(47) 
= -1 + cos^2(47) + cos^2(73) 
hence cos^2(47) + cos^2(73) = cos(73 - 47)cos(73 + 47) + 1 
cos^2(47) + cos^2(73) = cos(26)cos(120) + 1 
cos^2(47) + cos^2(73) = 1 - (1/2)cos(26) 

also, cos(73 - 47) + cos(73 + 47) 
= cos(73)cos(47) + sin(73)sin(47) + cos(73)cos(47) - sin(73)sin(47) 
= 2cos(73)cos(47) 
therefore cos(73)cos(47) = (1/2)[cos(73 - 47) + cos(73 + 47)] 
cos(73)cos(47) = (1/2)[cos(26) + cos(120)] 
cos(73)cos(47) = (1/2)[cos(26) - 1/2] 
cos(73)cos(47) = (1/2)cos(26) - 1/4 

using these identities, 
cos^2(73) + cos^2(47) + cos(73)cos(47) 
= 1 - (1/2)cos(26) + (1/2)cos(26) - 1/4 
= 1 - 1/4 
= 3/4 

hence cos^2(73) + cos^2(47) + cos(73)cos(47) = 3/4

Answer 2

Answer:

Step-by-step explanation: