Math, asked by Indahmhay7878, 1 year ago

Evaluate cos² 45° + cos² 135° + cos² 225° + cos² 315°.

Answers

Answered by abhi178
33
cos²45° + cos²135° + cos²225° + cos²315°

we know, cos45° = 1/√2

cos135° = cos(90° + 45°) = -sin45° = -1/√2

cos225° = cos(180° + 45°) = -cos45° = -1/√2

cos315° = cos(360° - 45°) = cos45° = 1/√2

now, cos²45° = (1/√2)² = 1/2

cos²135° = (-1/√2)² = 1/2

cos²225° = (-1/√2)² = 1/2

cos²315° = (1/√2)² = 1/2


now, cos² 45° + cos² 135° + cos² 225° + cos² 315° = 1/2 + 1/2 + 1/2 + 1/2 = 2
Answered by MaheswariS
9

Answer:

2

Step-by-step explanation:

cos² 45° + cos² 135° + cos² 225° + cos² 315°.


cos45°=1/√2


cos135°= cos(90°+45°)= -sin45°= -1/√2


cos225°= cos(180°+45°)= -cos45°= -1/√2


cos315°= cos(270°+45°)= sin45°= 1/√2


Now,

cos² 45° + cos² 135° + cos² 225° + cos² 315°.

= (1/2)+(1/2)+(1/2)+(1/2)

=2

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