Prove that cos 35° + cos 85° + cos 155° = 0
Answers
Answered by
73
Cos 35°+cos85°+cos155°
= 2cos{(35°+85°)/2}cos{(35°-85°)/2}+cos(180°-25°)
[cosC+cosD=2cos{(C+D)/2}cos{(C-D)/2}]
=2cos(120°/2)cos(50°/2)+cos{(90°×2)-25°}
=2cos60°cos25°+(-cos25°)
=2×(1/2)cos25°-cos25°
=0 (Proved)
if satisfied then mark it as branliest answer ☺️
Answered by
9
Answer:
- In context to the given we have to proof the given trigonometric function
- we have to proof; LHS = RHS
- LHS = cos 35° + cos 85° + cos 155°
- RHS = 0
Equation;
LHS = Cos 35°+ Cos85°+ Cos155°
as we know;
[cos A+ cos B= 2cos{(A+B)/2} cos{(A-B)/2}]
By applying property this in the first two terms,
we get;
⇒2cos {(35°+85°)/2} cos{(35°-85°)/2} + cos(180°-25°)
By simplifying, we get;
⇒ 2cos(120°/2)cos(50°/2) + cos{(90°×2)-25°}
⇒ 2cos60°cos25°+ (-cos25°)
(cos 60° = 1/2)
⇒2×(1/2)cos25°-cos25° [cos (180°-θ) = - cos θ]
= 0 = RHS
LHS = RHS (∴ Hence proved)
Similar questions